Methods for Solving Electrostatic and Magnetostatic Problems

Abstract

In the preceding chapters we have studied properties of the electrostatic and magnetostatic fields in vacuum and in the presence of matter. We have also seen how we can generate them and some techniques for calculating $\mathbf {E}$ and $\mathbf {B}$. However, it is not always possible to obtain a solution by means of these techniques for different reasons. For instance, when we wanted to calculate the electric field produced by a system of charges, we needed to know exactly its distribution and then apply Coulomb’s law. This method is very clear, but sometimes does not easily work. In fact, when the configuration of charges has a low symmetry the integral to be computed cannot be resolved in terms of elementary functions. The same occurs with the magnetic field. Even in case of symmetry its calculation may be of hight difficulty if the region where the field must be calculated is off the axis of revolution of the system (see Chap. 5). To solve these drawbacks different methods of analysis have been developed. In this chapter we try to introduce the reader to some of the most important analytical and numerical techniques for solving non-time dependent electromagnetic problems. We present a set of exercises with the aim to show the basic ideas so that the student can understand a further reading in specialized books about this subject. Because of the characteristics of the fields to be studied in this chapter, we focus our attention on the Laplace equation . As we will see many problems appearing in electromagnetics may be posed on the basis of this equation, or on the Poisson equality, depending on the existence of charges in the place where the potential must be obtained. The question is then to seek a solution subjected to some boundary conditions, which depend on the specific problem to be investigated.

Appendices

Solved Problems

Problems C

• 7.1 Consider two parallel conducting square plates located at $z_1=-d$ and $z_2=d$. If the potentials of the plates are $V_1$ and $V_2$, respectively, find: (a) The expression of the potential between plates. (b) The capacity of the capacitor.

Solution

This dispositive formed by two parallel thin metallic plates is known as a capacitor (see Chap. 2). This set-up is one of the most simple geometries we can have for a condenser. For solving this problem we suppose the distance between plates is much smaller than the length of the side, that is, $|z_2-z_1|<<L$, L being the length of the square sheet. By this way we can neglect end effects that capacitors have. Starting from this viewpoint, we can directly use the Laplace equation in cartesian coordinates. However, due to the symmetries of the device we may reduce the number of differential equations to be solved. In fact, as the ratio $\frac{2d}{L}<<1$, regardless of end effects, we can consider the plates as almost infinite, and then along the directions OX, and OY, translational symmetry holds. In this case, we do not need to solve the differential equations for all the variables. On the contrary, we must focus our attention only to z, where we have no symmetry. Employing (7.10) we can write

$$\begin{aligned} \frac{\partial ^2 V}{\partial z^2}=\frac{d^2 V}{d z^2}=0, \end{aligned}$$

(7.109)

which is a differential equation of only one variable. Integrating it gives

$$\begin{aligned} \frac{d V}{d z}=C_1, \end{aligned}$$

(7.110)

where $C_1$ is a constant. Integrating again

$$\begin{aligned} V=C_{1}z+C_{2}, \end{aligned}$$

(7.111)

$C_2$ being another constant. The problem is now to determine both constants. To this aim we use the boundary conditions, that is, the value of the potential on each plate of the capacitor. We impose the following conditions

$$\begin{aligned} V(z_1)=V(-d)=V_1, \end{aligned}$$

(7.112)

and

$$\begin{aligned} V(z_2)=V(d)=V_2, \end{aligned}$$

(7.113)

then we get

$$\begin{aligned} V(-d)=-C_{1}d+C_{2}=V_1, \end{aligned}$$

(7.114)

and

$$\begin{aligned} V(d)=C_{1}d+C_{2}=V_2. \end{aligned}$$

(7.115)

We have obtained a system of two linear equations with two unknowns, whose solutions is

$$\begin{aligned} C_{1}=\frac{V_2-V_1}{2d}, \end{aligned}$$

(7.116)

and

$$\begin{aligned} C_{2}=\frac{V_2+V_1}{2}, \end{aligned}$$

(7.117)

Introduction of these constants into (7.111) leads to

$$\begin{aligned} V(z)=\frac{V_2-V_1}{2d}\,z+\frac{V_2+V_1}{2}=\frac{1}{2}\left( \frac{V_2-V_1}{d}z+(V_2+V_1)\right) . \end{aligned}$$

(7.118)

As a particular case if $V_1=0$, and making $V_2=V$, we have

$$\begin{aligned} V(z)=\frac{V}{2}\left( \frac{z}{d}+1\right) . \end{aligned}$$

(7.119)

• 7.2 The capacitor of the figure consists of two metallic identical non-parallel plane plates a and b. The edge perpendicular to the plane of the Fig. 7.12 is $L_1$, and the other one $L_2$. The side $L_1$ is very large and $L_2$ is large enough so that we can neglect end effects. The angle between these plates is $\theta$ and the distance from the origin O to both edges parallel to OZ is c. The potential of the first one is $V_1$ [V], and of the upper plate $V_2$ [V]. Obtain: (a) The potential in the interior region of the plates. (b) The electric field. (c) The density of charge on the plates. (d) The total charge on a plate per unit length $L_1$.

Solution

As we can observe, both plates are located in such a manner that two of their edges are parallel to the OZ axis, then considering both sheets as very large in this direction the problem has translational symmetry along OZ. Due to this fact, we can suppress the z coordinate in (7.23) simplifying the problem, thus we may write

$$\begin{aligned} \triangle \equiv \frac{1}{\rho }\frac{\partial }{\partial \rho }\left( \rho \frac{\partial V(\rho ,\phi )}{\partial \rho }\right) +\frac{1}{\rho ^2}\frac{\partial ^2 V(\rho ,\phi )}{\partial \phi ^2}=0. \end{aligned}$$

(7.120)

Using the method of separation of variables we can try solutions of the form $V(\rho ,\phi )=R(\rho )\Phi (\phi )$. Introduction of this function into (7.120) leads to

$$\begin{aligned} \frac{1}{\rho }\frac{\partial }{\partial \rho }\left( \rho \frac{\partial (R(\rho )\Phi (\phi ))}{\partial \rho }\right) +\frac{1}{\rho ^2}\frac{\partial ^2 (R(\rho )\Phi (\phi ))}{\partial \phi ^2}=0, \end{aligned}$$

(7.121)

Fig. 7.12

Two thin metallic plates forming an angle $\alpha$. The surface of each plate is $L_1.L_2$

and computing the derivatives this equation may be expressed as

$$\begin{aligned} \frac{\rho }{R(\rho )}\frac{\partial }{\partial \rho }\left( \rho \frac{\partial R(\rho )}{\partial \rho }\right) +\frac{1}{\Phi (\phi )}\frac{\partial ^2 (R(\rho )\Phi (\phi ))}{\partial \phi ^2}=0. \end{aligned}$$

(7.122)

The first part of the last equation depends only on $\rho$ and the second only on $\phi$, therefore this equality holds if

$$\begin{aligned} \frac{\rho }{R(\rho )}\frac{\partial }{\partial \rho }\left( \rho \frac{\partial R(\rho )}{\partial \rho }\right) =-\frac{1}{\Phi (\phi )}\frac{\partial ^2 (R(\rho )\Phi (\phi ))}{\partial \phi ^2}=k^2, \end{aligned}$$

(7.123)

k being a constant. Resolving separately both equations we have

$$\begin{aligned} \frac{\rho }{R(\rho )}\frac{\partial }{\partial \rho }\left( \rho \frac{\partial R(\rho )}{\partial \rho }\right) =k^2. \end{aligned}$$

(7.124)

Here we have two possibilities, namely $k=0$ and $k\ne 0$. For the first case we get

$$\begin{aligned} \frac{\rho }{R(\rho )}\frac{\partial }{\partial \rho }\left( \rho \frac{\partial R(\rho )}{\partial \rho }\right) =0\Rightarrow \rho \frac{\partial R(\rho )}{\partial \rho }=A_1, \end{aligned}$$

(7.125)

where $A_1$ is a constant. The solution is

$$\begin{aligned} R(\rho )=A_{1}\ln \rho +A_{2}, \end{aligned}$$

(7.126)

for $A_2$ constant, too. If $k\ne 0$ the solution of (7.124) is

$$\begin{aligned} R(\rho )=B_{1}\rho ^{k}+B_{2}\rho ^{-k}. \end{aligned}$$

(7.127)

The equation for $\Phi (\phi )$ is

$$\begin{aligned} \frac{1}{\Phi (\phi )}\frac{\partial ^2 (R(\rho )\Phi (\phi ))}{\partial \phi ^2}=-k^2\Rightarrow \frac{\partial ^2 \Phi (\phi )}{\partial \phi ^2}+k^2\Phi (\phi )=0, \end{aligned}$$

(7.128)

which consist of two solutions as before for $R(\rho )$. In effect, when $k=0$

$$\begin{aligned} \frac{1}{\rho ^2}\frac{\partial ^2 V(\rho ,\phi )}{\partial \phi ^2}=0\Rightarrow \frac{\partial V(\rho ,\phi )}{\partial \phi }=C_1, \end{aligned}$$

(7.129)

and integrating for $C_1$ constant

$$\begin{aligned} V(\rho ,\phi )=C_{1}\phi +C_{2} \end{aligned}$$

(7.130)

If $k\ne 0$ the solution is

$$\begin{aligned} V(\rho ,\phi )=D_{1}\cos (k\phi )+D_{2}\sin (k\phi ). \end{aligned}$$

(7.131)

With all these results we must find what solution is valid for our case. Owing to the boundary conditions of the problem, that is, each of the plates are held a constant potential, the solution must not depend on the distance $\rho$. For this reason we, in principle, can exclude (7.126) and (7.127). From the other two referring to the angle, (7.131) does not work, because it cannot fulfil the boundary conditions of constant potential on each metallic plate of the capacitor. Thus, we have only (7.130). Imposing that $V(\phi =0)=V_1$ and $V(\phi =\theta )=V_2$, we easily get

$$\begin{aligned} V(\rho ,\phi )=\frac{(V_2-V_1)}{\theta }\,\phi +V_1. \end{aligned}$$

(7.132)

The electric field inside, neglecting end effects, may be calculated be means of the gradient, i.e.,

$$\begin{aligned} \nabla V=\frac{\partial V}{\partial \rho }+\frac{1}{\rho }\frac{\partial V}{\partial \phi }+\frac{\partial V}{\partial z}. \end{aligned}$$

(7.133)

As the potential only depends on $\phi$, we can write for the electric field

$$\begin{aligned} \mathbf {E}=-\nabla V=-\frac{1}{\rho }\frac{\partial V(\phi )}{\partial \phi }\,\mathbf {u}_{\phi }, \end{aligned}$$

(7.134)

which leads to

$$\begin{aligned} \mathbf {E}=-\nabla V=-\frac{1}{\rho }\frac{(V_2-V_1)}{\theta }\,\mathbf {u}_{\phi }. \end{aligned}$$

(7.135)

The surface charge density may be calculated by means of the following relation (see Chap. 4)

$$\begin{aligned} \sigma (\phi =0)=\varepsilon \,E=-\frac{1}{\rho }\frac{(V_2-V_1)}{\theta }\varepsilon , \end{aligned}$$

(7.136)

and for the upper plate

$$\begin{aligned} \sigma (\phi =\theta )=-\varepsilon \,E=\frac{1}{\rho }\frac{(V_2-V_1)}{\theta }\varepsilon . \end{aligned}$$

(7.137)

(d) For calculating the total charge per unit length $L_1$ on a plate we can employ the definition of surface charge density, i.e.

$$\begin{aligned} \sigma (x,y)=\frac{dQ(x,y)}{dS}. \end{aligned}$$

(7.138)

By integrating this equation we have

$$\begin{aligned} Q(x,y)=\int \int _{S}\sigma (x,y)\,dS. \end{aligned}$$

(7.139)

Let us suppose we choose the lower plate. Even though we have used at the beginning of the problem polar coordinates, since this conducting plate lies on the plane OXZ, for simplicity we can work with x and z coordinates. In this case the differential element of surface is $dS=dx\,dz$, then

$$\begin{aligned} Q(x,y)=\int \int _{S}\sigma (x,y)dxdz= -\int \int _{S}\frac{\varepsilon }{x}\frac{(V_2-V_1)}{\theta }dxdz= -\int _{0}^{L_1}dz\int _{c}^{c+L_2}\frac{\varepsilon }{x}\frac{(V_2-V_1)}{\theta }dx= \end{aligned}$$

(7.140)

$$ \frac{(V_2-V_1)\,L_1}{\theta }\ln \left( \frac{c+L_2}{c}\right) , $$

thus

$$\begin{aligned} \frac{Q(x,y)}{L_1}=\frac{(V_2-V_1)}{\theta }\ln \left( \frac{c+L_2}{c}\right) , \end{aligned}$$

(7.141)

where we suppose $L_2$ large in order to avoid end effects, but not infinite.¹⁴

• 7.3 Let us consider four metallic thin rectangular plates with sides $a\times L$ and $b\times L$ forming a prismatic system as shown in Fig. 7.13a. The edges of length L are much longer than the other. The side AD is held to a potential $V_0=200$ V, and the other parts are grounded. (a) Calculate the potential in the region $0<x<a$ and $0<y<b$, if $a=20$ cm and $b=10$ cm. (b) Obtain the electric field $\mathbf {E}(x,y)$.

Fig. 7.13

a Prismatic system formed by four metallic plates one of them is grounded. b Cross-sectional view of the system

Solution

As the edge of length L verifies that $L>>a$ and $L>>b$, we can consider that the system has translational symmetry with respect to the OZ axis. It then means that $V(x,y,z)=V(x,y,z+h)$ and $\mathbf {E}(x,y,z)=\mathbf {E}(x,y,z+h)$, h being a constant and neglecting end effects. For this reason we can work with the two-dimensional system depicted in Fig. 7.13b, which is easier. Bearing in mind the geometrical characteristics of the prismatic set of metallic plates, we can find the solution of the problem by employing the two dimensional Laplace equation in cartesian coordinates, that is,

$$\begin{aligned} \triangle V(x,y,z)\equiv \frac{\partial V}{\partial x}+\frac{\partial V}{\partial y}=0, \end{aligned}$$

(7.142)

constrained by the boundary conditions

$$\begin{aligned} V(0,y)=V_0=200, \end{aligned}$$

(7.143)

$$\begin{aligned} V(x,0)=0, \end{aligned}$$

(7.144)

$$\begin{aligned} V(a,y)=0, \end{aligned}$$

(7.145)

and

$$\begin{aligned} V(x,b)=0. \end{aligned}$$

(7.146)

The first difficulty to account for is that we do not have homogeneous boundary conditions, and if we would like to use the method of separation of variables we must have them. To solve this difficulty we first try solutions of the form

$$\begin{aligned} V(x,y,z)=X(x)\,Y(y), \end{aligned}$$

(7.147)

and later, we pursue to adjust some coefficients that will appear. Let us introduce (7.147) into (7.142)

$$\begin{aligned} Y(y)\frac{\partial ^2 X(x)}{\partial x^2}+X(x)\frac{\partial ^2 Y(y)}{\partial y^2}=0. \end{aligned}$$

(7.148)

Now, dividing by the product $X(x)\,Y(y)$ we have

$$\begin{aligned} \frac{1}{X(x)}\frac{\partial ^2 X(x)}{\partial x^2}+\frac{1}{Y(y)}\frac{\partial ^2 Y(y)}{\partial y^2}=0. \end{aligned}$$

(7.149)

The only possibility for a non-trivial solution is

$$\begin{aligned} \frac{1}{X(x)}\frac{d^2 X(x)}{d x^2}=-\frac{1}{Y(y)}\frac{d^2 Y(y)}{d y^2}=\pm k^2. \end{aligned}$$

(7.150)

k being a constant. The election of plus or minus in this equation depends on the conditions of the problem. So, in our case we know that the potential must be zero on sides AB, BC, and CD, then Y(y) must be zero at $y=0$ and $y=b$, and $X(x)=0$ at $x=a$, exclusively, because at $x=0$ it is non-homogeneous. Thus, one of these two possibilities does not work.

As we know, the general solution of the one dimensional differential equation $\frac{d^2 Q(s)}{d s^2}=k^{2}Q(s)$ is $Q(s)=A\exp (ks)+A\exp (-ks)$, which may also be written as an addition of hyperbolic sines and cosines. These kind of functions cannot be zero at two points, namely $y=0$ and $y=b$ at the same time. On the contrary, the solution of $\frac{d^2 Q(s)}{d s^2}=-k^{2}Q(s)$ is the combination of sinusoidal functions, $Q(s)=A\cos (ks)+A\sin (ks)$, which reach zero periodically. From these solutions we see that the only way for this problem to be solved is

$$\begin{aligned} \frac{1}{Y(y)}\frac{d^2 Y(y)}{d y^2}=-k^2,\,\,\,\,\,\,\,\,with\,\,\,\,Y(0)=Y(b)=0, \end{aligned}$$

(7.151)

and

$$\begin{aligned} \frac{1}{X(x)}\frac{d^2 X(x)}{d x^2}=k^2,\,\,\,\,\,\,\,\,with\,\,\,\,X(a)=0. \end{aligned}$$

(7.152)

The solution of the first equation must be zero at the first and last points of the interval, which may be accomplished by periodic functions. The second one (7.152) will be zero only at a point, and therefore a linear combination of hyperbolic functions holds.

The general solution of (7.151) is

$$\begin{aligned} Y(y)=A\cos ky+B\sin ky. \end{aligned}$$

(7.153)

Imposing the boundary conditions it yields

$$\begin{aligned} Y(0)=A=0, \end{aligned}$$

(7.154)

and for the extreme of the rectangle at $y=b$

$$\begin{aligned} Y(b)=B\sin kb=0\Rightarrow kb=n\pi \Rightarrow k=\frac{n\pi }{b}. \end{aligned}$$

(7.155)

$$\begin{aligned} Y(y)=B\sin \left( \frac{n\pi y}{b}\right) . \end{aligned}$$

(7.156)

Introduction of the values of k obtained in (7.155) into (7.153) gives

$$\begin{aligned} \frac{d^2 X(x)}{d x^2}=\left( \frac{n\pi }{b}\right) ^2 X(x). \end{aligned}$$

(7.157)

This is the differential equation for the function X(x) to be solved, with value $X(a)=0$. As we can observe, we only have homogeneous conditions at $x=a$, then this problem is not properly a boundary value problem as in the preceding case for Y(y). However, we will avoid this drawback later.

The general solution of this equation may be written in the form of real exponentials or as a combination of hyperbolic functions, that is,

$$\begin{aligned} X(x)=C\cosh \left( \left( \frac{n\pi }{b}\right) (x-L)\right) +D\sinh \left( \left( \frac{n\pi }{b}\right) (x-a)\right) . \end{aligned}$$

(7.158)

This solution must satisfy $X(a)=0$, thus

$$\begin{aligned} X(a)=C+0=0\Rightarrow C=0, \end{aligned}$$

(7.159)

and therefore

$$\begin{aligned} X(x)=D\sinh \left( \left( \frac{n\pi }{b}\right) (x-a)\right) . \end{aligned}$$

(7.160)

Knowing the expressions for X(x) and Y(y), we can put the solution as the product of (7.156) and (7.160)

$$\begin{aligned} V(x,y)=X(x)Y(y)=B\,D\sin \left( \frac{n\pi y}{b}\right) \sinh \left( \left( \frac{n\pi }{b}\right) (x-a)\right) . \end{aligned}$$

(7.161)

This function V(x, y) verify $V(0,y)=V_0=200$, but (7.161) does not fulfill this non-homogeneous boundary condition at all (it is not a boundary value problem). In order to avoid this trouble, we try with linear combinations of such functions (7.161). Denoting the product $BD=\alpha$, we can put

$$\begin{aligned} V(x,y)=X(x)\,Y(y)=\sum _{1}^{\infty } \alpha _{n}\sin \left( \frac{n\pi y}{b}\right) \sinh \left( \left( \frac{n\pi }{b}\right) (x-a)\right) . \end{aligned}$$

(7.162)

The objective is now to obtain the coefficients $\alpha _{n}$ of the series so that the solution meets the aforementioned condition at $x=0$. For calculating them we will employ the value of the potential on the side AD, that is,

$$\begin{aligned} V(0,y)=\sum _{1}^{\infty } \alpha _{n}\sin \left( \frac{n\pi y}{b}\right) \sinh \left( \frac{-n\pi a}{b}\right) =V_0. \end{aligned}$$

(7.163)

For obtaining the values of $\alpha _{n}$ we multiply both members by $\sin \left( \frac{m\pi y}{b}\right)$, m being an integer and we integrate in the interval from 0 to b for the variable y

$$\begin{aligned} \sum _{1}^{\infty } \int _{0}^{b}\alpha _{n}\sin \left( \frac{n\pi y}{b}\right) \sin \left( \frac{m\pi y}{b}\right) \sinh \left( \frac{-n\pi a}{b}\right) dy=\int _{0}^{b}V_0\sin \left( \frac{m\pi y}{b}\right) dy. \end{aligned}$$

(7.164)

As $\sinh (-x)=-\sinh (x)$, we have

$$\begin{aligned} -\sum _{1}^{\infty }\alpha _{n}\sinh \left( \frac{-n\pi a}{b}\right) \int _{0}^{b}\sin \left( \frac{n\pi y}{b}\right) \sin \left( \frac{m\pi y}{b}\right) dy=\int _{0}^{b}V_0\sin \left( \frac{m\pi y}{b}\right) dy. \end{aligned}$$

(7.165)

We must now compute the integral for all n and m. Taking into consideration that

$$\begin{aligned} \int \sin (py)\sin (qy)dy=\frac{\sin (p-q)y}{2(p-q)}-\frac{\sin (p+q)y}{2(p+q)}, \end{aligned}$$

(7.166)

we get

$$\begin{aligned} \int _{0}^{b}\sin \left( \frac{n\pi y}{b}\right) \sin \left( \frac{m\pi y}{b}\right) dy= {\left\{ \begin{array}{ll} 0,&{} \text {if } n\ne \text { m}, \\ \frac{b}{2},&{} \text {if }n=m. \end{array}\right. } \end{aligned}$$

(7.167)

$$\begin{aligned} -\alpha _{n}\frac{b}{2}\sinh \left( \frac{n\pi a}{b}\right) =\int _{0}^{b}V_0\sin \left( \frac{m\pi y}{b}\right) dy, \end{aligned}$$

(7.168)

thus

$$\begin{aligned} \alpha _{n}=-\frac{2}{b\sinh \left( \frac{n\pi a}{b}\right) }\int _{0}^{b}V_0\sin \left( \frac{n\pi y}{b}\right) dy=-\frac{2V_0}{b\sinh \left( \frac{n\pi a}{b}\right) }\frac{b}{n\pi }\left( -\cos (n\pi )+1 \right) , \end{aligned}$$

(7.169)

which depends on the values of n. In a general form it may be written¹⁵

$$\begin{aligned} \alpha _{n}= {\left\{ \begin{array}{ll} 0 &{} \text {if } n \text { even}, \\ -\frac{4V_0}{n\pi \sinh \left( \frac{n\pi a}{b}\right) } &{} n \text { odd}. \end{array}\right. } \end{aligned}$$

(7.170)

To be clear, right now we will denote these odd numbers by l instead of n. With all the data the final expression for the potential is

$$\begin{aligned} V(x,y)=\sum _{l=odd}^{\infty } -\frac{4V_0}{\sinh \left( \frac{l\pi a}{b}\right) }\sin \left( \frac{l\pi y}{b}\right) \sinh \left( \left( \frac{l\pi }{b}\right) (x-a)\right) . \end{aligned}$$

(7.171)

In Figs. 7.14, 7.15, and 7.16 the solution considering up to six terms of the expansion (7.171) is plotted. As we can see the potential reaches its maximum at $x=0$ and decrees progressively when x grows, being zero at $x=a$ and on sides $y=0$ and $y=b$. Due to the finite number of terms chosen (six), over the side (0, y) the solution oscillates not being $V_0$ exactly. To reproduce an almost constant potential on this side in this graphic, we would need more terms of the series. Figure 7.17a represents curves of constant potential for $n=1$ (Fig. 7.14a), and Fig. 7.17b the same in the case of $n=11$. As we can observe, both family of curves are similar to each other from $x=0.04$ m, being the major difference in the vicinity of the side AD. Therefore, in this problem the election of the number of terms of the series mostly affects the result near the edge not grounded.

Fig. 7.14

a Graphic of the first element of the expansion of (7.171). b This corresponds to the potential V(x, y) considering $n=1$ and $n=3$

(b) Once we have calculated the potential for any point interior to the region delimited by the four sides of the rectangle, the electric field may be obtained by means of the gradient. In fact,

$$\begin{aligned} \mathbf {E}=-\nabla V(x,y)=\sum _{l=odd}^{\infty }\frac{4V_0 l}{nb\sinh \left( \frac{l\pi a}{b}\right) }\sin \left( \frac{l\pi y}{b}\right) \cosh \left( \left( \frac{l\pi }{b}\right) (x-a)\right) \mathbf {u}_x+ \end{aligned}$$

(7.172)

$$ \sum _{l=odd}^{\infty }\frac{4V_0 l}{nb\sinh \left( \frac{l\pi a}{b}\right) }\cos \left( \frac{l\pi y}{b}\right) \sinh \left( \left( \frac{l\pi }{b}\right) (x-a)\right) \mathbf {u}_y. $$

Fig. 7.15

c Representation considering the three first terms. d Idem for four terms of the series.

Fig. 7.16

e Graphic with five terms. f Idem for six terms

Fig. 7.17

a Graphic of the first element of the expansion of (7.171). b This corresponds to the potential V(x, y) considering $n=1$ and $n=3$

Fig. 7.18

Graphic of the electric field with only the first element of the expansion of (7.172)

Fig. 7.19

This corresponds to the field $\mathbf {E}(x,y)$ with $n=1$ and $n=3$

Fig. 7.20

The calculation procedure is studied as two independent problems

Figures 7.18 and 7.19 show the electric field (red vectors) at every interior point of the cavity ABCD. Note that from 0.1 m, approximately, the electric field is very small. We can easily understand this result by looking at the graphics of Fig. 7.16 (the other pictures of Figs. 7.14 and 7.15 are also valid, but Fig. 7.16 is more precise). In fact, the function V(x, y) is very flat up $x=0.1$ m, even in the solution regarding only the first term. As a result, as the electrostatic field is related with the potential by a gradient, the electric field must be smaller on this part than in regions where the slope of the potential is larger (the region close to the side AD). Note: As we have commented in this problem, for correctly applying the method of separation of variables the partial differential equation and the boundary conditions must be linear and homogeneous. The importance of this exercise lies on the fact that other problems with the same symmetry, but with other boundary conditions, may be solved taking this result as a basis. For instance if we would have the same prismatic system but besides the side AD, the side AB held to a potential $V(a,y)=g(y)$, we can pose the problem as two independent problems, each of one has its boundary conditions, and at the end apply the principle of superposition. This is schematically represented in the picture (Fig. 7.20). Therefore the final solution V(x, y) of the problem may be obtained as the sum two problems in which only one side is held to a potential. We mean that

$$\begin{aligned} V(x,y)=V_1(x,y)+V_2(x,y). \end{aligned}$$

(7.173)

The same principle applies in the most general case if every side has a nonzero potential. For this case we can write

$$\begin{aligned} V(x,y)=V_1(x,y)+V_2(x,y)+V_3(x,y)+V_4(x,y). \end{aligned}$$

(7.174)

Fig. 7.21

Electric field $\mathbf {E}$ at two points on the same plane but with different coordinate $\phi$. Because of the rotational symmetry, its projections over the basis $\{\mathbf {u}_r,\mathbf {u}_\phi ,\mathbf {u}_\theta \}$ are the same when maintaining r and $\theta$ constant and $\phi$ varies to $(\phi +\phi ')$

• 7.4 A dielectric sphere of constant $\varepsilon _1$ and radius R is placed coinciding its centre with the origin of the coordinates system OXYZ. The sphere is surrounded by a system of permittivity $\varepsilon _2$. If before locating the sphere in the medium there was a homogeneous electric field $\mathbf {E}=E\mathbf {u}_z$, calculate the electric field inside and outside of the dielectric ball.

Solution

Due to the rotational symmetry of this problem we can employ spherical coordinates (see Fig. 7.21). Taking into consideration that we do not have free charges, the differential equation to be solved is that of the Laplace (7.29), thus

$$\begin{aligned} \triangle V\equiv \frac{1}{r^2}\frac{\partial }{\partial r}\left( r^2\frac{\partial V}{\partial r}\right) + \frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial V}{\partial \theta }\right) +\frac{1}{r^2\sin \theta }\frac{\partial ^2 V}{\partial \phi ^2}=0 \end{aligned}$$

(7.175)

In this problem the solution must not depend on the angle $\phi$, because of the revolution symmetry, that is, if the field at a point $P(r,\phi ,\theta )$ is $\mathbf {E}$, the components of $\mathbf {E}$ over the spherical unitary vectors basis $\mathbf {u}_r$, $\mathbf {u}_\phi$, and $\mathbf {u}_\theta$ displaced at any other point $P'(r,\phi +\phi ',\theta )$ must be the same. For this reason we can eliminate the derivative with respect to $\phi$ in the above equation, i.e.,

$$\begin{aligned} \frac{1}{r^2}\frac{\partial }{\partial r}\left( r^2\frac{\partial V}{\partial r}\right) + \frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial V}{\partial \theta }\right) =0. \end{aligned}$$

(7.176)

Considering the symmetry of the problem seem to be adequate, looking for solutions of the form shown in (7.30), that is, the spherical harmonics

$$\begin{aligned} V(r,\phi ,\theta )=\sum _{0}^{\infty }\left( a_{m}r^{m}+\frac{b_m}{r^{m+1}}\right) P_{m}(\cos \theta ), \end{aligned}$$

(7.177)

where $P_{i}^{m}(\cos \theta )$ are the Legendre polynomials. However, the solution will be only valid if it satisfies the following boundary conditions:

$$\begin{aligned} V_1(R,\phi ,\theta )=V_2(R,\phi ,\theta ), \end{aligned}$$

(7.178)

$$\begin{aligned} \varepsilon _1\left( \frac{\partial V_1}{\partial r}\right) = \varepsilon _2\left( \frac{\partial V_2}{\partial r}\right) , \end{aligned}$$

(7.179)

and

$$\begin{aligned} V_2(R,\phi ,\theta )\longrightarrow V, \,\,\,r\longrightarrow \infty . \end{aligned}$$

(7.180)

The first one means that the potentials $V_1$ and $V_2$ on the sphere surface must be the same. The second condition depicts that the normal components of the displacement vector $\mathbf {D}$ on the boundary are equal. The third shows that at large distances the potential takes the same values as before the dielectric sphere was placed into the electric field. Physically it is the same to say that the sphere has no influence on the potential far away of its loci; if an effect exists, it is in proximity to the sphere (Fig. 7.22).

Fig. 7.22

a Lines of electric field for $\varepsilon _1>\varepsilon _2$. b Field lines for $\varepsilon _1<\varepsilon _2$

To solve the problem we first study the potential inside of the dielectric ball. There holds

$$\begin{aligned} V_1(r,\theta )=a_{0}P_{0}(\cos \theta )+\frac{b_{0}}{r}P_{0}(\cos \theta )+a_{1}rP_{1}(\cos \theta )+\frac{b_{1}}{r^2}P_{1}(\cos \theta ) +a_{2} r^{2}P_{2}(\cos \theta )+\cdots \end{aligned}$$

(7.181)

The Legendre polynomials may be obtained by means of the following relations

$$\begin{aligned} P_{m}(\cos \theta )=\frac{1}{2^{m}m!}\frac{d^m}{d(\cos \theta )^m}(\cos \theta ^2-1), \end{aligned}$$

(7.182)

or through

$$\begin{aligned} P_{m+1}(\cos \theta )=\frac{(2m+1)\cos \theta \,P_{m}(\cos \theta )+mP_{m-1}(\cos \theta )}{(m+1)}. \end{aligned}$$

(7.183)

For the calculation it is not necessary to introduce the expanded expression of $P_{m}(\cos \theta )$. As we will see, we can work only with $P_{0}$ and $P_{1}$ at hand. These first two polynomials are $P_{0}(\cos \theta )=1$ and $P_{1}(\cos \theta )=\cos \theta$. Introduction of both terms into (7.181) gives

$$\begin{aligned} V_1(r,\theta )=a_{0}+\frac{b_{0}}{r}+a_{1}r \cos \theta +\frac{b_{1}}{r^2}\cos \theta +a_{2} r^{2}P_{2}(\cos \theta )+\frac{b_{2}}{r^3}P_{2}(\cos \theta )+\cdots \end{aligned}$$

(7.184)

This expression may be simplified if we take into account some properties that the potential inside of the sphere must have. So, the potential at the center of the ball must be well valued, but if we examine equality (7.181) we see that for $r=0$ it tends to infinity. To avoid this difficulty we make all $b_i=0$, for $i=0,1,2,\ldots$, and then we have

$$\begin{aligned} V_1(r,\theta )=a_{0}+a_{1}r \cos \theta +a_{2} r^{2}P_{2}(\cos \theta )+\cdots =\sum _{n=0}^{\infty }a_{n}r^n P_{n}(\cos \theta ) \end{aligned}$$

(7.185)

Outside of the sphere we have an expansion identical to (7.181) but with other coefficients (now with primes’), i.e.,

$$\begin{aligned} V_2(r,\theta )=a'_{0}+\frac{b'_{0}}{r}+a'_{1}r \cos \theta +\frac{b'_{1}}{r^2}\cos \theta +a'_{2} r^{2}P_{2}(\cos \theta )+\frac{b'_{2}}{r^3}P_{2}(\cos \theta )+\cdots = \end{aligned}$$

(7.186)

$$ =a'_{0}+\frac{b'_{0}}{r}+a'_{1}r \cos \theta +\frac{b'_{1}}{r^2}\cos \theta +\sum _{n>1}^{\infty }\frac{b'_{n}}{r^{n+1}}P_{n}(\cos \theta )+ \sum _{n>1}^{\infty } a'_{n}r^{n}P_{n}(\cos \theta ). $$

This equation must satisfy the boundary conditions too. As we commented before, far away from the ball the potential and the electric field must not depend on the presence of the dielectric sphere. It means that if the field is $\mathbf {E}=E\mathbf {u}_z$, then it holds for the potential

$$\begin{aligned} \mathbf {E}=-\nabla V\Rightarrow V(r,\theta )=-\int \mathbf {E}.d\mathbf {l}=-E\,z+V_0, \end{aligned}$$

(7.187)

$V_0$ being a constant. For an arbitrary point of space r, as $z=r\,\cos \theta$, (7.187) may be written as

$$\begin{aligned} V_2(r,\theta )=-E\,r\,\cos \theta +V_0. \end{aligned}$$

(7.188)

This is the form of that the potential has at infinity and this condition must be valid for (7.184), and therefore

$$\begin{aligned} \lim _{r\rightarrow \infty }V_2(r,\theta )=-E\,r\,\cos \theta +V_0= \end{aligned}$$

(7.189)

$$ =\lim _{r\rightarrow \infty }\left( a'_{0}+\frac{b'_{0}}{r}+a'_{1}r \cos \theta +\frac{b'_{1}}{r^2}\cos \theta +\sum _{n>1}^{\infty }\frac{b'_{n}}{r^{n+1}}P_{n}(\cos \theta )+ \sum _{n>1}^{\infty } a'_{n}r^{n}P_{n}(\cos \theta )\right) . $$

From this equality it follows

$$\begin{aligned} -E\,r\,\cos \theta +V_0=a'_{0}+a'_{1}r \cos \theta , \end{aligned}$$

(7.190)

thus

$$\begin{aligned} a'_{0}=V_0, \end{aligned}$$

(7.191)

$$\begin{aligned} a'_{1}=-E, \end{aligned}$$

(7.192)

and for $n>1$

$$\begin{aligned} a'_{n}=0, \end{aligned}$$

(7.193)

otherwise the potential at large distances would be divergent. Substitution of these constants into (7.184) leads to

$$\begin{aligned} V_2(r,\theta )=V_{0}+-E\,r \cos \theta +\frac{b'_{1}}{r^2}\cos \theta +\sum _{n\ne 1}^{\infty }\frac{b'_{n}}{r^{n+1}}P_{n}(\cos \theta ). \end{aligned}$$

(7.194)

Once we have the general expressions for the potential inside and outside of the sphere we can apply the other boundary conditions. The first one equates the potential on the ball surface, that is, $V_1(r,\phi ,\theta )=V_2(r,\phi ,\theta )$, then

$$\begin{aligned} \sum _{n=0}^{\infty } a_{n}r^{n}P_{n}(\cos \theta )|_{r=R}=V_{0}+\left( -E\,r+\frac{b'_{1}}{R^2}\right) \cos \theta +\sum _{n\ne 1}^{\infty }\frac{b'_{n}}{r^{n+1}}P_{n}(\cos \theta )|_{r=R}. \end{aligned}$$

(7.195)

Since the last equation must hold for any value of $\theta$ (remember that it does not depend on $\phi$), the only possibility we have is

$$\begin{aligned} a_{0}=V_0, \end{aligned}$$

(7.196)

$$\begin{aligned} \left( -E\,R+\frac{b'_{1}}{R^2}\right) =R\,a_{1}, \end{aligned}$$

(7.197)

and

$$\begin{aligned} \frac{b'_{n}}{R^{n+1}}=R^n\,a_{n},\,\,\,\,\,n>1. \end{aligned}$$

(7.198)

In order to apply (7.179) we take the derivative of (7.184) and (7.186), obtaining

$$\begin{aligned} \left( \frac{\partial V_1}{\partial r}\right) _{r=R}=\sum _{n>1}^{\infty }a_{n}r^{n+1} P_{n}(\cos \theta ) , \end{aligned}$$

(7.199)

and

$$\begin{aligned} \left( \frac{\partial V_2}{\partial r}\right) _{r=R}=-\left( E+\frac{2b'_{1}}{R^3}\right) \cos \theta -\sum _{n\ne 1}^{\infty }(n+1)\frac{b'_{n}}{r^{n+2}}P_{n}(\cos \theta ), \end{aligned}$$

(7.200)

then using (7.179)

$$\begin{aligned} \varepsilon _1\sum _{n>1}^{\infty }a_{n}r^{n+1} P_{n}(\cos \theta )=-\varepsilon _2\left( E+\frac{2b'_{1}}{R^3}\right) \cos \theta -\varepsilon _2\sum _{n\ne 1}^{\infty }(n+1)\frac{b'_{n}}{r^{n+2}}P_{n}(\cos \theta ) , \end{aligned}$$

(7.201)

which gives

$$\begin{aligned} \varepsilon _{1}a_{1}=-\varepsilon _2\left( E+\frac{2b'_{1}}{R^3}\right) , \end{aligned}$$

(7.202)

and for $n\ne 1$

$$\begin{aligned} \varepsilon _{1}n a_{n}R^{n-1}=-\varepsilon _{2}(n+1)\,\frac{b'_{n}}{R^{n+2}}. \end{aligned}$$

(7.203)

Combining (7.197) and (7.202) we get

$$\begin{aligned} a_{1}=-\frac{3\,\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\,E, \end{aligned}$$

(7.204)

and

$$\begin{aligned} b'_{1}=\frac{\varepsilon _1-\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\,R^3\,E. \end{aligned}$$

(7.205)

On the other hand (7.198) and (7.203) are only possible if $a_0=b'_0=0$ and $a_n=b'_n=0$ for $n>1$. By inspection of (7.196) we observe that as $a_0=0$, the potential $V_0=0$, that is, the potential at the center of the dielectric sphere vanishes (see (7.184) and note that $V_1(0,\theta )=0$). Once we have the value of the constants the potential and field inside and outside of the ball are, respectively

$$\begin{aligned} V_1(r,\theta )=-\frac{3\,\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\,E\,r\cos \theta , \,\,\,\,\,r<R, \end{aligned}$$

(7.206)

and

$$\begin{aligned} V_2(r,\theta )=-\left( 1-\frac{\varepsilon _1-\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\frac{R^3}{r^3}\right) \,r\,E\cos \theta , \,\,\,\,r>R. \end{aligned}$$

(7.207)

The electric field may be determined through the gradient. For $r<R$, we have

$$\begin{aligned} \mathbf {E}_1=-\nabla V=-\frac{\partial V_1}{\partial z}=-\frac{\partial V_2 }{\partial (r\cos \theta ) }\mathbf {u}_z=\frac{3\,\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\,E\,\mathbf {u}_z\,\,\,\,\,r<R. \end{aligned}$$

(7.208)

As it is known, when a dielectric material remains in the presence of an electric field, a local redistribution of its charges is produced and the material will polarize. As a consequence, in general, two kinds of bounded charges appear: a volume distribution of polarization charge $\rho _p=-\nabla \cdot \mathbf {P}$, and a surface polarization charge $\sigma _p=-\mathbf {P}\cdot \mathbf {n}$, where $\mathbf {P}$ and $\mathbf {n}$ are the polarization field and the unitary normal vector to the surface, respectively (see Chap. 2). In the case studied we see from (7.208) that inside the ball the electric field is uniform, hence the polarization $\mathbf {P}=\varepsilon _0\chi _e\mathbf {E}$ too, which means that we do not have $\rho _p$ but only $\sigma _p$. The apparition of these bounded surface charges behaves as a source of electric field modifying the electric properties inside of the sphere and in its outside surroundings. Physically it means that the field inside of the ball is the result of the addition of the exterior electric field $\mathbf {E}$ (generated before the ball was placed there) and the field created in the interior by the bound charges that appear on the surface of the dielectric sphere. Calling $\mathbf {E}_{\sigma _p}'$ the electric field created by $\sigma _p$ inside, it yields

$$\begin{aligned} \mathbf {E}_1=\mathbf {E}_{exterior}+\mathbf {E}_{\sigma _p}'\Rightarrow \mathbf {E}_{\sigma _p}'=\mathbf {E}_1-\mathbf {E}_{exterior}=\frac{3\,\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\,E\,\mathbf {u}_z-E\,\mathbf {u}_z= \left( \frac{\varepsilon _2-\varepsilon _1}{\varepsilon _1+2\varepsilon _2}\right) E\,\mathbf {u}_z. \end{aligned}$$

(7.209)

Outside we obtain the field in the same way

$$\begin{aligned} \mathbf {E}_2(r,\theta )=-\nabla V=-\frac{\partial V_2}{\partial r}=-\frac{\partial V_2}{\partial r }\mathbf {u}_r-\frac{1}{r}\frac{\partial V_2}{\partial \theta }\mathbf {u}_\theta =\left( 1+2\frac{\varepsilon _1-\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\frac{R^3}{r^3}\right) \,E\cos \theta \,\mathbf {u}_r \end{aligned}$$

(7.210)

$$ -\left( 1-\frac{\varepsilon _1-\varepsilon _2}{\varepsilon _1+2\varepsilon _2}\frac{R^3}{r^3}\right) \,E\sin \theta \,\mathbf {u}_\theta ,\,\,\,\,\,r>R. $$

As we can prove, contrary to what occurs inside, the field is not homogeneous and depends on the point examined. Equation (7.210) shows that in the surrounding material of the sphere we will observe the major modification of the electric field. However, a very different behavior of the field lines happens depending on the values of $\varepsilon _1$ and $\varepsilon _2$. So, from (7.208) it follows that

$$\begin{aligned} \mathbf {E}_1=\frac{3}{\left( 2+\frac{\varepsilon _1}{\varepsilon _2}\right) }\,E\,\mathbf {u}_z. \end{aligned}$$

(7.211)

and then, if $\varepsilon _1>\varepsilon _2$ the electric field inside is smaller than $\mathbf {E}$. On the contrary, when $\varepsilon _1<\varepsilon _2$ holds $\mathbf {E}_1>\mathbf {E}$ (the contrary occurs with the displacement vector $\mathbf {D}$). This case is very important in industrial production processes. In fact, the existence of air bubbles in a dielectric matrix creates an increment of the electric field inside of the microspheres. As the gas bubbles have a dielectric constant $\varepsilon _1\approx 1$, if $\varepsilon _1<\varepsilon _2$ (or $<<$), the electric field inside them can be very high, which may produce discharges into the system. For this reason, in most applications where it matters, a control in the material fabrication is needed in order to avoid this problem. As a particular case, if the permittivities are the same, that is $\varepsilon _1=\varepsilon _2$, then $\mathbf {E}_1=\mathbf {E}$ and the field are not affected.

The understanding of this study shown is very important in some fields of applied physics and technology. For example, sometimes researchers try to modify dielectric properties of materials by means of the inclusion of dielectric (and metallic too) balls into a dielectric matrix. In those cases the problem is to calculate the effective permittivity $\varepsilon _{eff}$ of the new system. To solve this question some formulae may be used. The two most important are the relationship of Maxwell-Wagner and the Rayleigh formula . The first one is

$$\begin{aligned} \varepsilon _{eff}(\alpha )=\left( \frac{2\,\varepsilon _2+\varepsilon _1+2\,\alpha \,(\varepsilon _1-\varepsilon _2)}{2\,\varepsilon _2+\varepsilon _1-\alpha \,(\varepsilon _1-\varepsilon _2)}\right) \,\varepsilon _2, \end{aligned}$$

(7.212)

where $\varepsilon _1$ and $\varepsilon _2$ are the permittivities of the spheres and of the medium (matrix), respectively. For the simplified relation of Rayleigh $(\alpha<<1)$ we have

$$\begin{aligned} \varepsilon _{eff}(\alpha )=\left( 1+3\,\alpha \frac{(\varepsilon _1-\varepsilon _2)}{(2\,\varepsilon _1+\varepsilon _2)}\right) \,\varepsilon _2. \end{aligned}$$

(7.213)

Recently, however, recent attention has shifted to the employment of inhomogeneous materials containing inclusions for modifying optical properties.¹⁶ Specifically, it is possible to induce surface anisotropy by different mechanisms by introducing inclusions adequately (dielectric or metallic-see next problem), leading to a change in the refractive index of the material (see Chapter 14). In fact, it can be shown that the anisotropic induced properties may be controlled by adjusting the concentration of the spheres and the thickness of the corresponding layer, which opens a large number of scientific and technical applications.

• 7.5 A metallic sphere of radius R is located with its center coinciding with the origin of the coordinates system OXYZ. The sphere is immersed into a dielectric substance of permittivity $\varepsilon _2$. If before locating the sphere in the medium there was a uniform electric field $\mathbf {E}=E\mathbf {u}_z$, obtain the electric field inside and outside of the sphere.

Solution

This problem is similar to the one explained previously, but now the ball is made of a metallic material instead of a dielectric. At first sight we can proceed in the same manner, however due to the characteristics of the conductors, some new information about the boundary conditions appears. In fact, the potential must fulfill the following conditions

$$\begin{aligned} V(R,\phi ,\theta )=C, \end{aligned}$$

(7.214)

C being a constant, and

$$\begin{aligned} V(R,\phi ,\theta )\longrightarrow V, \,\,\,r\longrightarrow \infty . \end{aligned}$$

(7.215)

The first condition shows that the potential at $r=R$ is constant. Actually this potential is constant in all the ball, because it is a conductor, hence the electric field inside vanishes. The second BC is the same as we studied before for the dielectric sphere and physically means that the effect of the metallic ball on the field is only important in its owing vicinity, and not far away from it.

The resolution of this problem again requires using the Laplace equation in spherical coordinates where the variable $\phi$ does not appear due to the symmetry of the system. However, instead of solving this equation together with the boundary conditions, we can proceed in an easier way if we consider the results of the preceding exercise. In effect, as the ball is made of a metallic material, we can consider that the constant $\varepsilon _1$ of the ball tends to infinity. If that is true, we could directly make the substitution $\varepsilon _1\rightarrow \infty$ in (7.206), (7.207), (7.208) and (7.209) for obtaining the solution. According to this idea we get

$$\begin{aligned} V_1(r,\theta )=0+C, \,\,\,\,\,r<R, \end{aligned}$$

(7.216)

and

$$\begin{aligned} V_2(r,\theta )=-\left( 1-\frac{R^3}{r^3}\right) \,r\,E\cos \theta , \,\,\,\,r>R. \end{aligned}$$

(7.217)

For the electric inside of the ball ($r<R$), we have

$$\begin{aligned} \mathbf {E}_1=0\,\,\,\,\,r<R, \end{aligned}$$

(7.218)

and outside ($r>R$)

$$\begin{aligned} \mathbf {E}_2(r,\theta )=\left( 1+2\,\frac{R^3}{r^3}\right) \,E\cos \theta \,\mathbf {u}_r- \left( 1-\frac{R^3}{r^3}\right) \,E\sin \theta \,\mathbf {u}_\theta ,\,\,\,\,\,r>R. \end{aligned}$$

(7.219)

In the same manner as we commented in the preceding problem, the inclusion of metallic balls in a dielectric matrix is of great practical interest. In this way we can modify the physical properties of the matrix varying the concentrations of the metallic balls in the dielectric. A simple formula that gives the modified permittivity is due to Bruggeman

$$\begin{aligned} \epsilon _{eff}(\alpha )=\frac{\epsilon _2}{(1-\alpha )^3}, \end{aligned}$$

(7.220)

where $\alpha$ is the volume concentrations of the spheres and $\epsilon _2$ is the permittivity of dielectric matrix.

• 7.6 A long wire of diameter R carries a current I. Find (a): The vector potential $\mathbf {A}$ for $\rho <R$. (b) The vector potential $\mathbf {A}$ for $\rho >R$. (c) The magnetic field $\mathbf {B}$.

Solution

(a) This problem could be solved using techniques we have seen in Chap. 5. In fact, by means of Ampère’s circuital law we can obtain the magnetic field everywhere and, once we know it, the relation between both fields, i.e., the potential vector and $\mathbf {B}$ is determined by the following relation

$$\begin{aligned} \oint _{\Gamma }\mathbf {A}d\mathbf {l}=\int \int _{S}\mathbf {B}\cdot d\mathbf {S}. \end{aligned}$$

(7.221)

However, taking into account the methods exposed in the introduction, we will try to seek a solution directly solving the corresponding differential equation.

For non-varying electromagnetic fields (see Chap. 5) it holds that

$$\begin{aligned} \nabla ^2\mathbf {A}=-\mu _0\mathbf {j}. \end{aligned}$$

(7.222)

Let us suppose we place the wire coinciding its length with the OZ axis. As the wire is very long and its cross section is constant, we have translational symmetry along OZ in the direction of the current and rotational symmetry, too. Starting from a cylindrical coordinate frame it means that the solution will depend neither on the coordinate z, nor the angle $\phi$. For this reason we can study the problem by eliminating the partial derivatives of these variables. Doing so, we have

$$\begin{aligned} \frac{1}{\rho }\frac{\partial }{\partial \rho }\left( \rho \frac{\partial A_z}{\partial \rho }\right) =\mu _0 j, \end{aligned}$$

(7.223)

whose solution may be obtained by direct integration

$$\begin{aligned} \rho \frac{\partial A_z}{\partial \rho }=-\mu _0\int j\rho d\rho =-\mu _0 j\frac{\rho ^2}{2}+C_1, \end{aligned}$$

(7.224)

and then

$$\begin{aligned} \frac{\partial A_z}{\partial \rho }=\frac{C_1}{\rho }-\mu _0 j\frac{\rho }{2}\Rightarrow A_z(\rho )=C_1\ln \rho -\mu _0 j\frac{\rho ^2}{4}+C_2, \end{aligned}$$

(7.225)

where $C_1$ and $C_2$ are two constants. As the vector potential must be finite in the region examined, to accomplish this condition, the only possibility we have is that $C_1=0$, otherwise the logarithm becomes infinity for $\rho =0$, with then yields

$$\begin{aligned} A_z(\rho )=-\mu _0 j\frac{\rho ^2}{4}+C_2. \end{aligned}$$

(7.226)

(b) In this case the calculation refers to exterior region of the wire, where there is no current. As a result, (7.222) adopts the form

$$\begin{aligned} \nabla ^2\mathbf {A}=0. \end{aligned}$$

(7.227)

Integrating again, we get

$$\begin{aligned} \frac{1}{\rho }\frac{\partial }{\partial \rho }\left( \rho \frac{\partial A_z}{\partial \rho }\right) =0\Rightarrow \frac{\partial }{\partial \rho }\left( \rho \frac{\partial A_z}{\partial \rho }\right) =0\Rightarrow \left( \rho \frac{\partial A_z}{\partial \rho }\right) =D_1, \end{aligned}$$

(7.228)

$D_1$ being a constant to be determined. An integration more gives

$$\begin{aligned} A_z=\int \frac{D_1}{\rho }d\rho +D_2=D_1\ln \rho +D_2. \end{aligned}$$

(7.229)

At this point we are tempted to put $D_1=0$ in (7.229), because for $\rho \longrightarrow \infty$ the $\ln \rho \longrightarrow \infty$, and making $D_1$ this constat zero this problem is avoided; however it would be wrong. The reason for this is that, contrary to what happens with the magnetic field, the vector potential can be unbounded if the region where the current extends is not confined in a finite domain of space, as in our case (the wire is infinite (mathematically) see Chap. 5). Therefore, the expression (7.229) is valid, in principle, it being necessary to seek other possibilities to determine the constants.

(c) The magnetic field can be computed by using the relation between B and A (5.11), i.e.

$$\begin{aligned} \mathbf {B(\mathbf {r})}=\nabla \times \mathbf {A(\mathbf {r})}. \end{aligned}$$

(7.230)

Considering that the only component of the vector potential we have is the z component, we can write

$$\begin{aligned} B_\phi =-\frac{\partial A_z}{\partial \rho }=\frac{1}{2}\mu _0 j\rho , \end{aligned}$$

(7.231)

which represents the general formula for B inside of the wire. In the same manner, calculating the rotational in a exterior region (5.11) leads to

$$\begin{aligned} B_\phi =-\frac{\partial A_z}{\partial \rho }=-\frac{D_1}{\rho }. \end{aligned}$$

(7.232)

Now, for obtaining $D_1$ and $D_2$, we impose the continuity of the magnetic field on the surface of the wire

$$\begin{aligned} B_{\phi }(1)_{\rho =R}=B_{\phi }(2)_{\rho =R}, \end{aligned}$$

(7.233)

yielding

$$\begin{aligned} \mu _0 j\frac{R}{2}=-\frac{D_1}{R}\Rightarrow D_1=-\mu _0 j\frac{R^2}{2}, \end{aligned}$$

(7.234)

thus

$$\begin{aligned} B_\phi =\frac{\mu _0 jR^2}{2\rho }. \end{aligned}$$

(7.235)

In relation with B, we do not need anything else. Observe that, even though we did not obtain constants $C_1$ and $D_2$ in (7.224) and (7.229), the magnetic field has been determined without difficulty. The reason of that is due to the fact that for computing B we must calculate a partial derivative, then eliminating $C_1$ and $D_2$, indirectly. Nevertheless, we can determine such constants by imposing the boundary conditions for the vector potential

$$\begin{aligned} A_{z}(1)_{\rho =R}=A_{z}(2)_{\rho =R}, \end{aligned}$$

(7.236)

which gives

$$\begin{aligned} -\mu _0 j\frac{R^2}{4}+C_2=-\mu _0 j\frac{R^2}{2}\ln R+D_2. \end{aligned}$$

(7.237)

This result has no unique solution, because we have two constants and only one equation. However, we can impose a normalization condition at the origin of coordinates. For instance, we can choose $A_{z}(1)_{\rho =0}=0$ which gives $C_2=0$ (or any other value), and then we get

$$\begin{aligned} -\mu _0 j\frac{R^2}{4}=-\mu _0 j\frac{R^2}{2}\ln R+D_2\Rightarrow D_2=\frac{1}{4}\mu _0 jR^2(\ln R^2-1). \end{aligned}$$

(7.238)

Introducing these values into (7.226) and (7.229), holds

$$\begin{aligned} A_z(\rho )=-\frac{1}{4}\mu _0 j\rho ^2, \end{aligned}$$

(7.239)

and

$$\begin{aligned} A_z(\rho )=-\frac{1}{2}\mu _0 jR^2\ln \rho +\frac{1}{4}\mu _0 jR^2(\ln R^2-1). \end{aligned}$$

(7.240)

• 7.7 Consider two charges of different signs separated by distance 2a over the OY axis. Show that if the absolute value of them is the same, the geometric points of zero potential are located on a plane.

Solution

As we explained in Sect. 7.4, when two charges of opposite signs, namely $q_1$ and $q_2$, are placed at a distance a, a family of different surfaces are obtained, each one corresponds to a constant C (7.48). In the example seen there, for $C=0$ we obtained a spherical surface whose centre was displaced with respect to the location of the charge $q_1$ (Fig. 7.23a). In the present problem the charges have the same absolute values, i.e. $|q_1|=|q_2|=q$, thus probably the surface of constant potential changes (see the example in Sect. 7.4).

Fig. 7.23

Potential due to two identical charges of opposite signs. a The straight line in the middle of this figure represents the points in two dimensions where the potential is zero. b In a three dimensional representation we have a plane

To demonstrate it, let us first suppose that the charges are located at points $P_1(0,-a,0)$ and $P_2(0,a,0)$. To see what are the loci of constant potential for this configuration we introduce $|q_1|=|q_2|=q$ into (7.48), and the distance between charges 2a, which leads to

$$\begin{aligned} (x^2+(y-a)^2+z^2)=(x^2+(y+a)+z^2). \end{aligned}$$

(7.241)

This equation is verified only in the case that $y=0$, which represents the equation of a plane (see Fig. 7.23b). This result is important and provides a means to other questions that may be solved by the method of images. In fact, it is tied to a problem that we could pose as follows. Let us imagine a point charge q placed at the point of coordinates P(0, a, 0) in front of a semi-infinite conducting block whose face coincides with the OXZ plane. If the metallic block is held to zero potential, calculate the potential at a generic point of space P(x, y, z), $y>0$. To stress this issue, we must find a charge disposition in such a way that the potential on the plane $y=0$ disappears. For this to occur we can place a fictitious charge $-q$ in front of the charge q, at a distance a with respect the origin of the coordinate frame. By this procedure we immediately see that if we add the potentials corresponding to both charges we again obtain

$$\begin{aligned} V(x,y,z)=V_1(x,y,z)+V_2(x,y,z)=\frac{1}{4\pi \varepsilon _0}\frac{q}{\sqrt{x^2+(y-a)^2+z^2}}+ \displaystyle \frac{1}{4\pi \varepsilon _0}\frac{-q}{\sqrt{x^2+(y+a)^2+z^2}}. \end{aligned}$$

(7.242)

As we can prove (7.242) fulfills the Laplace equation and the boundary conditions, namely $V(x,0,z)=0$. It means that, following the uniqueness theorem, expression (7.242) must be the solution for the potential created by a charge q in front of a metallic plane.

Once we know the potential, other questions as how does the density charge distribute on the metallic surface, or the force between the charge and the plane, can be answered easily.

For the new density of charge on the metallic plane as a result of the presence of the charge q at P(0, a, 0) we have

$$\begin{aligned} \sigma (x,0,z)=\epsilon _0\left( \frac{\partial V(x,y,z)}{\partial n}\right) _{y=0}=\epsilon _0\mathbf {n}\mathbf {E}, \end{aligned}$$

(7.243)

$\mathbf {n}$ being the normal unitary vector to the surface. The force is $\mathbf {F}=-q\nabla V_2(x,y,z)$, which coincides with the force that the image (fictitious) charge creates on the actual charge q.

Fig. 7.24

Metallic sphere in presence of a charge $q_2$. The charge $q_1$ is the corresponding image charge

• 7.8 A grounded conducting sphere of radius R has its center coinciding with the origin of coordinates O(0, 0, 0). A point charge $q_2$ is located at the point $P(0,d_0,0)$ with $d_0>R$. Find: (a) The potential at any point of space.(b) The induced surface charge density on the sphere (Fig. 7.24).

Solution

(a) Because the sphere is grounded, this problem is easily solved using the result seen in the theoretical introduction. In this introduction we have demonstrated that when two electric charges of different values are placed at a distance a from each other, the equipotential surface corresponding to the zero potential is a sphere whose center lies on the line connecting the charges. Thus, the problem can be understood as follows: given a metal sphere S and an external charge $q=q_2$, where an additional charge should be placed, such that the surface of zero potential obtained for the two charges does coincide with the spherical surface S?

To answer this question we can employ the equations seen in the theory (see Fig. 7.1). In particular those of most interest are

$$\begin{aligned} q_1=-\frac{R}{y_0+a}\,q_{2}=-\frac{R}{d_0}\,q_{2}, \end{aligned}$$

(7.244)

and

$$\begin{aligned} y_0=\frac{R^2}{d_0}, \end{aligned}$$

(7.245)

which represent the fictitious charge that together with $q_2$ produces zero potential on the surface of radius R, and the distance between $q_1$ and the center of the sphere, respectively.¹⁷ Physically it allows us to substitute the metallic sphere by a charge of magnitude given by (7.244), at $y_0$, because the effect is the same (it fulfills the boundary conditions). The potential created by this electric system outside of the sphere is

$$\begin{aligned} V(x,y,z) = \frac{1}{4\pi \varepsilon _0}\frac{R}{d_0}\frac{-q_2}{\sqrt{x^2+(y-\frac{R^2}{d_0})^2+z^2}}+ \frac{1}{4\pi \varepsilon _0}\frac{q_2}{\sqrt{x^2+(y-d_0)^2+z^2}}. \end{aligned}$$

(7.246)

The potential for points in the interior may be deduced by using the Gauss theorem shown in Chap. 2. As it was demonstrated the electric field in the interior of a conductor in equilibrium is zero. As a result, the potential is, in principle, a constant. In our case the sphere was grounded which means that the potential on the surface must be zero, and in the interior of the surface too, because the potential inside has to take the value zero when approaching the surface.

(b) The surface density charge may be calculated by means of the boundary condition in the proximity of the conductor

$$\begin{aligned} \sigma (x,y,z)=\epsilon _0\frac{\partial V(x,y,z)}{\partial n}=\epsilon _0\mathbf {n}\mathbf {E}. \end{aligned}$$

(7.247)

This means we have to obtain firstly the electric field, and then $\sigma$. The electric field is easy to calculate through the potential V(x, y, z), by using gradient, i.e. $\mathbf {E}=-\nabla V(x,y,z)$, obtaining

$$\begin{aligned} E(x,y,z)=\frac{q_2}{4\pi \varepsilon _0}\left( \frac{1}{|\mathbf {r}-\mathbf {r}_2|^{3}}(\mathbf {r}-\mathbf {r}_2) -\frac{R}{d_0|\mathbf {r}-\mathbf {r}_1|^{3}}(\mathbf {r}-\mathbf {r}_1)\right) \end{aligned}$$

(7.248)

where $\mathbf {r}_2$ and $\mathbf {r}_1$ refer to the position vectors of the charge $q=q_2$ and the image $q_1=-\frac{R}{d_0}q_2$, respectively.

Due to the symmetry of the problem, it seems to be adequate to choose the normal $\mathbf {n}$ to the surface in spherical coordinates, which coincides with $\mathbf {u}_{\rho }$.

$$\begin{aligned} E(x,y,z)=\frac{q_2}{4\pi \varepsilon _0}\left( \frac{1}{|\mathbf {r}-\mathbf {r}_2|^{3}}(r\mathbf {u}_r-d_0\mathbf {u}_y) -\frac{R}{d_0|\mathbf {r}-\mathbf {r}_1|^{3}}(r\mathbf {u}_r-\frac{R^2}{d_0}\mathbf {u}_y)\right) . \end{aligned}$$

(7.249)

In order to make the scalar product of the electric field with the normal vector we set $\mathbf {u}_y=\sin \theta \sin \phi \,\mathbf {u}_{\rho }+\cos \theta \sin \phi \,\mathbf {u}_{\theta }+\cos \theta \,\mathbf {u}_{\phi }$, obtaining,

$$\begin{aligned} \sigma (x,y,z)=\epsilon _0\mathbf {n}\mathbf {E}=\frac{q_2}{4\pi }\frac{1}{(x^2+(y-d_0)^2+z^2)^{3/2}}(R-d_0 \sin \theta \sin \phi ). \end{aligned}$$

(7.250)

$$ -\frac{q_2}{4\pi }\frac{R}{d_0(x^2+(y-\frac{R^2}{d_0})^2+z^2)^{3/2}}(R-\frac{R^2}{d_0}\sin \theta \sin \phi ) $$

Obviously, the induced surface charge density is not homogeneous (Figs. 7.25 and 7.26). This is logical because the presence of the charge $q_2$ in front of the metallic sphere modifies the charge distribution on its surface. Observe that, for a fixed distance between the charge and the sphere, the accumulation of induced charge is a maximum in the direction $(\phi ,\theta )=(\pi /2,\pi /2)$ with respect to the coordinate frame chosen.

Fig. 7.25

Density of charge versus the angles $\phi$ and $\theta$ corresponding to spherical coordinates. The maximum absolute value of $\sigma (x,y)$ is reached at $(\phi ,\theta )=(\pi /2,\pi /2)$. The difference between the graphics in (a) and (b) corresponds only to the sign of the charge induced

Fig. 7.26

Curves of constant density charge in the space of the angles $\phi$ and $\theta$

• 7.9 A metallic sphere of radius R has its center coinciding with the origin of coordinates O(0, 0, 0). A point charge $q_2$ is located at the point $P(0,d_0,0)$ with $d_0>R$. If the sphere is connected to a constant potential $V_0$, find: (a) The potential for all points of $\mathfrak {R}^3$. (b) The electric field. (c) The surface charge density on the surface. (d) The force exerted on the charge (Fig. 7.27).

Solution

If we wish to solve this problem directly as in the preceding case we have some difficulties, because the results obtained in the introduction were developed for the case of zero potential. At first glance we could think in solving the (7.48) for an arbitrary value of C, and proceed like in the previous exercise, that is, adjusting an image charge and its location so that the spherical surface of constant potential obtained coincides with the potential V. However this is not possible. In fact, if we solve the (7.48) for an arbitrary value of the constants C, it can be demonstrated that the surface of constant potential is not a sphere, which means that it is not possible to solve this problem by only using one image charge (Fig. 7.28).

To face this problem we can divide the system into more subproblems, but being careful that the boundary conditions of the initial system are maintained. To do so, we will employ the principle of superpositions of fields. This procedure is justified because of the linearity of the Laplace equation.¹⁸ In fact, if $V_1(\mathbf {r})$ and $V_2(\mathbf {r})$ separately verify this equation, the function constructed by the sum, i.e. $V(\mathbf {r})=V_1(\mathbf {r})+V_2(\mathbf {r})$ is also a solution. On the other hand, if we obtain a solution by means of a configuration of fictitious charges, due to the uniqueness theorem, the solution sought is the same as the actual problem (Fig. 7.29).

Fig. 7.27

Metallic sphere held to a constant potential $V_0$ in front of a charge $q_2$

Fig. 7.28

a Surface of constant potential $V(x,y,z)=C=0.25$ of two charges $q_1$ and $q_2=-\frac{R}{d_0}q_1$, for $\frac{R}{d_0}=0.5$. b Two dimensional cross-section. Observe that only changing the potential from $V=C=0$ to $V=C=0.25$, the equipotential surface obtained does not correspond to a sphere

Fig. 7.29

Considering the principle of superposition the problem is equivalent to the sum of the two problems. If we find the charge distribution that creates a potential $V_0$ on a surface of radius R, the solution may be expressed as sum of the potential due to two charges $q_1$ and $q_2$, so that $q_1=-\frac{R}{d_0}q_2$ (we obtain zero potential), and the potential produced by the other system of charges

If we remember the preceding problem, we obtained a spherical surface of zero potential. As in our case the problem is to simulate the metallic sphere that has a potential $V_0$, we must only add a configuration of charges in such a way that its corresponding potential coincides with $V_0$. Usually it would be a very, if not impossibly, difficult task. However, due to the spherical symmetry of the problem, it seems to be plausible to seek a geometrical disposition of charges that fulfills the requirements. In fact, as the potential generated by an electric charge has spherical symmetry, the easiest solution one may think is to locate a charge Q at the center of the sphere, whose exact value is at the beginning unknown. In order to obtain the same behavior for the system, the value and sign of Q must be adjusted so that the potential on the sphere surface is $V_0$. To calculate its magnitude we need only employ the expression for the potential of an electric charge Q, i.e.

$$\begin{aligned} V(r)=\frac{1}{4\pi \varepsilon _0}\frac{Q}{r}, \end{aligned}$$

(7.251)

and then the potential $V_0$ may be obtained from this latter equation as follows

$$\begin{aligned} Q=4\pi \varepsilon _{0}RV(R)=4\pi \varepsilon _{0}RV_0. \end{aligned}$$

(7.252)

Applying the commented principle of superposition, the potential at any point of space may be regarded as the sum of the potential generated by two charges $q_1$ and $q_2$ (see the former problem) and the potential due to a point charge at the origin of the coordinate frame

$$\begin{aligned} V(x,y,z) = \displaystyle \frac{1}{4\pi \varepsilon _0}\frac{R}{d_0}\frac{-q_2}{\sqrt{x^2+(y-\frac{R^2}{d_0})^2+z^2}}+ \displaystyle \frac{1}{4\pi \varepsilon _0}\frac{q_2}{\sqrt{x^2+(y-d_0)^2+z^2}}+\frac{RV_0}{\sqrt{x^2+y^2+z^2}}\,. \end{aligned}$$

(7.253)

For points $r<R$ we proceed using Gauss’ theorem. In effect, we know that the electric field inside of the conducting sphere is zero, and the potential is a constant magnitude. However, contrary to the preceding exercise, the potential now is $V_0$ if $r<R$, which fulfils the boundary condition on the surface.

(b) The electric field may be directly calculated by means of the gradient of the potential,

$$\begin{aligned} \mathbf {E}=-\nabla V(x,y,z)=\frac{q_2}{4\pi \varepsilon _0}\left( \frac{1}{|\mathbf {r}-\mathbf {r}_2|^{3}}(\mathbf {r}-\mathbf {r}_2) -\frac{R}{d_0|\mathbf {r}-\mathbf {r}_1|^{3}}(\mathbf {r}-\mathbf {r}_1)\right) +\frac{RV_0}{r^3}\mathbf {r}, \end{aligned}$$

(7.254)

where $\mathbf {r}_2=d_0\,\mathbf {u}_y$, and $\mathbf {r}_1=\frac{R^2}{d_0}\mathbf {u}_y$. Introduction of these values in the latter equality leads to

$$\begin{aligned} E(x,y,z) = \frac{q_2}{4\pi \varepsilon _0}\left( \frac{1}{|\mathbf {r}-\mathbf {r_2}|^{3}}(r\mathbf {u}_r-d_0\mathbf {u}_y) -\frac{R}{d_0|\mathbf {r}-\mathbf {r_1}|^{3}}(r\mathbf {u}_r-\frac{R^2}{d_0}\mathbf {u}_y)\right) +\frac{RV_0}{r^2}\mathbf {u}_r. \end{aligned}$$

(7.255)

In the interior of the sphere the electric field is zero.

(c) Using the formula (7.247), the surface charge density is

$$\begin{aligned} \sigma (x,y,z)=\epsilon _0 \frac{\partial V(x,y,z)}{\partial n}, \end{aligned}$$

(7.256)

and then

$$ \sigma (x,y,z)=\epsilon _0\mathbf {n}\mathbf {E}=\frac{1}{4\pi }\left( \frac{1}{(x^2+(y-d_0)^2+z^2)^{3/2}}(R-d_0 \sin \theta \sin \phi )\right) $$

$$\begin{aligned} -\frac{1}{4\pi }\left( \frac{R}{d_0(x^2+(y-\frac{R^2}{d_0})^2+z^2)^{3/2}}(R-\frac{R^2}{d_0}\sin \theta \sin \phi )\right) +\frac{\varepsilon _0RV_0}{r^3}. \end{aligned}$$

(7.257)

(d) The force on the charge by the conducting sphere is very hard to compute. However we have found an electrical configuration which, in some extent (see next problem), is equivalent to the system in Fig. 7.27. For this reason the calculus may be simplified if we take into consideration that this force is the same as that which would be produced by the two image charges. The force is

$$\begin{aligned} \mathbf {F}=\frac{-1}{4\pi \varepsilon _0}\frac{Rd_0q_{2}^2}{(d_0^2-R^2)^2}\mathbf {u}_r+ \frac{RV_0q_2}{d_0^2}\mathbf {u}_r. \end{aligned}$$

(7.258)

• 7.10 For the geometrical disposition of the preceding problem, let us suppose that initially the charge $q_2$ was far away of the sphere. If the sphere is maintained at constant potential $V_0$, calculate the work necessary to bring the charge $q_2$ from its initial position to the point $d_0$.

Solution

Applying the definition of work we will calculate it. Starting from $P_1=\infty$ and finishing at $P_2=d_0$, we have

$$\begin{aligned} W=\int _{P_1}^{P_2}\mathbf {F}(\mathbf {r}).d\mathbf {r}=\lim _{P_1\rightarrow \infty }\int _{P_1}^{P_2} q_{2}\,\mathbf {E}(\mathbf {r}).d\mathbf {r}=\lim _{P_1\rightarrow \infty }\int _{P_1}^{P_2} -q_{2}\,\nabla V(x,y,z).d\mathbf {r}= \end{aligned}$$

(7.259)

$$ =-q_2\,(V(P_2)-V(\infty ))=-q_2V(P_2), $$

that is,

$$\begin{aligned} W=\frac{-1}{8\pi \varepsilon _0}\frac{Rq_{2}^2}{(d_0^2-R^2)}+\frac{RV_0q_2}{d_0}. \end{aligned}$$

(7.260)

• 7.11 On the same metallic sphere of the preceding exercises a charge Q is placed on its surface. A point charge q is located at the point P(0, b, 0) where $b>R$. Determine: (a) The potential for all points of space. (b)The electric field. (c) The surface charge density of the conducting sphere.

Solution

This problem can be resolved with the help of the results obtained in the preceding exercises. As we have shown in the problem (7.8), by means of an image charge $q_1$ with respect to an external charge $q_2$, we can only obtain a spherical surface if the potential is zero $(V=C=0)$, otherwise it is necessary to introduce changes in the system in order to reach the desired potential. In a similar manner, when a charge Q is placed on a metallic sphere, the only thing we know is that the conducting surface will be a sphere of constant potential, but not necessarily zero. The condition we have is that the total charge enclosed in the surface is Q, then employing the Gauss theorem it may be written

$$\begin{aligned} \oint _S\mathbf {E}\cdot d\mathbf {S}=\frac{Q}{\varepsilon _0}. \end{aligned}$$

(7.261)

Due to the spherical symmetry of the problem, the idea consists of introducing another charge $q_3$ at the centre of the sphere whose value must be determined, and adjusting it so that we obtain the constant potential desired. To find out $q_3$, we can apply (7.261) as follows. Let us first thing of a charge $q_2$ in the exterior and a grounded spherical surface. For understanding this system we can put an fictitious image charge $q_1$ inside of the sphere whose effect together with $q_2$ is to bring the metallic sphere to zero potential. If now we have a actual total charge Q on the surface the potential changes. In this situation, applying (7.261) holds

$$\begin{aligned} \oint _S\mathbf {E}\cdot d\mathbf {S}=\frac{Q}{\varepsilon _0}=\frac{q_1+q_3}{\varepsilon _0}. \end{aligned}$$

(7.262)

Physically this latter result means that the final effect of having a charge Q on the sphere is equivalent to those generated by two fictitious charges at points $y_0=\frac{R^2}{b}$, and $y=0$. From (7.262) we have

$$Q=q_1+q_3\Rightarrow q_3=Q-q_1,$$

thus introducing in this expression the value of $Q_1$ (7.244) calculated previously it yields

$$\begin{aligned} q_3=Q+q_2\frac{R}{b}, \end{aligned}$$

(7.263)

and then the potential is

$$\begin{aligned} V(x,y,z) = \frac{1}{4\pi \varepsilon _0}\left( \frac{R}{b}\frac{-q_2}{\sqrt{x^2+(y-\frac{R^2}{b})^2+z^2}}+ \frac{q_2}{\sqrt{x^2+(y-b)^2+z^2}} +\frac{Q+q_2\frac{R}{b}}{\sqrt{x^2+y^2+z^2}}\right) \end{aligned}$$

(7.264)

For the interior of the metallic sphere, as we know, the potential is uniform, but due to the boundary conditions imposed it must not be necessarily the same as we saw in latter problems. In fact, the potential depends only on the net charge enclosed by the surface, and this is the charge Q located externally and the image charge that appears as a consequence of $q_2$. Thus, we have

$$\begin{aligned} V(x,y,z) = \frac{1}{4\pi \varepsilon _0}\frac{Q+q_2\frac{R}{b}}{\sqrt{x^2+y^2+z^2}}=\frac{1}{4\pi \varepsilon _0}\frac{Q+q_2\frac{R}{b}}{R} \end{aligned}$$

(7.265)

(b) The electric field may be directly calculated by means of the gradient

$$\begin{aligned} \mathbf {E}(x,y,z) = \frac{1}{4\pi \varepsilon _0}\left( {\frac{q_2}{|\mathbf {r}-\mathbf {r}_2|^{3}}(r\mathbf {u}_r-b\mathbf {u}_y) -{\frac{q_2R}{b|\mathbf {r}-\mathbf {r}_1|^{3}}(r\mathbf {u}_r-\frac{R^2}{b}\mathbf {u}_y)}}\ +\frac{Q+q_2\frac{R}{b}}{R^2}\mathbf {u}_r\right) \end{aligned}$$

(7.266)

(c) The surface charge density is (7.247)

$$\begin{aligned} \sigma (x,y,z)=\epsilon _0 \frac{\partial V(x,y,z)}{\partial n}, \end{aligned}$$

(7.267)

thus

$$\begin{aligned} \sigma (x,y,z)=\epsilon _0\mathbf {n}\mathbf {E}=\frac{1}{4\pi }\left( \frac{q_2}{(x^2+(y-b)^2+z^2)^{3/2}}(R-b \sin \theta \sin \phi )\right) - \end{aligned}$$

(7.268)

$$ \frac{1}{4\pi }\left( \frac{q_2R}{b(x^2+(y-\frac{R^2}{b})^2+z^2)^{3/2}}(R-\frac{R^2}{b}\sin \theta \sin \phi ) +\frac{Q+q_2\frac{R}{b}}{R^2}\right) . $$

• 7.12 The system of the figure is composed by a metallic sphere of radius $R_1$ and a circular conducting wire of radius $R_2$. The coil is charged with a total charge $Q_2$ (Fig. 7.30). Find the electric field at any point P on the OZ axis of a cartesian coordinate frame.

Fig. 7.30

A metallic sphere with a circular conducting ring

Fig. 7.31

Planar view of the system. a The charges $q'_{1}$, $q''_{1}$, $q'''_{1}$,..., $q_{1}^{'n}$ are the respective images of the charges $q'_{2}$, $q''_{2}$, $q'''_{2}$,..., $q_{2}^{'n}$ located in the conducting wire. b The circle in red corresponds to the metallic sphere of radius $R_1$

Solution

To face this problem we can use the results obtained in the Exercise 7.8, to help us. In fact, we will first focus our attention on the potential generated by a system composed of a metallic sphere and a charge $q_2$. As we are going to see, the solution found is directly applicable to our present problem. To explain it, look at the Fig. 7.31a. Let us suppose we examine only one of the charges located over the conducting ring of radius $R_2$, namely $q_2$. Setting aside the other charges, to $q_2$ corresponds an image charge $q_1$ located inside of the sphere. Now, if we fix our attention only to another charge $q'_{2}$ infinitely close to $q_2$, as it would be alone, we can again find a fictitious charge $q'_{1}$. Both image charges obtained are placed at distinct points in the space, but at the same distance to the surface S. It seems to be logical that, if we repeat this reasoning with all charges composing the metallic ring we will obtain a set of image charges $q_{1}^{'n}$ whose location forms another circle of radius $r=y_0=\frac{R_1^2}{R_2}$ (7.52). This new circle represents the image of the external circular conducting wire, and having a total charge $Q_1=\sum _{i=1}^{n}q_{1}^{'n}$. From the viewpoint of the method of images, this reasoning shown leads to an equivalent pose of the problem in the following terms: calculate the electric field on the axis OZ, generated by two concentric and parallel metallic rings of radius¹⁹ $R_2$ and $R_1$, charged with $Q_2$ and $Q_1$ coulombs, respectively.

To respond to this question we can directly use (7.244). Introducing the notation chosen it yields

$$\begin{aligned} Q_{1}=-\frac{R_1}{R_2}Q_{2}. \end{aligned}$$

(7.269)

In Chap. 2 we studied the electric field produced by a circular conducting wire charged by a homogeneous lineal charge density $\lambda =2\pi R_2Q_2$. The formula is

$$\begin{aligned} \mathbf {E}=\frac{1}{4\pi \varepsilon _0}\frac{zq}{(R^2+z^2)^\frac{3}{2}}\mathbf {u}_z \end{aligned}$$

(7.270)

Taking into account this result, substituting $R=R_2$ and $q=Q_2$, we get the field $\mathbf {E}_w$ produced by the metallic wire

$$\begin{aligned} \mathbf {E}_w=\frac{1}{4\pi \varepsilon _0}\frac{zQ_2}{(R_2^2+z^2)^\frac{3}{2}}\mathbf {u}_z. \end{aligned}$$

(7.271)

In the same way, setting $r=\frac{R_1^2}{R_2}$ and (7.269) into (7.270) we get the electric field generated by the image ring

$$\begin{aligned} \mathbf {E}_R=\frac{1}{4\pi \varepsilon _0}\frac{zQ_1}{(r^2+z^2)^\frac{3}{2}}\mathbf {u}_z= \frac{-1}{4\pi \varepsilon _0}\frac{zR_1Q_2}{R_2((\frac{R_1^2}{R_2})^2+z^2)^\frac{3}{2}}\mathbf {u}_z. \end{aligned}$$

(7.272)

Applying the principle of superposition

$$\begin{aligned} \mathbf {E}=\mathbf {E}_w+\mathbf {E}_R=\frac{zQ_2}{4\pi \varepsilon _0}\left( \frac{1}{(R_2^2+z^2)^\frac{3}{2}}- \frac{R_1}{R_2((\frac{R_1^2}{R_2})^2+z^2)^\frac{3}{2}}\right) \mathbf {u}_z, \end{aligned}$$

(7.273)

which represents the field at any point on the revolution axis of the system.

• 7.13 Let us suppose that the Neumann boundary conditions apply. Demonstrate that the election of $\frac{\partial G}{\partial n}=0$ for the Green function on the surface yields to a contradiction.

Solution

In the development of the theory we have seen for the definition of Neumann’s function imposed that the normal derivative on the surface S must be a constant (7.37). Taking into consideration the definition of Green’s function it seems more logical to choose this derivative to be zero because it makes the solution easier. However, if we analyze the problem in more detail we conclude that this choice leads to a contradiction. In fact, by applying Gauss’ theorem on the left side of (7.36) we obtain

$$\begin{aligned} \int \int \int _D\triangle N(\mathbf {r,r'})dV'=\int \int \int _D\nabla \cdot \nabla N(\mathbf {r,r'})dV'=\int \int _{S'}(\frac{\partial N(\mathbf {r,r'})}{\partial \mathbf {n'}})dS'=0. \end{aligned}$$

(7.274)

Taking volume integrals on the right side of (7.36),it results

$$\begin{aligned} \int \int \int _D\triangle N(\mathbf {r,r'})dV'=\int \int \int _D \delta ({r,r'})dV'=1, \end{aligned}$$

(7.275)

that is $1=0$, which is a contradiction.

• 7.14 Show that in the case of the Neumann problem, the constant value chosen for the normal derivative on the surface $S'$ is not arbitrary, and calculate it.

Solution

By applying Gauss’s theorem to (7.36), and taking into consideration (7.37), we have

$$\begin{aligned} \int \int \int _D\triangle N(\mathbf {r,r'})dV'=\int \int \int _D\nabla \cdot \nabla N(\mathbf {r,r'})dV'=\int \int _{S'}(\frac{\partial N(\mathbf {r,r'})}{\partial \mathbf {n'}})dS'=C\int \int _{S'}=CS. \end{aligned}$$

(7.276)

On the other hand, the first integral on the left may be calculated

$$\begin{aligned} \int \int \int _D\triangle N(\mathbf {r,r'})dV'=\int \int \int _D \delta ({r,r'})dV'=1, \end{aligned}$$

(7.277)

then,

$$\begin{aligned} C=\frac{1}{S}, \end{aligned}$$

(7.278)

where S is the boundary surface of the region D.

• 7.15 When studying the conformal mapping in the theoretical introduction we proposed a formulation for defining a complex potential under some circumstances. By using the Maxwell equations show another procedure to introduce the complex potential.

Solution

As we saw in the introduction of this chapter, sometimes a problem in three dimensions can be reduced to two variables because of symmetries. This is the case, for example, of problems in which the fields are the same on every section perpendicular to an axis. As we shall show in the next problems, for these specials problems it is possible to use methods that are based on complex variables to find the solution. To show this, consider a region of space free of charge, then

$$\begin{aligned} \nabla \cdot \mathbf {E}=0, \end{aligned}$$

(7.279)

$$\begin{aligned} \nabla \times \mathbf {E}=0. \end{aligned}$$

(7.280)

Equation (7.279) means that a scalar potential $V(\mathbf {r})$ for $\mathbf {E}$ may be found, i.e.

$$\begin{aligned} \mathbf {E}=-\nabla V(\mathbf {r}). \end{aligned}$$

(7.281)

On the other hand, as in the case of the magnetic field $(\nabla \cdot \mathbf {B}=0)$, (7.280) allows us to introduce a vector potential for $\mathbf {E}$. In effect, let this potential be $\mathbf {\Pi }$, then

$$\begin{aligned} \mathbf {E}=\nabla \times \mathbf {\Pi }. \end{aligned}$$

(7.282)

To study a plane electric (or magnetic) field we only need two variables, namely (x, y). In this case, the third component of $\mathbf {E}$ does not appear, and then component z of the rotational (7.282) must be zero, i.e.

$$\begin{aligned} E_z=\frac{\partial \Pi _y}{\partial x}-\frac{\partial \Pi _x}{\partial y}=0. \end{aligned}$$

(7.283)

The simplest possibility to fulfill (7.283) is to choose $\Pi _x=\Pi _y=0$, hence the vector potential $\mathbf {\Pi }$ has only the component along the OZ axis,

$$\begin{aligned} \mathbf {E}=\frac{\partial \Pi _z}{\partial y}\mathbf {u}_x-\frac{\partial \Pi _z}{\partial x}\mathbf {u}_y. \end{aligned}$$

(7.284)

Simultaneously, from (7.279) we have for the x and y components of the electric field $\mathbf {E}$,

$$\begin{aligned} \mathbf {E}=-\frac{\partial V}{\partial x}\mathbf {u}_x-\frac{\partial V}{\partial y}\mathbf {u}_y, \end{aligned}$$

(7.285)

and comparing (7.284) with (7.285), we have

$$\begin{aligned} E_x=-\frac{\partial V}{\partial x}=\frac{\partial \Pi _z}{\partial y}, E_y=-\frac{\partial V}{\partial y}=-\frac{\partial \Pi _z}{\partial x}. \end{aligned}$$

(7.286)

These (7.286) have the same structure as the Cauchy–Riemann relations of complex analysis, and as we have commented in Sect. 7.5.4, we can form a function of complex variable z which physically represents the potential. In fact, we can write

$$\begin{aligned} f(z)=\Pi (x,y)+iV(x,y). \end{aligned}$$

(7.287)

This equation is similar to (7.69), but its real and imaginary parts have been interchanged. Now the curves of constant potential $V(x,y)=C_1$ correspond to the imaginary part of (7.287), and the lines of force $\Pi (x,y)=C_2$ take the place of the imaginary part of f(z). This result affects neither the key idea of the method nor the basic procedure for solving problems. Actually, the choice of V(x, y) and $\Pi (x,y)$ as real or imaginary parts of the potential is completely arbitrary. We could have develop in the same manner and with the same validity the technique exposed in the theory with an expression like (7.287), but the formulae seen there changed some signs and constants. For instance, considering the potential in the form of (7.287) for the electric field holds

$$\begin{aligned} E(z)=-\frac{\partial V(x,y)}{\partial x}-i\frac{\partial V(x,y)}{\partial y}=-\frac{\partial V(x,y)}{\partial x}-i\frac{\partial \Pi (x,y)}{\partial x}=-\left( \frac{\partial V(x,y)}{\partial x}+i\frac{\partial \Pi (x,y)}{\partial x}\right) = \end{aligned}$$

(7.288)

$$ -i\left( \frac{\partial \Pi (x,y)}{\partial x}-i\frac{\partial V(x,y)}{\partial x}\right) =-i\overline{f'(z)}. $$

As we can see the only difference with (7.74) is the factor i.

• 7.16 In Chap. 2 we have seen the electric field created by a very long conducting wire charged with a homogeneous density of charge $\lambda$. Find its complex potential and show the potential level curves and the force lines.

Solution

Let us suppose, for simplicity, that the wire is located along the OZ axis coinciding with the origin of coordinates of the cartesian system. As the wire is very large, we have translational symmetry along OZ. This means that we can reduce the calculation to a plane perpendicular to the OZ direction, with the result we will obtain the same for every plane parallel to the OXY plane (see theory).

For finding the complex potential we start with the formula of the electric field generated by a large metallic wire holding a density of charge per unit length $\lambda$. As it was demonstrated in Chap. 2, the expression of $\mathbf {E}$ is

$$\begin{aligned} \mathbf {E}=\frac{\lambda }{2\pi \varepsilon _0\rho }\mathbf {u}_\rho , \end{aligned}$$

(7.289)

where $\rho$ is the distance from the point where the field is calculated to the wire, and $\varepsilon _0$ is the permittivity. Setting $\mathbf {u}_\rho$ as functions of its cartesian components we obtain

$$\begin{aligned} \mathbf {E}=\frac{\lambda }{2\pi \varepsilon _0\sqrt{x^2+y^2}}(\cos \phi \,\mathbf {u}_x+\sin \phi \,\mathbf {u}_y)= \frac{\lambda }{2\pi \varepsilon _0\sqrt{x^2+y^2}}(\frac{x}{\sqrt{x^2+y^2}}\mathbf {u}_x+\frac{y}{\sqrt{x^2+y^2}}\mathbf {u}_y)= \end{aligned}$$

(7.290)

$$ \frac{\lambda }{2\pi \varepsilon _0}\left( \frac{x}{(x^2+y^2)}\mathbf {u}_x+\frac{y}{(x^2+y^2)}\mathbf {u}_y\right) . $$

Now, we construct a complex number in such a way that its real and imaginary parts coincide with the components of (7.290)

$$\begin{aligned} E=E_x+iE_y=\frac{\lambda }{2\pi \varepsilon _0}\frac{x+iy}{(x^2+y^2)}. \end{aligned}$$

(7.291)

Transforming this equation to z coordinates, we have

$$\begin{aligned} E=\frac{\lambda }{2\pi \varepsilon _0}\,\frac{z}{z\,\overline{z}}=\frac{\lambda }{2\pi \varepsilon _0}\frac{1}{\overline{z}}. \end{aligned}$$

(7.292)

Introduction of this last result into (7.75) gives

$$\begin{aligned} f(z)=-\int \overline{E}(z)\,dz+a=-\int \frac{\lambda }{2\pi \varepsilon _0}\frac{1}{z}\,dz+a= -\frac{\lambda }{2\pi \varepsilon _0}\int \frac{1}{z}\,dz+a=\frac{\lambda }{2\pi \varepsilon _0}\ln \frac{1}{z}+a, \end{aligned}$$

(7.293)

a being a constant. This result represents the complex potential corresponding to the electric field (7.289). Taking into consideration that $z=\rho \exp \,i\theta$ (7.293) yields²⁰

$$\begin{aligned} f(z)=\frac{\lambda }{2\pi \varepsilon _0}(\ln \rho +i\,\theta ), \end{aligned}$$

(7.294)

Fig. 7.32

a Real part of the complex potential. Observe that the circles represent the curves of constant potential. b Imaginary part of f(z). The level curves are straight lines starting from the origin of coordinates, where the wire is located

Fig. 7.33

Electric field created by the charged wire. The field lines are perpendicular to the level curves of the potential (real part). The direction coincides with the force lines shown in Fig. 7.32a

then

$$\begin{aligned} u(\rho ,\theta )=\frac{\lambda }{2\pi \varepsilon _0}\ln \rho , \end{aligned}$$

(7.295)

and

$$\begin{aligned} v(\rho ,\theta )=\frac{\lambda }{2\pi \varepsilon _0}\,\theta . \end{aligned}$$

(7.296)

As we commented in the theory, the level lines $u=C_1$ (Fig. 7.32a) represent the equipotential curves, and $v=C_2$ (Fig. 7.32b) the constant lines of force. If $\lambda >0$ the lines of force are directed away of the singularity , and for $\lambda <0$ the field lines are pointed toward $z=0$. The electric field is depicted in Fig. 7.33. Note that these vectors are perpendicular to the curves $u=C_1$.

• 7.17 Let us suppose a very large metallic wire carries a homogeneous density current j. Give the corresponding complex potential and study its real and imaginary parts.

Solution

As we explained in Chap. 5 (see Problem 5.1), the magnetostatic field produced by a very large metallic wire only has a tangential component, i.e.

$$\begin{aligned} \mathbf {H}(\rho )=\frac{I}{2\pi \rho }\mathbf {u}_\phi . \end{aligned}$$

(7.297)

By introducing the expression of the unitary vector $\mathbf {u}_\phi =(-\sin \phi \,\mathbf {u}_x+\cos \phi \,\mathbf {u}_y)$ (7.297) leads to

$$\begin{aligned} \mathbf {H}=\frac{I}{2\pi \sqrt{x^2+y^2}}(-\sin \phi \,\mathbf {u}_x+\cos \phi \,\mathbf {u}_y)= \frac{I}{2\pi \sqrt{x^2+y^2}}(\frac{-y}{\sqrt{x^2+y^2}}\mathbf {u}_x+\frac{x}{\sqrt{x^2+y^2}}\mathbf {u}_y)= \end{aligned}$$

(7.298)

$$ \frac{I}{2\pi }\left( \frac{-y}{x^2+y^2}\mathbf {u}_x+\frac{x}{x^2+y^2}\mathbf {u}_y\right) . $$

With this last formula the magnetic field expressed as a complex number adopts the following form

$$\begin{aligned} H=H_x+iH_y=\frac{I}{2\pi }\frac{-y+ix}{(x^2+y^2)}=\frac{I}{2\pi }\frac{i(x+iy)}{(x^2+y^2)}= \frac{I\,i}{2\pi }\frac{z}{z\overline{z}}=\frac{I\,i}{2\pi \overline{z}}. \end{aligned}$$

(7.299)

Once the field is known, the complex potential may be obtained by means of (7.87)

$$\begin{aligned} F(z)=-\int \overline{H}(z)\,dz+b=-\int \frac{-I\,i}{2\pi z}dz+b= \frac{I\,i}{2\pi }\int \frac{1}{z}\,dz+b=\frac{I\,i}{2\pi }\ln z=\frac{I\,i}{2\pi }(\ln \rho +i\,\theta )+b, \end{aligned}$$

(7.300)

where b is a constant which for simplicity we consider zero. Separating in real and imaginary parts $F(z)=\frac{I}{2\pi }(-\theta +i\ln \rho )=u+iv$, we have

$$\begin{aligned} u(\rho ,\theta )=-\frac{I}{2\pi }\theta , \end{aligned}$$

(7.301)

and

$$\begin{aligned} v(\rho ,\theta )=\frac{I}{2\pi }\ln \rho . \end{aligned}$$

(7.302)

As we can observe the result is similar to the former problem, but the value of the real and imaginary parts are switched. In fact, the curves of constant potential in a reference frame of polar coordinates are now $-\frac{I}{2\pi }\theta =C_1$, which represent straight lines as depicted in Fig. 7.32b for the electric field of a wire. On the contrary, $\frac{I}{2\pi }\ln \rho =C_2$ are the line forces and as the reader can probe there are perpendicular to the isolines $u(\rho ,\theta )=Constant$ at each point. The field lines, i.e. the magnetostatic field generated by the wire currying an intensity I has the form shown in Fig. 7.34.

Fig. 7.34

Magnetic field generated by a very large wire. The direction of the field coincides with the lines of force represented by the imaginary part of the complex magnetic potential F(z)

• 7.18 The attached figure shows a system composed by two very long conducting cylinders of radii R and r, respectively $(r<R)$, whose revolution axes are parallel to each other, but they do not coincide. The inner cylinder is held to zero potential and the other one to $V_0$ volts. Determine the potential in the region between cylinders and the capacitance per unit length of the system.

Solution

The problem in its presented form is very difficult to solve. It would be easier if we had two exact concentric cylinders as depicted in Fig. 7.35, because in such a case the solution is well known. In order to obtain these new geometric configuration, we must find a mapping for which our problem is transformed. With this aim, we will investigate the characteristics we must have. In the introductory theory given about this subject (Sect. 7.5.3), we saw that the fractional transformation (Möbius mapping) has properties very adequate for this case. For instance, circles in the z-plane are mapped into circles in the w-plane, then it seems to be logical to try using these kind of functions for solving the question. Actually, we must find a transformation that maps the exterior circle onto itself, and the center of the small cylinder transforms into the origin of coordinates in the w-plane. Thus, we would have two concentric circles in w, as we wanted. In this way, by means of the bilinear transformation, points symmetric in the z-plane with respect to a circle are mapped into points symmetric with respect to the transformed circle in the plane w. Taking it into account we see that the circles will be concentric if we find two points in z-plane that are symmetric with respect to both circles, simultanously. These two symmetric points will be transformed into two symmetric points in w with respect to the new geometry. Considering the disposition shown in Fig. 7.35, and labelling $x_1$ and $x_2$ the coordinates over the OX axis in the z-plane (because of the disposition), the equations that must be fulfilled in order to find a transformation are the following

$$\begin{aligned} x_1\cdot x_2=R^2, \end{aligned}$$

(7.303)

Fig. 7.35

Two metallic cylinders whose revolution axes are parallel to each other

and

$$\begin{aligned} (x_1-d)\cdot (x_1-d)=r^2. \end{aligned}$$

(7.304)

The first one gives the symmetric point with respect to the big cylinder, and (7.304) allows us to compute the same for the small circle. Observe that if we would only have a centered circle (in this case of radius R), the symmetric point to (0, 0) is the point $z=\infty$. However, because of the two conditions (7.303), and (7.304), the solution is different. Combining both equations, we obtain a new expression where only one variable appears,

$$\begin{aligned} d\,x_{1}^{2}+(r^2-R^2-d^2)\,x_1+d\,R^2=0. \end{aligned}$$

(7.305)

The solution of (7.305) gives two roots that solve the problem. The bilinear transformation we find has the following form

$$\begin{aligned} w=f(z)=\alpha \frac{z-x_1}{z-x_2}, \end{aligned}$$

(7.306)

where $x_1$ and $x_2$ are yet unknown. The parameter $\alpha$ is, in general, a complex number that does not affect the final result. Introductions of the values $r=0.2$, $d=0.2$, and $R=1$ into (7.305) $0.2\,x_{1}^{2}-x_1+0.2=0$ yields $x_1=0.21$ and $x_2=4.79$, therefore

$$\begin{aligned} f(z)=\frac{z-0.21}{z-4.79}=w. \end{aligned}$$

(7.307)

The mapping (7.307) transforms the two eccentrical circles of the z-plane into two centered circles in the w-plane. The new radii of the cylinders can be calculated by means of the (7.307). In fact, $w_1=|f(0)|\approx 0.04$ and $w_2=|f(R=1)|\approx 0.2$. The expression of the potential between two concentric metallic cylinders held to potentials $V_1=0$ and $V_2=v_0$ in the coordinates of the plane w is²¹

$$\begin{aligned} \varphi (w)=\frac{V_0}{\ln \left( \frac{w_2}{w_1}\right) }\ln \left( \frac{|w|}{w_1}\right) , \end{aligned}$$

(7.308)

where $|w|=\sqrt{u^2+v^2}$ represents in that plane the same as $\rho$ in the z-plane. The solution is determined by mapping (7.308) back to the z-plane. It can be immediately performed by substituting (7.307) into (7.308)

$$\begin{aligned} V(x,y)=\frac{V_0}{\ln (5)}\ln \left| \frac{25(z-0.21)}{(z-4.79)}\right| = \frac{V_0}{\ln (5)}\left[ \ln (25)+\frac{1}{2}((x-0.21)^2+y^2)-\frac{1}{2}((x-4.79)^2+y^2))\right] . \end{aligned}$$

(7.309)

This result is represented in Fig. 7.36a. Figure 7.36b shows the curves of constant potential between the two cylinders. The imaginary part of the complex potential which gives information of the line forces can be computed with the (7.70) by means of a simple integration. The result is depicted in Figs. 7.37 and 7.38.

Fig. 7.36

a Graphic of the potential V(x, y). b Curves of constant potential on the OXY plane

Fig. 7.37

The geometry of the problem in the w-plane corresponds to two concentric cylinders of radii $w_1$ and $w_2$

Fig. 7.38

Line forces between the two circles in the z-plane. Observe that this vector field is perpendicular to the level potential curves

• 7.19 The cross section of a square prismatic metallic system (see Problem 7.3) is formed by four sides of length L. The depth h of the system is larger than L $(L<<h)$ and each of its sides is held to potentials A, B, C, and D, respectively. Calculate the potential at any interior point of the system in the following cases: (a) $A=200$, $B=0$, $C=0$, and $D=0$. (b) $A=500$, $B=0$, $C=0$, and $D=0$. (c) $A=500$, $B=100$, $C=0$, and $D=0$. (d) $A=500$, $B=100$, $C=300$, and $D=400$.

Solution

Due to the side of the square section being much smaller than the length of the prismatic hole $(L<<h)$, we can study the system as a two dimensional problem (Fig. 7.39). The only solution we are going to find differs from the actual one in the vicinity of the top and bottom of the prismatic geometry, where the end effects are present. Accepting this viewpoint, we will work with a cross-section of the four metallic plates as depicted in Fig. 7.40. For obtaining the electric field inside of the system we have previously successfully employed the technique of separation of variables. However, now we want to apply the method of finite differences with the aim to explain the procedure. As we will see, by examining only a few points we are able to know roughly the electric field between the metallic sheets. By this method we want to obtain a numerical solution of the laplace equation inside of the region of interest. Because of its characteristics, it follows that we can perform the calculation only at a finite number of points, and therefore we must first construct a discretized model adapted to the specifications of the problem to be solved. The points selection over the domain of interest depends on the geometry and resolution we want to reach.

Fig. 7.39

Square metallic guide

Fig. 7.40

Square sections with the 36 points to be studied

We begin to construct a grid of points with the intersections of crossing parallel lines to the coordinate frame. The more resolution, the more points, but also more computation steps and time consuming. As we have seen in Sect. 7.6.1, the basic idea consists in determining an approximate solution at any node of the mesh by averaging of its nearest four neighbors. Boundary points have information of the boundary conditions given, and their values and the mesh size must be introduced in the equations correctly. Otherwise the solution propagates with errors which are amplified at every step.

In order to have all possibilities for the different Dirichlet boundary conditions on all the sides, we suppose that the potential on the edges of the square section are A, B, C and D, respectively. It means that the potential for the points over these lines are known. We start at point 8 of the Fig. 7.40, which depends on $V_2$, $V_7$, $V_9$, and $V_{14}$, one of them is located on the boundary ($V_2$), and continue in order to the point 29. The general system of equations for the 16 points inside of the square is the following

$$\begin{aligned} 4V_8-V_9-V_{14}=A+D\end{aligned}$$

(7.310)

$$\begin{aligned} 4V_9-V_8-V_{10}-V_{15}=A\end{aligned}$$

(7.311)

$$\begin{aligned} 4V_{10}-V_9-V_{11}-V_{16}=A\end{aligned}$$

(7.312)

$$\begin{aligned} 4V_{11}-V_{10}-V_{17}=A+B\end{aligned}$$

(7.313)

$$\begin{aligned} 4V_{14}-V_{8}-V_{20}-V_{15}=D\end{aligned}$$

(7.314)

$$\begin{aligned} 4V_{15}-V_{9}-V_{14}-V_{16}-V_{21}=0\end{aligned}$$

(7.315)

$$\begin{aligned} 4V_{16}-V_{10}-V_{15}-V_{17}-V_{22}=0\end{aligned}$$

(7.316)

$$\begin{aligned} 4V_{17}-V_{11}-V_{16}-V_{23}=B\end{aligned}$$

(7.317)

$$\begin{aligned} 4V_{20}-V_{14}-V_{21}-V_{26}=D\end{aligned}$$

(7.318)

$$\begin{aligned} 4V_{21}-V_{15}-V_{20}-V_{22}-V_{27}=0\end{aligned}$$

(7.319)

$$\begin{aligned} 4V_{22}-V_{16}-V_{21}-V_{23}-V_{28}=0\end{aligned}$$

(7.320)

$$\begin{aligned} 4V_{23}-V_{17}-V_{22}-V_{29}=B\end{aligned}$$

(7.321)

$$\begin{aligned} 4V_{26}-V_{20}-V_{27}=C+D\end{aligned}$$

(7.322)

$$\begin{aligned} 4V_{27}-V_{21}-V_{26}-V_{28}=C\end{aligned}$$

(7.323)

$$\begin{aligned} 4V_{28}-V_{22}-V_{27}-V_{29}=C\end{aligned}$$

(7.324)

$$\begin{aligned} 4V_{29}-V_{23}-V_{28}=C+B. \end{aligned}$$

(7.325)

It is interesting to note that for Dirichlet boundary conditions, the procedure does not begin at points over the boundary. We start with points placed over the next line. This situation differs from that corresponding to Neumann. In that case the conditions are over the normal derivative on the surface, which leads to one additional system of equations taking the points over the boundary.

Setting the (7.310)–(7.325) in matrix form we get

$$\begin{aligned} \left[ \begin{array}{cccccccccccccccc} 4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0 \\ \ -1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ \ 0&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0 \\ \ 0&{}0&{}-1&{}4&{}0&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ \ -1&{}0&{}0&{}0&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0&{}0 \\ \ 0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0&{}0\\ \ 0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0&{}0&{}0 \\ \ 0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}0&{}0&{}0&{}-1&{}0&{}0&{}0&{}0\\ \ 0&{}0&{}0&{}0&{}-1&{}0&{}0&{}0&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0&{}0 \\ \ 0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0\\ \ 0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0 \\ \ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}0&{}0&{}0&{}-1\\ \ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}0&{}4&{}-1&{}0&{}0 \\ \ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0\\ \ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1 \\ \ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4 \end{array}\right] \left[ \begin{array}{c} V_8\\ V_9\\ V_{10}\\ V_{11}\\ V_{14}\\ V_{15}\\ V_{16}\\ V_{17} \\ V_{20}\\ V_{21}\\ V_{22}\\ V_{23}\\ V_{26}\\ V_{27}\\ V_{28}\\ V_{29} \end{array}\right] =\left[ \begin{array}{c} A+D\\ A\\ A\\ A+B\\ D\\ 0\\ 0\\ B\\ D\\ 0\\ 0\\ B\\ C+D\\ C\\ C\\ C+B \end{array}\right] \end{aligned}$$

(7.326)

The problem reduces to calculating the inverse matrix and multiplying it by the vector containing a combination of values of boundary conditions. Specifically for the first case (case (a)) the side A is held to potential $V=200$ v, then introducing in (7.326) $A=200$, we obtain

$$\begin{aligned} \left[ \begin{array}{c} V_8\\ V_9\\ V_{10}\\ V_{11}\\ V_{14}\\ V_{15}\\ V_{16}\\ V_{17} \\ V_{20}\\ V_{21}\\ V_{22}\\ V_{23}\\ V_{26}\\ V_{27}\\ V_{28}\\ V_{29} \end{array}\right] =\left[ \begin{array}{c} 90.9\\ 118.9\\ 118.9\\ 90.9\\ 44.7\\ 65.9\\ 65.9\\ 44.7\\ 22.0\\ 34.1\\ 34.1\\ 22.0\\ 9.1\\ 14.4\\ 14.4\\ 9.1 \end{array}\right] . \end{aligned}$$

(7.327)

Figure 7.41 represents the scalar surface corresponding to potential. Observe that even if the mesh resolution chosen is bad, the result is enough to see the behavior of the potential (Fig. 7.42).

(b) Starting with the same scheme we substitute $A=500$, $B=0$, $C=0$, and $D=0$ into (7.326), obtaining the following vector with values of the potential for all points studied, i.e.

Fig. 7.41

Surface representing the potential inside of the prismatic system for $A=200$, $B=0$, $C=0$, and $D=0$. Observe the level curves in both graphics

Fig. 7.42

Three dimensional representation of V(x, y) for $A=500$, $B=0$, $C=0$, and $D=0$

$$\begin{aligned} \left[ \begin{array}{c} V_8\\ V_9\\ V_{10}\\ V_{11}\\ V_{14}\\ V_{15}\\ V_{16}\\ V_{17} \\ V_{20}\\ V_{21}\\ V_{22}\\ V_{23}\\ V_{26}\\ V_{27}\\ V_{28}\\ V_{29} \end{array}\right] =\left[ \begin{array}{c} 227.3\\ 297.3\\ 297.3\\ 227.3\\ 111.7\\ 164.8\\ 164.8\\ 111.7\\ 54.9\\ 85.2\\ 85.2\\ 54.9\\ 22.7\\ 36.0\\ 36.0\\ 22.7 \end{array}\right] \end{aligned}$$

(7.328)

The diagram for this case is exhibited in Sect. 7.6.1. The figure has the same basic characteristics like the picture presented in Fig. 7.41. The most important difference appears in the potential on the side A, because of the initial condition here. The rest of the interior points take also other values, but the form of the surface has the same structure.

(c) In the same manner that we have seen in the foregoing sections, if two sides of the system are held to potentials $A=500$, $B=100$, and on the edges C and D are zero, we put such a values in (7.326) again yielding

$$\begin{aligned} \left[ \begin{array}{c} V_8\\ V_9\\ V_{10}\\ V_{11}\\ V_{14}\\ V_{15}\\ V_{16}\\ V_{17} \\ V_{20}\\ V_{21}\\ V_{22}\\ V_{23}\\ V_{26}\\ V_{27}\\ V_{28}\\ V_{29} \end{array}\right] =\left[ \begin{array}{c} 231.8\\ 308.3\\ 319.7\\ 272.7\\ 118.9\\ 181.8\\ 197.7\\ 171.2\\ 62.1\\ 102.2\\ 118.2\\ 114.4\\ 27.3\\ 47.0\\ 58.3\\ 68.2 \end{array}\right] \end{aligned}$$

(7.329)

The potential versus the coordinates x and y is shown in Fig. 7.43. Now, the level curves²² are not similar to that which we saw in Fig. 7.17 for the rectangular plate. This follows from the introduction of the potential on side B, which breaks the symmetry with respect to an axis passing across the geometrical center and parallel to B (Fig. 7.44).

Fig. 7.43

Two perspectives of the surface representing the potential inside of the prismatic system for $A=500$, $B=100$, $C=0$, and $D=0$. Note that in this case the curves of constant potential are not symmetric like Fig. 7.17

Fig. 7.44

Graphic when $A=500$, $B=100$, $C=300$, and $D=400$. a Lateral view of the function V(x, y). b Same representation but viewed from the right

(d) Introduction of $A=500$, $B=100$, $C=300$, and $D=400$ into (7.326) leads to

$$\begin{aligned} \left[ \begin{array}{c} V_8\\ V_9\\ V_{10}\\ V_{11}\\ V_{14}\\ V_{15}\\ V_{16}\\ V_{17} \\ V_{20}\\ V_{21}\\ V_{22}\\ V_{23}\\ V_{26}\\ V_{27}\\ V_{28}\\ V_{29} \end{array}\right] =\left[ \begin{array}{c} 427.3\\ 419.3\\ 385.2\\ 304.5\\ 389.8\\ 364.8\\ 317.0\\ 233.0\\ 367.0\\ 333.0\\ 285.2\\ 210.2\\ 345.5\\ 314.8\\ 280.7\\ 222.7 \end{array}\right] . \end{aligned}$$

(7.330)

The picture corresponding to this vector is represented below.

• 7.20 A big ferromagnetic material has a hole inside with the shape of a square prism. In the centre of it a very large metallic wire carrying a current $I=1$ A is located (see figure attached). Supposing that $\mu$ is very high and the edge of the hole has length L, using the finite difference method (FDM), determine the magnetostatic potential in the square region showed in the figure and sketch the magnetic field lines.

Solution

Before beginning the calculation we are going to analyze the symmetries, because it shows us how to simplify the problem.

As we can observe, the system of Fig. 7.45 has translational symmetry in the direction of the axis perpendicular to the plane view. Let us choose this direction as the OZ axis. Because of this characteristic we can consider the system as a two dimensional problem, and therefore the solution we will find to be the same for all parallel planes to that figure. On the other hand, according to its square cross section, we also have a four order rotational symmetry with respect to OZ, which means that we do not need to calculate the potential directly at every point of the grid, but only at those locations inside a triangle of side L, because the solution for the other parts are connected together. Now, let us follow with the scalar potential.

Fig. 7.45

Rectangular grid. The first region corresponds to the triangle without an asterisk. The other regions on the left, on the right and on the upper triangle are depicted with one asterisk $*$, two $**$ and three $***$, respectively

Fig. 7.46

a Open curve with extremes a and b. b Closed curve. The initial point a on the plane OXY is the same as the final point in the integration

In a region of the space where there are no currents $\nabla \times \mathbf {H}=0$, so in a way similar to what we have seen for the electric field, we can define a magnetostatic potential $V_m$ so that

$$\begin{aligned} \mathbf {H}=-\nabla V_m. \end{aligned}$$

(7.331)

By virtue of that equation the linear integral between two points a and b not passing throughout the current depends on such endpoints, i.e.,

$$\begin{aligned} \int _{a}^{b}\mathbf {H}\,d\mathbf {l}=-\int _{a}^{b}\nabla V_m\,d\mathbf {l}=V_m(a)-V_m(b). \end{aligned}$$

(7.332)

In principle, it seems to be the same as we studied for the electric potential, however there are some differences. The most important difference between both potentials deals with the region where the magnetostatic potential is defined. To explain them, let us suppose a point a as shown in Fig. 7.46, from which we begin to construct a closed curve. If we go step by step we have an open curve, but at the end of the trace we reach the same point a, then we draw a closed line. Applying Ampère’s law we have for the magnetic field $\mathbf {H}$

$$\begin{aligned} \oint _{\Gamma }\mathbf {H}\,d\mathbf {l}=I, \end{aligned}$$

(7.333)

which means that the integral over a closed curve $\Gamma$ is not zero. Now, if we combine this result with (7.332) it yields

$$\begin{aligned} V_m(a)-V_m(a)=I\,\,\,(!!!), \end{aligned}$$

(7.334)

a result that seems to be illogical because $V_m(a)-V_m(a)$ should be zero, and not I. The only possibility to make (7.334) feasible is that the value of the function $V_m$ at point a when starting to follow the curve $\Gamma$ is not the same as the value it reaches when arriving again at a after going around the curve; in other words it means that the potential is a multivalued function of position, that is to say, that $V_m$ maps different images at the same point when closing the curve. In order to be clear and distinguish the two values, we will label with an asterisk $*$ the second potential, then the (7.333) converts to

$$\begin{aligned} V_m(a)-V^{*}_{m}(a)=I. \end{aligned}$$

(7.335)

Mathematically this equation means that a discontinuity exists when completing a whole closed circulation, and it is due to the fact that the region where we have the potential is multiply connected (see Chap. 1).²³ In such a case we can understand this behavior for $V_m$ as if it displaces on another sheet when passing across the discontinuity (see Fig. 1.30 from Chap. 1).²⁴ However we can prevent this problem by introducing a cut barrier which avoids having a closed curve in such a way to link currents. Once we have located the cut (there are infinite possibilities for that), the potential in this constrained region is single-valued, then the curvilinear integral of H over any closed curve will be zero.

To determine the scalar potential of this problem we must use the (5.60) Problem 5.1 of the magnetic field created by a very large metallic wire currying a current I that we have seen in Chap. 5, i.e.,

$$\begin{aligned} \mathbf {B}=\frac{\mu _0I}{2\pi \rho }\mathbf {u}_\phi \Rightarrow \mathbf {H}=\frac{I}{2\pi \rho }\mathbf {u}_\phi . \end{aligned}$$

(7.336)

Taking into account the symmetry of the system, by virtue of (7.331) and using cylindrical coordinates we can write

$$\begin{aligned} \mathbf {H}=-\nabla V_{m}\mathbf {u}_\phi =-\frac{1}{\rho }\frac{\partial V_m}{\partial \phi }\mathbf {u}_\phi =\frac{I}{2\pi \rho }\mathbf {u}_\phi , \end{aligned}$$

(7.337)

thus

$$\begin{aligned} \frac{1}{\rho }\frac{\partial V_m}{\partial \phi }=-\frac{I}{2\pi \rho }\Rightarrow V_m(\phi )=-\int \frac{I}{2\pi }d\phi +D=-\frac{I}{2\pi }\phi +D, \end{aligned}$$

(7.338)

D being an arbitrary constant. For this problem we choose $D=0$, which does not affect to the final result in any way.

Once we know the function $V_m(\phi )$ for the wire alone, we can employ the finite difference method for calculating the scalar potential of the system represented in Fig. 7.45. As we have seen in the previous problem, for applying the FDM we first divide the region to be studied in subdomains by means of a mesh formed by parallel lines, which intersect at points where we will calculate the solution of the differential equation. However, in the present problem, we must not compute the calculations for all the points we see in Fig. 7.45. On the contrary, due to the symmetry of the rectangular cavity, we can perform the calculations in one of the triangles depicted, and then the results for the other points may be computed by using the relation between triangles which, as we are going to see, differ in quantities determined by (7.338).

At the starting point let us divide the square region into four identical triangles (Fig. 7.45). We will denote with asterisks the triangles on the left, on the right and on the upper part of the square, and the first one on the lower region without signs. For applying the finite difference method we mesh the region of interest by parallel lines and the intersections of them are labelled with numbers. In this problem, with the aim to be didactic and show the procedure, we have drawn in the square only 64 points. To explain the calculations it is enough and it does not lose generality. It only has influence on the resolution of the final result, but it does not affect the idea of the technique. Once we have identified the points, we choose one of the triangular regions of the square. We begin with the lower triangle where every point has a number from 1 to 15. Observe that at the corner we have written 0, then we have 16 points to be studied in each triangle. All points in this subregion have their corresponding point in the other parts, which corresponds to the same number but labelled with asterisks (see Fig. 7.45). Now, we will apply (7.95), however as it can be seen some problems occur. As we have explained in Sect. 7.6.1, using FDM requires the knowing of the potential values at the neighbors of the points where the potential will be computed. As a result in our case we see that points such as 1, 2 or 5 need data from region $*$, which are not known a priori. This difficulty may be solved by employing (7.338). In fact, as the magnetic potential grows linearly with angle $\phi$, then every point $i^{*}$ of the region b will differ by a quantity $\frac{I}{4}$, that is to say, if the potential at point 4 is $V_4$ its equivalent in the region b, $V^{*}_{4}=V_4+\frac{I}{4}$. The same occurs for domains c and d. For these subregions it holds that $V^{**}_{j}=V_j-\frac{I}{4}$ and $V^{***}_{j}=V_j+\frac{I}{2}$ for $j=1,\ldots N$, N being the number of points. With these relations, the values of the potential at set points of a domain are connected with the other ones. The second important thing refers to the conditions on the boundaries. Let us suppose two media as shown in Fig. 7.47, of permeabilities $\mu _1$ and $\mu _2$, respectively. If the angles of incidence with respect the normal to the surface are $\alpha _1$ and $\alpha _2$, from the figure it yields

$$\begin{aligned} \frac{\tan \alpha _1}{\tan \alpha _2}=\frac{\left( \frac{B_{1t}}{B_{1n}}\right) }{\left( \frac{B_{2t}}{B_{2n}}\right) }, \end{aligned}$$

(7.339)

Fig. 7.47

a Magnetic field in the proximity of a boundary surface. b If $\mu _2>>\mu _2$ the field lines almost do not penetrate into the medium 2

but as $B_{1n}=B_{2n}$, $H_{1t}=H_{2t}$ and $B_{it}=\mu _iH_{it}$ ($i=1,2$), this equation becomes

$$\begin{aligned} \frac{\tan \alpha _1}{\tan \alpha _2}=\frac{B_{1t}}{B_{2t}}= \frac{\left( \frac{H_{1t}}{\mu _1}\right) }{\left( \frac{H_{2t}}{\mu _2}\right) }=\frac{\mu _1}{\mu _2}. \end{aligned}$$

(7.340)

This expression gives the behavior of the magnetic lines in the proximity of the boundary of two media. If one of them has a high permeability (as in our case), for instance material 2, $\mu _2>>\mu _1$ separation, and therefore from (7.340) holds

$$\begin{aligned} \frac{\tan \alpha _1}{\tan \alpha _2}=\frac{\mu _1}{\mu _2}\approx 0\Rightarrow \alpha _1\approx 0. \end{aligned}$$

(7.341)

This important result means that the magnetic lines in the region of lower permeability are practically perpendicular to the boundary surface (Fig. 7.47). As a consequence the separation surface corresponding to the domain of high $\mu$ may be approximately considered as an equipotential surface. Taking into account that the material of the problem is ferromagnetic, we cannot exactly speak about the magnetic permittivity, because of its non-linear behavior. We mean that $\mu$ is a function of the magnetic field and not a constant (see Chap. 6). However, even though in case of low $\mathbf {H}$ fields, the corresponding $\mu (H)$ is usually larger than $\mu _0$ (vacuum or air), hence we can employ the aforementioned conclusions. Thus, choosing for instance that the potential at all points on the surface domain A are zero (at 0, 10, 11, 12, 13, 14 and 15),²⁵ and applying (7.94) to the points of the first triangle, we have

$$\begin{aligned} 4V_1-V_{4}^{*}-V_{1}^{*}-V_3-V_{1}^{**}=0\Rightarrow 4V_1-\left( V_{4}+\frac{I}{4}\right) -\left( V_{1}+\frac{I}{4}\right) -V_3-\left( V_{1}-\frac{I}{4}\right) =0\end{aligned}$$

(7.342)

$$\begin{aligned} 4V_2-V_3-V_6-V_{4}^{*}-V_{9}^{*}=0\Rightarrow 4V_2-V_3-V_6-\left( V_{4}+\frac{I}{4}\right) -\left( V_{9}+\frac{I}{4}\right) =0\end{aligned}$$

(7.343)

$$\begin{aligned} 4V_{3}-V_1-V_{2}-V_{4}-V_{7}=0\end{aligned}$$

(7.344)

$$\begin{aligned} 4V_{4}-V_{3}-V_{2}^{**}-V_8-V_{1}^{**}=0\Rightarrow 4V_{4}-V_{3}-\left( V_{2}-\frac{I}{4}\right) -V_8-\left( V_{1}-\frac{I}{4}\right) =0\end{aligned}$$

(7.345)

$$\begin{aligned} 4V_{5}-V_{6}-V_{15}^{*}-V_{10}-V_{9}^{*}=0\Rightarrow 4V_{5}-V_{6}-\left( 0+\frac{I}{4}\right) -V_{10}-\left( V_{9}+\frac{I}{4}\right) =0\end{aligned}$$

(7.346)

$$\begin{aligned} 4V_{6}-V_{2}-V_{5}-V_{7}-V_{11}=0 \Rightarrow 4V_{6}-V_{2}-V_{5}-V_{7}-0=0\end{aligned}$$

(7.347)

$$\begin{aligned} 4V_{7}-V_{3}-V_{6}-V_{8}-V_{12}=0\Rightarrow 4V_{7}-V_{3}-V_{6}-V_{8}-0=0\end{aligned}$$

(7.348)

$$\begin{aligned} 4V_{8}-V_{4}-V_{7}-V_{9}-V_{13}=0\Rightarrow 4V_{8}-V_{4}-V_{7}-V_{9}-0=0\end{aligned}$$

(7.349)

$$\begin{aligned} 4V_{9}-V_{8}-V_{2}^{**}-V_{5}^{**}-V_{14}=0\Rightarrow 4V_{9}-V_{8}- \left( V_{2}-\frac{I}{4}\right) -\left( V_{5}-\frac{I}{4}\right) -0=0. \end{aligned}$$

(7.350)

Simplifying these equations we obtain

$$\begin{aligned} 2V_1-V_3-V_4=\frac{I}{4}\end{aligned}$$

(7.351)

$$\begin{aligned} 4V_2-V_3-V_4-V_{6}-V_{9}=\frac{I}{2}\end{aligned}$$

(7.352)

$$\begin{aligned} 4V_{3}-V_1-V_{2}-V_{4}-V_{7}=0\end{aligned}$$

(7.353)

$$\begin{aligned} 4V_{4}-V_{1}-V_{2}-V_{3}-V_8=-\frac{I}{2}\end{aligned}$$

(7.354)

$$\begin{aligned} 4V_{5}-V_{6}-V_{9}=\frac{I}{2}\end{aligned}$$

(7.355)

$$\begin{aligned} 4V_{6}-V_{2}-V_{5}-V_{7}=0\end{aligned}$$

(7.356)

$$\begin{aligned} 4V_{7}-V_{3}-V_{6}-V_{8}=0\end{aligned}$$

(7.357)

$$\begin{aligned} 4V_{8}-V_{4}-V_{7}-V_{9}=0\end{aligned}$$

(7.358)

$$\begin{aligned} 4V_{9}-V_{2}-V_{5}-V_{8}=-\frac{I}{2}, \end{aligned}$$

(7.359)

which form a system of 9 linear equation with 9 unknowns.

$$\begin{aligned} \left[ \begin{array}{ccccccccc} 2&{}0&{}-1&{}-1&{}0&{}0&{}0&{}0&{}0 \\ \ 0&{}4&{}-1&{}-1&{}0&{}-1&{}0&{}0&{}-1\\ \ -1&{}-1&{}4&{}-1&{}0&{}0&{}-1&{}0&{}0 \\ \ -1&{}-1&{}-1&{}4&{}0&{}0&{}0&{}-1&{}0\\ \ 0&{}0&{}0&{}0&{}4&{}-1&{}0&{}0&{}-1 \\ \ 0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0&{}0\\ \ 0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1&{}0 \\ \ 0&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4&{}-1\\ \ 0&{}-1&{}0&{}0&{}-1&{}0&{}0&{}-1&{}4 \end{array}\right] \left[ \begin{array}{c} V_1\\ V_2\\ V_{3}\\ V_{4}\\ V_{5}\\ V_{6}\\ V_{7}\\ V_{8}\\ V_{9} \end{array}\right] =\left[ \begin{array}{c} \frac{I}{4}\\ \frac{I}{2}\\ 0\\ {-\frac{I}{2}}\\ \frac{I}{2}\\ 0\\ 0\\ 0\\ -\frac{I}{2} \end{array}\right] \end{aligned}$$

(7.360)

Calculating the inverse and multiplication by the column on the right side introducing $I=1$ we get

$$\begin{aligned} \left[ \begin{array}{c} V_1\\ V_2\\ V_{3}\\ V_{4}\\ V_{5}\\ V_{6}\\ V_{7}\\ V_{8}\\ V_{9} \end{array}\right] =\left[ \begin{array}{c} 0.125\\ 0.125\\ 0.055\\ -0.055\\ 0.125\\ 0.069\\ 0.025\\ -0.025\\ -0.069 \end{array}\right] . \end{aligned}$$

(7.361)

Knowing the potential for all points in the subdomain a, we can determine the other values by using the relations between homologous points $V^{**}_{j}=V_j-\frac{I}{4}$, $V^{*}_{j}=V_j+\frac{I}{4}$ and $V^{***}_{j}=V_j+\frac{I}{4}$, the potential for the other points in b, c and d may be determined. The results for these potentials are

$$\begin{aligned} \left[ \begin{array}{c} V^{*}_1\\ V^{*}_2\\ V^{*}_{3}\\ V^{*}_{4}\\ V^{*}_{5}\\ V^{*}_{6}\\ V^{*}_{7}\\ V^{*}_{8}\\ V^{*}_{9} \end{array}\right] =\left[ \begin{array}{c} 0.375\\ 0.375\\ 0.305\\ 0.195\\ 0.375\\ 0.319\\ 0.275\\ 0.225\\ 0.181 \end{array}\right] , \end{aligned}$$

(7.362)

$$\begin{aligned} \left[ \begin{array}{c} V^{**}_1\\ V^{**}_2\\ V^{**}_{3}\\ V^{**}_{4}\\ V^{**}_{5}\\ V^{**}_{6}\\ V^{**}_{7}\\ V^{**}_{8}\\ V^{**}_{9} \end{array}\right] =\left[ \begin{array}{c} -0.125\\ -0.125\\ -0.195\\ -0.305\\ -0.125\\ -0.181\\ -0.225\\ -0.275\\ -0.319 \end{array}\right] , \end{aligned}$$

(7.363)

Fig. 7.48

Different perspective views of the reconstructed scalar function $V_m(x,y)$. a On the lower graphic we can identify the points where the potential has been calculated by means of the FDM. The upper side represents the same function after smoothing. b Idem from a viewpoint parallel to the cut line $C^{*}$. Observe the curves of constant potential on the plane OXY

and

$$\begin{aligned} \left[ \begin{array}{c} V^{***}_1\\ V^{***}_2\\ V^{***}_{3}\\ V^{***}_{4}\\ V^{***}_{5}\\ V^{***}_{6}\\ V^{***}_{7}\\ V^{***}_{8}\\ V^{***}_{9} \end{array}\right] =\left[ \begin{array}{c} 0.625\\ 0.625\\ 0.555\\ 0.445\\ 0.625\\ 0.569\\ 0.525\\ 0.475\\ 0.431 \end{array}\right] . \end{aligned}$$

(7.364)

These data allow reconstruction of the function $V_m(x,y)$ in the region LxL. Figure 7.48 show the mathematical surface corresponding to the aforementioned values. Observe that this function is similar to the graphic (1.30) studied in Chap. 1 (this similitude is not fortuitous-see Problem 1.17).

Fig. 7.49

Magnetic field at points studied. Note that on the right side and on the upper row do not appear vectors because the derivative needs two points to be performed. Over the edge of triangle C we can calculate the partial derivative with respect the coordinate y, but not x. The contrary occurs for the points over the upper side

The magnetic field may be sketched approximating the partial derivatives that appear in $\mathbf {H}=-\nabla V_m(x,y)$ by means of the following formulae

$$\begin{aligned} \frac{\partial V_m(x,y)}{\partial x}\approx \frac{\triangle V_{\text {x-direction}}}{\triangle x}, \end{aligned}$$

(7.365)

and

$$\begin{aligned} \frac{\partial V_m(x,y)}{\partial y}\approx \frac{\triangle V_{\text {y-direction}}}{\triangle y}, \end{aligned}$$

(7.366)

where $\triangle V_{\text {x-direction }}$ represents the difference between two contiguous potentials in the direction of the x coordinate, and $\triangle V_{\text {y-direction }}$ the same for y. For example, the gradient at point 3 in the domain A (first triangle) is

$$\begin{aligned} \mathbf {H}=-\nabla V_m(x,y)\approx -\frac{V_4-V_3}{a}\mathbf {u}_x-\frac{V_1-V_3}{a}\mathbf {u}_y. \end{aligned}$$

(7.367)

Computing all the differences, the magnetic field H in the hole has the form represented in Fig. 7.49. Obviously a part of the vector field depicted is wrong. As we commented at the beginning of this problem, we have a symmetry of fourth order, but this cannot be seen in all regions of this figure. The result seems to be correct on the domains A, B and D, and fails for some points of C. Specifically, at points $1^{**}$, $4^{**}$, $9^{**}$, $1^{***}$, $2^{***}$, and $5^{***}$, the magnetic field does not correspond to the actual H. Because of the symmetry, at each of these points the field should theoretically have the same modulus as its complementary point of the other parts by changing the sign of the components adequately. For instance, at point 4 the field is $\mathbf {H}_4=\frac{0.07}{a}\mathbf {u}_x+\frac{0.07}{a}\mathbf {u}_y$, and at $4^{**}$ it would be $\mathbf {H}_{4^{**}}=-\frac{0.07}{a}\mathbf {u}_x+\frac{0.07}{a}\mathbf {u}_y$. The reason for this lies in the fact that for calculating the partial derivatives at these points we need points over the cut line $C^{*}$(see Fig. 7.45).²⁶ It leads to a great error because the points on this line are multivalued, and then the result must be wrong. On the other hand, if we analyze the values of the well behaved points in domain A, we observe that they do not exactly maintain the foreign principle of symmetry stated. At point 3 the field is $\mathbf {H}_3=\frac{0.11}{a}\mathbf {u}_x-\frac{0.07}{a}\mathbf {u}_y$, and at point 4 it reaches $\mathbf {H}_4=\frac{0.07}{a}\mathbf {u}_x+\frac{0.07}{a}\mathbf {u}_y$ instead of $\mathbf {H}^{'}_{4}=\frac{0.11}{a}\mathbf {u}_x+\frac{0.07}{a}\mathbf {u}_y$. The reason may be found in the precision of the grid elements we have chosen. Our election was very rough with the only objective to present the method didactically without addressing other concerns. If much smaller elements would be selected, the difference between the field components at such points would be negligent. The easiest way to generate the magnetic field H the rest of the points in regions B, C and D, is to take as a basis the results obtained in the triangle A, and then determine their field by means of a rotation around one axis coinciding with the wire axis. So, in this sense, for calculating $\mathbf {H}$ at points belonging to the domain C, we apply the following transformation

$$\begin{aligned} R_z(\phi )=\left[ \begin{array}{cc} \cos \phi &{} -\sin \phi \\ \sin \phi &{} \cos \phi \end{array}\right] _{\phi =\frac{\pi }{2}}=\left[ \begin{array}{cc} 0&{} -1\\ 1&{} 0 \end{array}\right] . \end{aligned}$$

(7.368)

For instance, let us suppose we want to compute the field at point $3^{**}$. Applying this matrix we have

$$\begin{aligned} \mathbf {H}_{**}^{3}=R_{z}(\frac{\pi }{2})\mathbf {H}_3=\left[ \begin{array}{cc} 0&{} -1\\ 1&{} 0 \end{array}\right] \left[ \begin{array}{c} \displaystyle \frac{0.11}{a} \\ \displaystyle -\frac{0.07}{a} \end{array}\right] = \left[ \begin{array}{c} \displaystyle \frac{0.07}{a} \\ \displaystyle \frac{0.11}{a} \end{array}\right] . \end{aligned}$$

(7.369)

For the domain B, the following transformations holds

$$\begin{aligned} \left[ \begin{array}{cc} -1&{} 0\\ 0&{} -1 \end{array}\right] , \end{aligned}$$

(7.370)

and for D

$$\begin{aligned} \left[ \begin{array}{cc} 0&{} 1\\ -1&{} 0 \end{array}\right] . \end{aligned}$$

(7.371)

A hand sketch of the magnetic lines has been represented in Fig. 7.50.

Fig. 7.50

Sketch of the lines of magnetic field $\mathbf {H}$

• 7.21 Let a closed curve be the cross section of a conducting cylindrical shell, whose cross section has one axis of symmetry (Fig. 7.51). Let this shell be divided into four parts by two planes at right angles, the line of intersection of the planes being parallel to the generator and one of the planes containing the symmetry axis. Show that the direct capacitance, per unit length of the cylinder, between opposing parts of the shell due to the field inside is a constant of value $C_0=2,8$ pFm$^{-1}$. This result is known as the Thompson-Lampard theorem.

Fig. 7.51

Plane view of a cross section of a system with an axis of symmetry

Fig. 7.52

Transformation of the region delimited by the curve $\Gamma$ in the z-plane onto the circle (region $\Omega$) of the w-plane

Fig. 7.53

Transformation of the region D belonging to the z-plane into the w-plane

Solution

To demonstrate this theorem we calculate the cross capacitance between two segments of the Fig. 7.51. Let us consider, for example, the arcs (a, b) and (c, d). The cross capacitance $C_{13}$ is the ratio of the negative of the charge $Q_3$ to the potential V on the surface $S_1$. For simplicity we suppose that the surface $S_1$ is held at unity potential and the rest of the system is held at zero. The calculation of charge for a generic geometry with an axis of symmetry like that shown in Fig. 7.51 may be very difficult. In order to make it easier, we can use the Riemann theorem. According to this theorem, any simply connected region D of the plane can be transformed conformally into a unit circle (Fig. 7.52). Thus in our case, we can map any cylinder cross section onto a circle in which each segment of the original geometry ((a, b),(b, c), etc.) over the curve $\Gamma$ has its corresponding arc over $\gamma$ (circle) in the w-plane (Fig. 7.52). Using an adequate phase in the conformal mapping we can get the segment (a, c) over the OX-axis to be transformed into the diameter of the circle (A, C) (Fig. 7.53). The problem reduces to calculating the cross capacitance of the arcs (A, B) and (C, D) in the w-plane. As we have seen in the theory, if the boundary conditions are those of Dirichlet, the values of the potential in the boundary $\Gamma$ remain unaltered on $\gamma$ when the region D is transformed into $\Omega$. The potential $\Phi (\rho ,\phi )$ inside the circle is well known and may be found by using the following series

$$\begin{aligned} \Phi (\rho ,\phi )=\sum _{n=1}^{\infty } \rho ^n[a_n\,\cos (n\phi )+b_n\,\sin (n\phi )], \end{aligned}$$

(7.372)

where the coefficients of (7.372) for the unit circle $(\rho =1)$ are

$$\begin{aligned} a_n=\frac{1}{\pi }\int _{0}^{2\pi }\Phi (1,\phi )\,\cos (n\phi )d\phi , \end{aligned}$$

(7.373)

$$\begin{aligned} b_n=\frac{1}{\pi }\int _{0}^{2\pi }\Phi (1,\phi )\,\sin (n\phi )d\phi , \end{aligned}$$

(7.374)

respectively. In our case

$$\begin{aligned} a_n=\frac{1}{\pi }\int _{0}^{2\pi }\Phi (1,\phi )\,\cos (n\phi )d\phi = \frac{1}{\pi }\int _{0}^{\alpha }\,\cos (n\phi )d\phi =\frac{1}{n\pi }\,\sin \,n\alpha , \end{aligned}$$

(7.375)

and

$$\begin{aligned} b_n=\frac{1}{\pi }\int _{0}^{2\pi }\Phi (1,\phi )\,\sin (n\phi )d\phi = \frac{1}{\pi }\int _{0}^{\alpha }\,\cos (n\phi )d\phi =\frac{1}{n\pi }[1-\cos \,n\alpha ]. \end{aligned}$$

(7.376)

As we have commented, for calculating the cross capacitance $C_{13}$ it is necessary to know the charge on the arc (C, D), which may be obtained by means of the following expression

$$\begin{aligned} Q_3=\int _{S_3}\sigma \,dS= \varepsilon _0\int _{S_3}\frac{\partial \Phi }{\partial n_3}\,dS. \end{aligned}$$

(7.377)

Introducing (7.372) into (7.377), and knowing that due to the circular symmetry $\frac{\partial \Phi }{\partial n_3}=\frac{\partial \Phi }{\partial \rho }$, it yields for $V=1$

$$\begin{aligned} C_{13}=\frac{Q_3}{V}=\frac{Q_3}{\varepsilon _0}=\sum _{n=1}^{\infty } \int _{\pi }^{2\pi -\alpha }n[a_n\,\cos (n\phi )+b_n\,\sin (n\phi )]=\nonumber \\ \sum _{n=1}^{\infty }\left\{ n\frac{1}{n\pi }\,\sin \,n\alpha \int _{\pi }^{2\pi -\alpha }\cos (n\phi )d\phi +n\frac{1}{n\pi }[1-\cos \,n\alpha ]\int _{\pi }^{2\pi -\alpha } \sin (n\phi )d\phi \right\} =\nonumber \\ \sum _{n=1}^{\infty }\left\{ \frac{1}{n\pi }\,\sin \,n\alpha \,\,\sin [(2\pi -\alpha )n]+\frac{1}{n\pi }[1-\cos \,n\alpha ] [-\cos (2\pi -\alpha )n+cos\,n\pi ]\right\} =\nonumber \\ \sum _{n=1}^{\infty }\left\{ -\frac{1}{n\pi }\,\sin ^2n\alpha -\frac{1}{n\pi }[1-\cos \,n\alpha ][\cos \,n\alpha -(-1)^{n}]\right\} . \end{aligned}$$

(7.378)

However, taking into account that $-(-1)^{n}=(-1)^{n-1}$ we have

$$\begin{aligned} \frac{-Q}{\varepsilon _0}= \sum _{n=1}^{\infty }\left\{ -\frac{1}{n\pi }\,\sin ^2n\alpha -\frac{1}{n\pi } [\cos \,n\alpha -\cos ^2n\alpha -(-1)^{n-1}\cos \,n\alpha +(-1)^{n-1}]\right\} =\nonumber \\ -\sum _{n=1}^{\infty }\frac{1}{n\pi }(-1)^{n-1}- \sum _{n=1}^{\infty }\frac{1}{n\pi }\left\{ \sin ^2n\alpha -\cos ^2n\alpha +\cos \,n\alpha [1-(-1)^{n-1}]\right\} =\nonumber \\ -\sum _{n=1}^{\infty }\frac{1}{n\pi }(-1)^{n-1}- \sum _{n=1}^{\infty }\frac{1}{n\pi }\left\{ -\cos \,2n\alpha +\cos \,n\alpha [1-(-1)^{n-1}]\right\} . \end{aligned}$$

(7.379)

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\cdots . \end{aligned}$$

(7.380)

This is an alternating series whose sum may be calculated by means of different procedures. Perhaps the easiest way is to compare it with the Taylor series of a known function whose coefficients coincide with those appearing in (7.380) (if possible). In this case we know that the terms of the logarithmic function are alternately positive and negative, and the denominator of the fractions grow with the number of terms n, then we try the expansion

$$\begin{aligned} \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}\cdots , \end{aligned}$$

(7.381)

which is convergent in the interval $-1<x\le 1$. If we put $x=1$ in (7.381), we have

$$\begin{aligned} \ln (2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\cdots , \end{aligned}$$

(7.382)

which is the same as (7.380). Thus we can write

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n}=\ln \,2 \end{aligned}$$

(7.383)

$$\begin{aligned} \frac{-Q}{\varepsilon _0}=-\ln \,2+ \frac{1}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\,\cos \,2n\alpha - \frac{1}{\pi }\sum _{n\,even}^{\infty }\frac{2}{n}\,\cos \,n\alpha =&\nonumber \\ -\ln \,2+ \frac{1}{\pi }\left\{ \cos \,2\alpha +\frac{1}{2}\cos \,4\alpha +\frac{1}{4}cos\,6\alpha +\cdots -\cos \,2\alpha - \frac{1}{2}\cos \,4\alpha -\frac{1}{4}\cos \,6\alpha -\cdots \right\} .&\end{aligned}$$

(7.384)

If we observe this expression for the charge on the arc $(\gamma ,\delta )$ we see that the terms in the brackets cancel each other out and result in a constant, hence we can write

$$\begin{aligned} \frac{Q}{\varepsilon _0}=\ln \,2. \end{aligned}$$

(7.385)