Static Electric Field in Vacuum

Abstract

This chapter introduces forces between charges at rest, which are not supposed to be inside any media (Coulomb’s law). Concepts such as electric field or electric potential are introduced, as well as its calculus when produced by different charge distributions, including conductive materials. Gauss’ law and its use to calculate electric field caused by certain charge distributions is also seen.

Appendices

Solved Problems

Problems A

2.1 :

In a cartesian coordinate system, with the axis in meters, two point charges are considered, one of them positive of $1\,$nC, located at the origin of coordinates, and the other one, negative of $-20\,$nC, located at A(0, 1). Determine the resulting field at B(2, 0) and the necessary work to take a positive charge of $3\,\upmu$C from B(2, 0) to C(4, 2).

Solution

Applying the superposition principle for electrostatic field, field at B will be the vectorial addition of the fields due to each charge (Fig. 2.4), this is:

$$ \mathbf {E}_B=\mathbf {E}_{OB}+\mathbf {E}_{AB}. $$

To calculate the field due to each charge, (2.15) is applied. Point B’s coordinates are $\mathbf {r}=2\mathbf {u}_x$. In the case for the charge at O, position vector is $\mathbf {r\,'}=0$, and we obtain $\mathbf {r}-\mathbf {r\,'}=\mathbf {OB}=2\mathbf {u}_x$ and $|\mathbf {r}-\mathbf {r\,'}|=2$. Position vector of charge at A is $\mathbf {r\,'}=\mathbf {u}_y\,$, so $\mathbf {r}-\mathbf {r\,'}=\mathbf {AB}=2\mathbf {u}_x-\mathbf {u}_y$ and $|\mathbf {r}-\mathbf {r\,'}|=\sqrt{5}$. If (2.15) is applied to the charge located at O, the result is,

Fig. 2.4

Field produced by point charges

$$ \mathbf {E}_{OB}=\frac{1}{4\pi \varepsilon _{0}}\frac{q'}{|\mathbf {r}-\mathbf {r\,'}|^{2}}\, \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|}=9\cdot 10^9\frac{10^{-9}}{4}\mathbf {u}_x=\frac{9}{4}\mathbf {u}_x. $$

And for the charge located at A,

$$\begin{aligned} \mathbf {E}_{AB}&=\frac{1}{4\pi \varepsilon _{0}}\frac{q'}{|\mathbf {r}-\mathbf {r\,'}|^{2}}\, \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|}=9\cdot 10^9\frac{(-20 \cdot 10^{-9})}{5}\left( \frac{2}{\sqrt{5}}\mathbf {u}_x-\frac{1}{\sqrt{5}}\mathbf {u}_y\right) \\&=\frac{36}{\sqrt{5}}\left( -2\mathbf {u}_x+\mathbf {u}_y\right) . \end{aligned}$$

The field could be obtained without using position vectors, and bearing in mind Fig. 2.4. Since there is a negative charge at A, field $E_{AB}$ is pointed to A. If we use (2.15) to calculate the electric field magnitude and if we project it, we obtain the field expressed as,

$$ \mathbf {E}_{AB}=9\cdot 10^9\frac{20 \cdot 10^{-9}}{5}(-\cos \alpha \mathbf {u}_x+\sin \alpha \mathbf {u}_y)=36\left( \frac{-2}{\sqrt{5}}\mathbf {u}_x+\frac{1}{\sqrt{5}}\mathbf {u}_y\right) . $$

The resulting field will be

$$ \mathbf {E}_B=\mathbf {E}_{OB}+\mathbf {E}_{AB}=\left( \frac{9}{4}-\frac{72}{\sqrt{5}}\right) \mathbf {u}_x+\frac{36}{\sqrt{5}}\mathbf {u}_y=-29.95\mathbf {u}_x+16.10\mathbf {u}_y. $$

Since the electrostatic field is conservative, the work necessary to take a charge from B to C is equal to the variation of potential energy between B and C (2.20), with negative sign. From (2.21) which relates electrostatic potential to the potential energy,

$$ W_{BC}=-(E\mathrm{p}_B-E\mathrm{p}_C)=-q(V_B-V_C), $$

and potentials at points B and C due to the charge system of the problem must be calculated. To obtain the potential on each point, the superposition principle is applied, this is, the potential to be the added due to each charge, $V_B=V_{OB}+V_{AB}$ and $V_C=V_{OC}+V_{AC}$. If (2.24) (potential produced by a point charge) is applied

$$ V=\frac{1}{4 \pi \varepsilon _0}\frac{q'}{|\mathbf {r}-\mathbf {r}\,'|}, $$

potentials due to each charge on each point $V_{OB}, V_{AB}, V_{OC}$ and $V_{AC}$ can be determined. It is necessary to define the terms ($|\mathbf {r}-\mathbf {r\,'}|$), which are the distances from each charge to points B and C. On the case of point B the distances have already been determined for the calculus of the electric field, so potential in B will be

$$ V_B=V_{OB}+V_{AB}=9\cdot 10^9\left( \frac{10^{-9}}{2}+\frac{(-20 \cdot 10^{-9})}{\sqrt{5}}\right) =-76\,\text {V}. $$

For point C, its position vector is $\mathbf {r}=4\mathbf {u}_x+2\mathbf {u}_y$, and it becomes that $\mathbf {r}-\mathbf {r\,'}$ values are $\mathbf {OC}=4\mathbf {u}_x+2\mathbf {u}_y$ for the charge at O and $\mathbf {AC}=(4\mathbf {u}_x+2\mathbf {u}_y)-\mathbf {u}_y=4\mathbf {u}_x+\mathbf {u}_y$ for the charge at A. Potential is

$$ V_C=V_{OC}+V_{AC}=9\cdot 10^9\left( \frac{10^{-9}}{\sqrt{20}}+\frac{(-20 \cdot 10^{-9})}{\sqrt{17}}\right) =-41.6\,\text {V}. $$

And circulation from point B to C is

$$ W_{BC}=-q(V_B-V_C)=-3\cdot 10^{-6}(-76+41.6)=103.2\cdot 10^{-6}\mathrm {J}=103.2\,\upmu \mathrm {J}. $$

2.2 :

In the space region defined by $y>0$, a charge density $\rho =cy$ exists, with $c=2\,\upmu$C/m$^4$ and y the distance (in meters) from any point to plane XOZ. Calculate: (a) The flux of the electrostatic field through the prism’s surface in Fig. 2.5. (b) Divergence of electrostatic field in the prism’s faces which are parallel to plane XOZ.

Fig. 2.5

Prism of Problem 2.2

Solution

Fig. 2.6

Differential element to calculate internal charge

(a) Gauss’ law (2.28) states the flux of the electric field due to a charge distribution,

$$ \Phi _E=\oint _{\partial V} \mathbf {E}\cdot d\mathbf {S} =\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _V \rho dV. $$

Therefore we must calculate the charge $q_\mathrm{in}$ inside the prism in Fig. 2.5. Since charge density $\rho$ only depends on coordinate y, every point located at the same distance y have the same density. Let’s consider the infinitesimal volume drawn in Fig. 2.6, $dV=3\cdot 3\,dy=9\,dy$. At every point in it the density is cy, so the volume integral becomes a simple integral:

$$ q_\mathrm{in}=\int _V \rho dV=\int _1^5 cy\,9\,dy= \left. \frac{9 cy^2}{2}\right| _1^5= 216\,\upmu \mathrm{Cm}^{-3}. $$

And therefore the flux is

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{216\cdot 10^{-6}}{8.85\cdot 10^{-12}}=24.4\cdot 10^{6}\,\mathrm{N C}^{-1}\mathrm{m}^2. $$

(b) We obtain the divergence of the electric field from (2.29). If we particularize for each of the prism faces, we obtain:

$$\begin{aligned} {\text {div}}\,\mathbf {E}=\frac{\rho }{\varepsilon _0}=\frac{cy}{\varepsilon _0}\left\{ \begin{array}{ll} y=1, &{} \displaystyle {\text {div}}\,\mathbf {E}= \frac{2\cdot 10^{-6}\cdot 1}{8.85\cdot 10^{-12}} =2.26 \cdot 10^{5}\,\mathrm{NC}^{-1}\mathrm{m}^{-1}\\ \\ y=5, &{} \displaystyle \text {div}\,\mathbf {E}= \frac{2\cdot 10^{-6}\cdot 5}{8.85\cdot 10^{-12}} =1.13 \cdot 10^{6}\,\mathrm{NC}^{-1}\mathrm{m}^{-1}\end{array}\right. \end{aligned}$$

2.3 :

Determine the electric field produced on any point in space by a very long line (infinite) charged with a uniform density $\lambda$.

Solution ⁸

This problem has cylindrical symmetry. Hence all the points at the same distance to the line have the same electric field magnitude, and $\mathbf {E}$ is perpendicular to the line.⁹ The problem can be solved by applying Gauss’ law (2.28). We take as a Gaussian surface $\partial V$ a closed cylindrical surface, with any length L, with the axis on the line and with radius r making the surface to pass through point P, the field desired to be calculated (discontinuous line in Fig. 2.7). The electric field at any point on the Gaussian surface is radial (perpendicular to the cylinder lateral surface), outward pointed, if we suppose the line as positively charged,¹⁰ and has the same magnitude at every point on the lateral surface. It should be observed that the Gaussian surface is a closed surface and therefore to calculate the flux the cylinder lateral surface where the point P is and the cylinder bases, where the magnitude of the field is not constant and is different from the one at the lateral surface, should be considered. However, this is not a problem, since the flux through the bases is null due to the fact that $d\mathbf {S}$ and $\mathbf {E}$ are perpendicular at any point. Calculating the flux through the lateral surface, where $d\mathbf {S}$ and $\mathbf {E}$ are parallel:

$$ \Phi _E=\oint _{\partial V}\mathbf {E}\cdot d\mathbf {S} =\int _{S_\mathrm{lat}} E\,{ {dS}}=E\int _{S_\mathrm{lat}}dS=E2\pi rL, $$

where $S_\mathrm{lat}$ is the cylinder lateral surface. On another side, if Gauss’ law is applied, it is obtained

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{\lambda L}{\varepsilon _0}. $$

Equating both expressions,

$$ E=\frac{\lambda }{2\pi \varepsilon _0 r}, $$

shows that the field varies inversely with the distance to the line. This expression is the same as the one calculated by integration in Problem 2.4. Using vectors:

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

(2.36)

where $\mathbf {u}_\rho$ is the radial unit vector for cylindrical coordinates.

Fig. 2.7

Gaussian surface and field vectors to apply Gauss’ law to an infinite line

Problems B

2.4 :

Determine the electric field, at any point in space, produced by a line with length L, which has been uniformly charged by a total charge Q: (a) directly; (b) from electrostatic potential. Apply the result for the particular cases of the field produced by an infinitely long line with density $\lambda$ and to the field produced by a semi-infinite line with density $\lambda$ at a point located on the perpendicular to its extreme.

Solution

(a) The problem above is drawn in Fig. 2.8, where a cartesian coordinate system has been defined, just to simplify the calculus. To achieve this, we choose a point P where we want to calculate the field (field point), and we define a plane XY as drawn. The X axis is on the charged line and the origin is at one extreme of the line. The coordinates of point P will be (x, y) and the electric field $\mathbf {E}$ will be on plane XY.

To solve the problem we calculate the field due to an element $dq'$ from the line, and the superposition principle is applied. The field produced by an element $dq'$ is given by (2.15)

$$ d\mathbf {E}(\mathbf {r})=\frac{dq'}{4 \pi \varepsilon _{0}|\mathbf {r}-\mathbf {r\,'}|^{2}}\, \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|}, $$

where $\mathbf {r}=x\mathbf {u}_x+y\mathbf {u}_y$ is the position vector of point P and $\mathbf {r\,'}=x'\mathbf {u}_x$ is the position of each charge element (source point) $dq'$. Charge $dq'$ is obtained from lineal charge density definition $\lambda$ (2.5) as

$$ dq'=\lambda dl=\lambda dx'. $$

$\lambda$ is obtained from (2.6), bearing in mind that, since it is a uniform charge, $\lambda$ is constant:

$$ \lambda =\frac{Q}{L}. $$

With $\mathbf r$ and $\mathbf {r\,'}$ values we obtain vector $\mathbf {r}-\mathbf {r\,'}$ and the distance from $dq'$ to point P:

$$ \mathbf {r}-\mathbf {r\,'}=(x-x')\mathbf {u}_x+y\mathbf {u}_y\,,\quad d=|\mathbf {r}-\mathbf {r\,'}|=\sqrt{(x-x')^2+y^2}. $$

If we substitute the electric field produced by $dq'$ at point P, we have

$$ d\mathbf {E}(\mathbf {r})=\frac{dq'}{4 \pi \varepsilon _{0}|\mathbf {r}-\mathbf {r\,'}|^{2}}\, \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|}=\frac{1}{4 \pi \varepsilon _{0}}\frac{\lambda dx'}{\left[ (x-x')^2+y^2\right] }\, \frac{(x-x')\mathbf {u}_x+y\mathbf {u}_y}{\sqrt{(x-x')^2+y^2}}. $$

The total field is obtained by applying the superposition principle (2.16):

$$ \mathbf {E}=\frac{\lambda }{4 \pi \varepsilon _{0}}\int _L \frac{dx'}{\left[ (x-x')^2+y^2\right] }\frac{(x-x')\mathbf {u}_x+y\mathbf {u}_y}{\sqrt{(x-x')^2+y^2}}. $$

Fig. 2.8

Field produced by an finite line at any point

Solving using integrals, where the only variable is $x'$, allows us to obtain the total field’s value. A way of solving this integral is by introducing a variable change, depending on the angle $\varphi$ in Fig. 2.8,

$$ \cos \varphi =\frac{x-x'}{d},\quad \quad \sin \varphi =\frac{y}{d},\quad \quad \cot \varphi =\frac{\cos \varphi }{\sin \varphi }=\frac{x-x'}{y}. $$

If we take the derivative of the last expression, the result is

$$ \frac{d\varphi }{\sin ^2\varphi }=\frac{dx'}{y}. $$

If we obtain $dx'$ and d values,

$$ dx'=\frac{y\,d\varphi }{\sin ^2\varphi },\quad d=\frac{y}{\sin \varphi }, $$

and we substitute into the integral, and substitute $\sqrt{(x-x')^2+y^2}$ for d, it follows

$$\begin{aligned} \mathbf {E}= & {} \frac{\lambda }{4 \pi \varepsilon _{0}}\int _L \frac{dx'}{d^2}\frac{(x-x')\mathbf {u}_x+y\mathbf {u}_y}{d}\\ {}= & {} \frac{\lambda }{4\pi \varepsilon _0}\left( \int _{\varphi }\frac{\sin ^2\varphi }{y^2}\frac{y\,d\varphi }{\sin ^2\varphi }\,\cos \varphi \mathbf {u}_x+\int _{\varphi }\frac{\sin ^2\varphi }{y^2}\frac{y\,d\varphi }{\sin ^2\varphi }\,\sin \varphi \mathbf {u}_y\right) \\= & {} \frac{\lambda }{4\pi \varepsilon _0 y}\left( \int _{\varphi _i}^{\varphi _f}\cos \varphi d\varphi \mathbf {u}_x+\int _{\varphi _i}^{\varphi _f}\sin \varphi d\varphi \mathbf {u}_y\right) . \end{aligned}$$

where $\varphi _i$ and $\varphi _f$ are the angles measured from the initial and final positions of the line, as shown in Fig. 2.8. It should be observed that y, the generic ordinate of a point field, is not a variable in the integral. If we solve it, it results

$$\begin{aligned} \mathbf {E}=\frac{\lambda }{4\pi \varepsilon _0 y}\left( (\sin \varphi _f-\sin \varphi _i)\mathbf {u}_x+(\cos \varphi _i-\cos \varphi _f)\mathbf {u}_y\right) .\end{aligned}$$

(2.37)

The result can be written in Cartesian coordinates:

$$\begin{aligned} \mathbf {E}=\frac{Q}{4\pi \varepsilon _0 y L}&\left[ \left( \frac{y}{\sqrt{(L-x)^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}\right) \mathbf {u}_x\right. \nonumber \\&\,\,+\left. \left( \frac{x}{\sqrt{x^2+y^2}}+\frac{L-x}{\sqrt{(L-x)^2+y^2}}\right) \mathbf {u}_y\right] , \end{aligned}$$

(2.38)

where $\lambda$’s value has been replaced by Q / L.

(b) To calculate the electric field at P from the potential V, we must obtain a potential generic expression at any point (for example P(x, y)) and then obtain the field from (2.18). It is important to notice that this procedure can be done because the potential at all the points is known, and therefore its gradient can be calculated. To obtain the potential at point P, the potential produced by a charge element $dq'$ is expressed, taking the potential reference at infinity (2.24),

$$ dV=\frac{1}{4 \pi \varepsilon _{0}}\frac{dq'}{|\mathbf {r}-\mathbf {r\,'}|}=\frac{\lambda }{4\pi \varepsilon _0}\frac{dx'}{\sqrt{(x-x')^2+y^2}}, $$

and the superposition principle for electrostatic potential is applied (2.25):

$$ V=\frac{\lambda }{4\pi \varepsilon _0}\int _0^L\frac{dx'}{\sqrt{(x-x')^2+y^2}}. $$

Solving this integral,

$$ V=-\frac{\lambda }{4\pi \varepsilon _0}\ln \left( x-x'+\sqrt{(x-x')^2+y^2} \right) \bigg |_0^L=\frac{\lambda }{4\pi \varepsilon _0}\ln \frac{x+\sqrt{x^2+y^2}}{x-L+\sqrt{(x-L)^2+y^2}}. $$

If we apply now (2.18),

$$ \mathbf {E}=-\nabla V=-(\partial V/\partial x)\mathbf {u}_x-(\partial V/\partial y)\mathbf {u}_y, $$

and if we solve the indicated partial derivative, the total field $\mathbf {E}$ at P is obtained (2.38).

Let’s consider the case of an infinite line (very long line). Since the line has an infinite length, (2.38) is not obvious. It is easier to use (2.37). It should be observed in Fig. 2.8 that if the line is infinite, initial and final angles are $\varphi _i=0$ and $\varphi _f=\pi$. If we substitute in (2.37), it results

$$ \mathbf {E}=\frac{\lambda }{2\pi \varepsilon _0 y}\mathbf {u}_y. $$

Let’s consider that the line begins at O and is very long (semi-infinite line) and P is over the perpendicular to the line at O. Point P coordinates are P(0, y). If we consider (2.37), we observe in Fig. 2.8 that, since P(0, y), $\varphi _i=\pi /2$ and $\varphi _f=\pi$. Then,

$$ \mathbf {E}=\frac{\lambda }{4\pi \varepsilon _0 y}(-\mathbf {u}_x+\mathbf {u}_y), $$

We should be cautious when applying the used procedure in section (b) to obtain the field when the line is infinitely long. We observe the potential for this charge distribution as

$$ V=\lim _{L\longrightarrow \infty }\frac{\lambda }{4\pi \varepsilon _0} \ln \frac{x+\sqrt{x^2+y^2}}{x-L+\sqrt{(x-L)^2+y^2}}\longrightarrow \infty . $$

So it is not possible to obtain the field from this potential. Infinite potential has been obtained because for charge distributions that spread in an infinite region, it can never be certain that this potential converges. However, the field from the potential can be obtained as follows: firstly we calculate the finite line potential, then its gradient, and then we make the line’s length to infinity. This difficulty will also appear in the magnetostatic chapter.

2.5 :

Determine the electric field and the potential at any point in space produced by a spherical crown where the internal radius is $R_1$ and the external one $R_2$, with a total charge Q, for the following cases: (a) non conducting and an uniform charge distributed throughout the volume; and (b) metallic and on electrostatic equilibrium. Particularize the results for a solid sphere with radius R.

Solution

Fig. 2.9

Gaussian surfaces and field vectors to apply Gauss’ law to a spherical crown

(a) In this case the spherical crown has a uniform volume charge density $\rho$ at every point between $R_1$ and $R_2$, that can be calculated by applying (2.2):

$$ Q=\int _V \rho dV=\rho \int _V dV=\rho V = \rho \frac{4}{3} \pi (R_2^3-R_1^3) \Rightarrow $$

$$ \rho =\frac{Q}{V}=\frac{3Q}{4\pi (R_2^3-R_1^3)}. $$

Because of its symmetry, Gauss’ law (2.28 ) is applied, considering a spherical Gaussian surface $\partial V$ (Fig. 2.9), concentric with the charge distribution, and passing through the point where we want to calculate the field. At every point inside a sphere whose radius is $r\le R_1$, field is null since internal charge is zero:

$$ E_{( r\le R_1)}=0. $$

Due to the charge distribution symmetry, the electric field at any other point on the Gaussian surface will be radial,¹¹ outward¹² and with the same magnitude at every point on the surface. Two different expressions are obtained for the electric field, depending on the point to be studied, if it is outside or inside the spherical crown (P and $P'$ in Fig. 2.9), since internal charge to the Gaussian surface has a different expression.

Let’s firstly consider point P, outside the crown. The flux through the Gaussian surface passing through P is

$$ \Phi _E=\oint _{\partial V}\mathbf {E}\cdot d\mathbf {S}=\oint _{\partial V} E\,dS=E\oint _{\partial V}dS= E 4\pi r^2, $$

where $4\pi r^2$ is the spherical surface’s area, with radius r. We observe that the total charge within the Gaussian surface is all the charge of the spherical crown (grey in figure). If Gauss’ law is applied, the flux is

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{Q}{\varepsilon _0} =\frac{\rho 4\pi (R_2^3-R_1^3)}{3\varepsilon _0}. $$

If we equate the two previous expressions for the flux and we solve it, we obtain the field at an external point P,

$$ E_{(r\ge R_2)}=\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r^2}=\frac{Q}{4\pi \varepsilon _0 r^2}, $$

expression that coincides with the field produced by a point charge Q located at the spherical crown centre. In fact, the field produced by a point charge, known from Coulomb’s law, can be calculated by applying Gauss’ law to a random spherical surface whose centre is on the charge.

To know the field at a point $P'$ (Fig. 2.9), inside the spherical crown, the same procedure is followed. We set up a spherical concentric surface, with radius r, passing through $P'$. The expression of the flux through a surface of radius r is the same, but since $R_1\le r\le R_2$, internal charge to the Gaussian surface is now not the total charge of the crown charge, but only the dark grey region in the figure. If we calculate

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{\rho 4\pi (r^3-R_1^3)}{3\varepsilon _0}, $$

and then, from $\Phi _E=E 4\pi r^2$, the field at a point $P'$ inside the spherical crown is obtained,

$$ E_{(R_1\le r\le R_2)}=\frac{\rho (r^3-R_1^3)}{3\varepsilon _0 r^2}=\frac{Q}{4\pi \varepsilon _0 r^2}\,\frac{(r^3-R_1^3)}{(R_2^3-R_1^3)}. $$

Joining both results and expressing the field as a vector, it results

$$\begin{aligned} \displaystyle \mathbf {E}= \left\{ \begin{array}{ccccc} 0&{}&{} &{}\quad &{} r\le R_1,\\ \\ \displaystyle \frac{Q}{4\pi \varepsilon _0 r^2}\,\frac{(r^3-R_1^3)}{(R_2^3-R_1^3)}\mathbf {u}_r&{}=&{} \displaystyle \frac{\rho (r^3-R_1^3)}{3\varepsilon _0 r^2}\mathbf {u}_r&{}\quad &{}R_1\le r\le R_2,\\ &{} &{} &{} &{}\\ \displaystyle \frac{Q}{4\pi \varepsilon _0 r^2}\mathbf {u}_r&{}=&{}\displaystyle \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r^2}\mathbf {u}_r&{}\quad &{}r\ge R_2, \end{array}\right. \end{aligned}$$

(2.39)

where $\mathbf {u}_r$ is the radial unit vector for spherical coordinates. The field at a point on the outside spherical surface can be calculated by using either one of the expressions, making $r=R_2$:

$$ \mathbf {E}_{(r=R_2)}=\frac{Q}{4\pi \varepsilon _0 R_2^2 }\mathbf {u}_r=\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 R_2^2}\mathbf {u}_r. $$

If the sphere were solid, with radius R, results can be obtained by replacing $R_1=0$ and $R_2=R$:

$$\begin{aligned} \displaystyle \mathbf {E}= \left\{ \begin{array}{ccccc}\displaystyle \frac{Q}{4\pi \varepsilon _0}\,\frac{r}{R^3}\mathbf {u}_r&{}=&{} \displaystyle \frac{\rho r}{3\varepsilon _0}\mathbf {u}_r&{}\quad &{}r\le R,\\ &{} &{} &{} &{}\\ \displaystyle \frac{Q}{4\pi \varepsilon _0 r^2}\mathbf {u}_r&{}=&{}\displaystyle \frac{\rho R^3}{3\varepsilon _0 r^2}\mathbf {u}_r&{}\quad &{}r\ge R, \end{array}\right. \end{aligned}$$

(2.40)

And the field at a point on the surface would be:

$$ \mathbf {E}_{(r=R)}=\frac{Q}{4\pi \varepsilon _0 R^2 }\mathbf {u}_r=\frac{\rho R}{3\varepsilon _0}\mathbf {u}_r. $$

To determine the potential at any point, we calculate the potential difference between that point and infinity, with null potential. Since the circulation of the electrostatic field is path-independent, we take the radial direction from the point, where $\mathbf {E}$ and $d\mathbf {l}$ are parallel,¹³ as shown in Fig. 2.10.

For an external point P we have, if we circulate the field $\mathbf {E}$ between P and infinity (Fig. 2.10), that

$$\begin{aligned}V_P-V_\infty =V_P\,= & {} \,\int _P^\infty \mathbf {E}\cdot d\mathbf {l}=\int _r^\infty E_{(r\ge R_2)} dr\\= & {} \,\int _r^\infty \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r^2}dr=\\= & {} \,-\left. \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r}\right| _r^\infty .\end{aligned}$$

Then,

$$ V_P=\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r}=\frac{Q}{4\pi \varepsilon _0 r}. $$

It can be observed how the potential given by the charged spherical crown at an external point is the same as a point charge, located at the centre, with the same charge, would create.

If the point $P'$ is inside the spherical crown, from which we need to circulate from $P'$ to infinity, field expressions are different depending on where we circulate, inside or outside the spherical crown. Potential difference is obtained from

$$\begin{aligned}V_{P'}-V_\infty =V_{P'}\,= & {} \,\int _{P'}^\infty \mathbf {E}\cdot d\mathbf {l}=\int _r^{R_2} E_{(r\le R_2)} dr+\int _{R_2}^\infty E_{(r\ge R_2)}dr\\ \!\!= & {} \int _r^{R_2} \frac{\rho (r^3-R_1^3)}{3\varepsilon _0 r^2}dr+\int _{R_2}^\infty \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r^2}dr=\\= & {} \,\left. \frac{\rho r^2}{6\varepsilon _0}\right| _r^{R_2}+\left. \frac{\rho R_1^3}{3\varepsilon _0 r}\right| _r^{R_2}-\left. \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r}\right| _{R_2}^\infty . \end{aligned}$$

The potential will be

$$ V_{P'}=\frac{\rho (R_2^2-r^2)}{6\varepsilon _0}+\frac{\rho R_1^3}{3\varepsilon _0}\left( \frac{1}{R_2}-\frac{1}{r}\right) +\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 R_2}. $$

Fig. 2.10

Potential calculation of a charged spherical crown

If the point ($P''$) is in the hole (Fig. 2.10), so $r\le R_1$, given the field at the hole is null, circulation from point $P''$ to the inner radius of the crown $R_1$ is also null. Point $P''$ potential is the same as the one at point $P'$ located on the inner spherical surface, with $r=R_1$. If the previous result is specified for $r=R_1$ the result is:

$$ V_{P''}=\frac{\rho (R_2^2-R_1^2)}{6\varepsilon _0}+\frac{\rho R_1^3}{3\varepsilon _0}\left( \frac{1}{R_2}-\frac{1}{R_1}\right) +\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 R_2}. $$

If all the results are joined,

$$\begin{aligned} V= \left\{ \begin{array}{ccc}\displaystyle \frac{\rho (R_2^2-R_1^2)}{6\varepsilon _0}+\frac{\rho R_1^3}{3\varepsilon _0}\left( \frac{1}{R_2}-\frac{1}{R_1}\right) +\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 R_2} &{}\quad &{}r\le R_1,\\ &{} &{} \\ \displaystyle \frac{\rho (R_2^2-r^2)}{6\varepsilon _0}+\frac{\rho R_1^3}{3\varepsilon _0}\left( \frac{1}{R_2}-\frac{1}{r}\right) +\frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 R_2} &{}\quad &{}R_1< r\le R_2,\\ &{} &{} \\ \displaystyle \frac{\rho (R_2^3-R_1^3)}{3\varepsilon _0 r}=\displaystyle \frac{Q}{4\pi \varepsilon _0 r}&{}\quad &{}r\ge R_2. \end{array}\right. \end{aligned}$$

(2.41)

(b) If the spherical crown is metallic and has electrostatic balance, charge is only distributed on its external surface, as it can be deduced from conductor properties seen in Sect. 2.8. To obtain null electric field inside the conductor, the crown must be charged on its external surface with a uniform superficial density. This density can be obtained from (2.4):

$$ Q=\int _S \sigma dS= \sigma \int _S dS = \sigma 4\pi R_2^2\Rightarrow \sigma =\frac{Q}{4\pi R_2^2}. $$

The field at internal points ($r<R_2$) is null, due to the charge distribution symmetry. To obtain the electric field produced by the crown at external points to it, and due to the fact that the problem’s symmetry is analogous to the one in section (a), we proceed as it was done in that section. We use the same Gaussian surfaces (Fig. 2.9) but just changing that the charge Q is only on the surface. The flux through the spherical surface of radius r is, as it happened with the previous case,

$$ \Phi _E=E 4\pi r^2. $$

If Gauss’ law is applied to an external point P to the spherical crown,¹⁴ ($r>R_2$), results in

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{Q}{\varepsilon _0}=\frac{4\pi R_2^2\sigma }{\varepsilon _0}. $$

In solving it, the result for an external point P is

$$ E_{(r>R_2)}=\frac{Q}{4\pi \varepsilon _0 r^2}=\frac{\sigma R_2^2}{\varepsilon _0 r^2}, $$

the same expression depending on the total charge Q that in section (a). If it is expressed with vector notation, the results are:

$$\begin{aligned} \mathbf {E}= \left\{ \begin{array}{ccccc}\displaystyle &{} 0&{} &{}\quad &{}r<R_2,\\ \displaystyle \frac{Q}{4\pi \varepsilon _0 r^2}\mathbf {u}_r&{}=&{}\displaystyle \frac{\sigma R_2^2}{\varepsilon _0 r^2}\mathbf {u}_r&{}\quad &{}r>R_2, \end{array}\right. \end{aligned}$$

(2.42)

where $\mathbf {u}_r$ is the radial unit vector for spherical coordinates.

Since the field at external points to the metallic crown is the same as the one in section (a), if the same reference is taken (infinity), the potential is also the same for points outside the crown. For points inside it, given the field is null at these points, potential does not change and has the same value that on the crown’s surface. Then,

$$\begin{aligned} \displaystyle V= \left\{ \begin{array}{ccccc}\displaystyle \frac{Q}{4\pi \varepsilon _0 R_2}&{}=&{}\displaystyle \frac{\sigma R_2}{\varepsilon _0}&{}\quad &{}r \le R_2,\\ &{} &{} &{} &{}\\ \displaystyle \frac{Q}{4\pi \varepsilon _0 r}&{}=&{}\displaystyle \frac{\sigma R_2^2}{\varepsilon _0 r}&{}\quad &{}r\ge R_2. \end{array}\right. \end{aligned}$$

(2.43)

2.6 :

Determine the electric field produced by a very long charged cylinder at any point, with inner radius $R_1$ and external radius $R_2$, with a charge per unit of length q, for the following cases: (a) non conductor and uniformly charged; and (b) metallic and in electrostatic balance. Specify the results for a solid cylinder with radius R.

Solution

(a) In this case the cylinder has a uniform volumetric charge density $\rho$, at every point between $R_1$ and $R_2$, that can be calculated by applying (2.2), and considering the finite cylinder length L,

$$ Q=\int _V \rho dV=\rho \int _V dV=\rho V = \rho \pi (R_2^2-R_1^2)L \Rightarrow $$

$$ \Rightarrow \rho =\frac{Q}{V}=\frac{Q/L }{\pi (R_2^2-R_1^2)}=\frac{q}{\pi (R_2^2-R_1^2)}. $$

Fig. 2.11

Gaussian surface and field vectors to apply Gauss’ law to an infinite cylinder

The problem has cylindrical symmetry: at points inside the cylinder ($r<R_1$), the field is null, due to that symmetry and, for the other zones, all the points at the same distance to the cylinder axis have the same electric field magnitude, with direction perpendicular to that axis.¹⁵ The problem can be solved by applying Gauss’ law (2.28), the same as it was done for Problem 2.3. For this we take as a Gaussian surface $\partial V$ a cylindrical surface, with any length L, with the same axis as the cylinder and radius r, making the surface pass through point P (or $P'$) where we want to calculate the field (discontinuous line in Fig. 2.11).

The flux through the bases of the Gaussian surface, is null, since $d\mathbf {S}$ and $\mathbf {E}$ are perpendicular at any given point. Calculating the flux through the lateral surface, where $d\mathbf {S}$ and $\mathbf {E}$ are parallel:

$$\begin{aligned} \Phi _E=\oint _{\partial V}\mathbf {E}\cdot d\mathbf {S} =\int _{S_\mathrm{lat}} E\,dS=E\int _{S_\mathrm{lat}}dS=E2\pi rL, \end{aligned}$$

(2.44)

where $S_\mathrm{lat}$ is the cylinder lateral surface. Applying Gauss’ law we have:

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}, $$

where $q_\mathrm{in}$’s value depends on the point where the field is calculated, whether it’s inside ($P'$) or outside (P) the charged cylinder. If we calculate the field at an external point P, the entire charged cylinder (with height L, light grey coloured in Fig. 2.11) remains inside the Gaussian surface, and it results

$$ \Phi _{E(r>R_2)}=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _{R_1}^{R_2}\rho dV=\frac{\rho }{\varepsilon _0}\pi (R_2^2-R_1^2)L=\frac{qL}{\varepsilon _0}, $$

If the point $P'$ is inside the charged cylinder, there is a part of the charge outside the Gaussian surface, which produces no flux. Then the only inner charge that remains is the dark grey coloured one in Fig. 2.11, and it results

$$ \Phi _{E(R_1<r<R_2)}=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _{R_1}^{r}\rho dV=\frac{\rho }{\varepsilon _0}\pi (r^2-R_1^2)L. $$

Equating these expressions with (2.44) it results, at external points as P,

$$ E(r>R_2)=\frac{\rho \pi (R_2^2-R_1^2)L}{\varepsilon _0 2\pi rL}=\frac{\rho (R_2^2-R_1^2)}{2 \varepsilon _0 r}, $$

And at internal points as $P'$,

$$ E(R_1<r<R_2)=\frac{\rho \pi (r^2-R_1^2)L}{\varepsilon _0 2\pi rL}=\frac{\rho (r^2-R_1^2)}{2 \varepsilon _0 r}. $$

If we combine both results and we express them using vectors, it results

$$ \displaystyle \mathbf {E}=\left\{ \begin{array}{ccccc} 0&{}&{}&{}\quad &{} r\le R_1,\\ \\ \displaystyle \frac{\rho (r^2-R_1^2)}{2 \varepsilon _0 r}\mathbf {u}_\rho &{}=&{} \displaystyle \frac{q}{2 \pi \varepsilon _0 r}\frac{(r^2-R_1^2)}{(R_2^2-R_1^2)}\mathbf {u}_\rho &{}\quad &{}R_1\le r\le R_2,\\ \ &{} &{} &{} &{} \\ \displaystyle \frac{\rho (R_2^2-R_1^2)}{2 \varepsilon _0 r}\mathbf {u}_\rho &{}=&{}\displaystyle \frac{q}{2 \pi \varepsilon _0 r}\mathbf {u}_\rho &{}\quad &{}r\ge R_2, \end{array}\right. $$

where $\mathbf {u}_\rho$ is the radial unit vector for cylindrical coordinates. The field at any point on the cylinder outside surface can be calculated by using either expressions, and making $r=R_2$.

$$ \mathbf {E}_{(r=R_2)}=\frac{\rho (R_2^2-R_1^2)}{2 \varepsilon _0 R_2}\mathbf {u}_\rho =\frac{q}{2 \pi \varepsilon _0 R_2}\mathbf {u}_\rho . $$

If the cylinder were solid, with radius R, the field can be obtained by replacing $R_1=0$ and $R_2=R$:

$$ \displaystyle \mathbf {E}= \left\{ \begin{array}{ccccc}\displaystyle \frac{\rho r}{2 \varepsilon _0}\mathbf {u}_\rho &{}=&{} \displaystyle \frac{q}{2 \pi \varepsilon _0 r}\frac{r^2}{R^2}\mathbf {u}_\rho &{}\quad &{} r\le R,\\ &{} &{} \\ \displaystyle \frac{\rho R^2}{2 \varepsilon _0 r}\mathbf {u}_\rho &{}=&{}\displaystyle \frac{q}{2 \pi \varepsilon _0 r}\mathbf {u}_\rho &{}\quad &{} r\ge R. \end{array}\right. $$

And at any point on the surface

$$ \mathbf {E}_{(r=R)}=\frac{\rho R}{2 \varepsilon _0}\mathbf {u}_\rho =\frac{q}{2 \pi \varepsilon _0 R}\mathbf {u}_\rho . $$

(b) If the cylinder is metallic (conductor) and in electrostatic balance, charge is only distributed on the external surface and its distribution is uniform.¹⁶ Charge superficial density $\sigma$ can be calculated from (2.4), considering a cylinder with length L:

$$ Q=\int _S \sigma dS= \sigma \int _S dS = \sigma 2\pi R_2 L\Rightarrow \sigma =\frac{Q/L}{2\pi R_2}=\frac{q}{2\pi R_2}. $$

The field at points inside this surface ($r<R_2$) is null, due to the symmetry of the charge distribution. To calculate the field at external points ($r\ge R_2$) we do the same as in section (a). Calculating the flux through the Gaussian surface, the same (2.44) is obtained. Applying Gauss’ law, it results:

$$ \Phi _{E(r>R_2)}=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _S\sigma dS=\frac{\sigma }{\varepsilon _0} S_\mathrm{lat}=\frac{\sigma }{\varepsilon _0}2\pi R_2L=\frac{qL}{\varepsilon _0}, $$

where $S_\mathrm{lat}$ is the lateral surface area of the external cylindrical surface, where the charge is distributed. If we calculate with (2.44) it results

$$ E(r>R_2)=\frac{\sigma 2\pi R_2L}{\varepsilon _0 2\pi rL}=\frac{\sigma R_2}{\varepsilon _0 r}=\frac{q}{\varepsilon _0 2\pi r}. $$

Combining all results,

$$ \displaystyle \mathbf {E}= \left\{ \begin{array}{ccc} 0 &{}\quad &{} r<R_2,\\ \\ \displaystyle \frac{\sigma R_2}{\varepsilon _0 r}\mathbf {u}_\rho =\frac{q}{\varepsilon _0 2\pi r}\mathbf {u}_\rho &{}\quad &{}r>R_2, \end{array}\right. $$

where $\mathbf {u}_\rho$ is the radial unit vector for cylindrical coordinates. The field at a closed point to the external surface of the cylinder can be calculated with $r=R_2$:

$$ \mathbf {E}_{(r=R_2)}=\frac{\sigma }{\varepsilon _0}\mathbf {u}_\rho , $$

which is the known value for the field at points near the conductor surface.

2.7 :

Determine the electric field and the potential produced at any point by a very large plate, with thickness d, on the following cases: (a) non conductor and uniformly charged with density $\rho$; and (b) metallic and with electrostatic balance, with charge density $\sigma$. Particularize these results to the case in which the plate thickness is null (infinite plane). Note: take as the potential reference its central plane.

Solution

(a) Let’s firstly consider the case in which charge is uniformly distributed on the plate. Due to symmetry of the charge distribution, every point at the same distance from the central plane of the plate and far away from the ends has the same electric field value, which is also perpendicular to the plate.¹⁷ The problem can be solved by applying Gauss’ law (2.28). As a Gaussian surface $\partial V$, and due to the symmetry, it can be chosen a straight cylinder (any parallelepiped surface would also be valid), with its axis perpendicular to the plate, with one of its bases passing through point where the field is calculated, and the other base symmetric to the previous one, referring to the central plane of the plate (Fig. 2.12). Two different expressions are obtained for the electric field, depending on the point to be studied, if it is outside or inside the plate (P and $P'$ in Fig. 2.12), since charge inside the Gaussian surface has a different expression.

Fig. 2.12

Gaussian surfaces and field vectors to apply Gauss’ law to a uniformly charged plate

Electric field at any point on the Gaussian surface is perpendicular to the plate, outward if we suppose the plate positively charged, and with the same magnitude at every point of the two cylinder bases. The flux through the lateral surface of either gaussian cylinder $\partial V$ is null, since $\mathbf {E}$ and $d\mathbf {S}$ are perpendicular at any point of this surface. There is only flux through the bases, it is

$$\begin{aligned} \Phi _E=\oint _{\partial V }\mathbf {E}\cdot d\mathbf {S}=\int _{B_\mathrm{upp}}EdS+\int _{B_\mathrm{low}}EdS=E2S. \end{aligned}$$

(2.45)

It should be noticed that S is the area of the upper base $B_\mathrm{upp}$ and that of the lower base $B_\mathrm{low}$.

If we apply Gauss’ law, it is obtained for an external point P,

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0} =\frac{\rho Sh}{\varepsilon _0}. $$

Charge inside the Gaussian surface is only inside the cylinder of height h (marked in light grey in Fig. 2.12). That’s the reason that the charged volume is just Sh. Equating this expression with (2.45), and solving, it results

$$ E_\mathrm{ext}=\frac{\rho h}{2\varepsilon _0}. $$

If we apply Gauss’ law for an internal point $P'$, inside the plate, the flux is

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0} =\frac{\rho 2rS}{\varepsilon _0}\, $$

The cylinder intersection with the plate is the entire cylinder with height 2r (marked in dark grey in Fig. 2.12). Equating this expression with (2.45) and solving, it results

$$ E_\mathrm{in}=\frac{\rho r}{\varepsilon _0}. $$

If it is expressed with vector notation, using the distance r to the central plane of the plate or Cartesian coordinates, it results

$$\begin{aligned} \displaystyle \mathbf {E}= \left\{ \begin{array}{ccc}\displaystyle \frac{\rho r}{\varepsilon _0}\,\frac{\mathbf r}{r}=\frac{\rho r}{\varepsilon _0}\,\text{ sgn }(z)\,\mathbf {u}_z&{}\quad &{} r\le h/2 \quad (|z|\le h/2),\\ \\ \displaystyle \frac{\rho h}{2\varepsilon _0}\,\frac{\mathbf r}{r}=\frac{\rho h}{2\varepsilon _0 }\,\text{ sgn }(z)\,\mathbf {u}_z&{}\quad &{}r\ge h/2 \quad (|z|\ge h/2), \end{array}\right. \end{aligned}$$

(2.46)

where (x, y, z) are the coordinates of field point P and sgn(z) the signum function of z, which indicates that the direction of electric field is downward at points under the central plane of the plate.

To determine the potential at any point P, and due to the fact that the potential reference is on the central plane of the plate, the circulation of the electric field must be calculated from the point P to any point of the central plane of the plate. Since circulation is independent of the chosen path, and any point of the central plane has null potential, we take as the circulation line the perpendicular one from the point to the plane, for which $\mathbf {E}$ and $\displaystyle d\mathbf {l}=dr\frac{\mathbf r}{r}$ are parallel,¹⁸ as shown in Fig. 2.13. If we consider an internal point $P'$ in Fig. 2.13 at a distance r from the central plane of the plate and if O is the point of the central plane perpendicular to $P'$, which has null potential, it results,

$$ V_{P'}=V_{P'}-V_O=\int _{P'}^O \mathbf {E}_\mathrm{in}\cdot d\mathbf {l}=\int _r^0 \displaystyle \frac{\rho }{\varepsilon _0}r dr= \left. \frac{\rho }{2\varepsilon _0}r^2 \right| _r^0=-\frac{\rho r^2}{2\varepsilon _0}. $$

To determine the potential at an external point P, since the electric field expression is different depending on whether the point is inside or outside the plate, it is necessary to circulate $\mathbf {E}$ from P to a point $P_s$ on the surface using the expression for an external field, and then from this point $P_s$ to the centre of the plate, using the expression for the field at internal points. The result is

$$\begin{aligned}V_P=V_P-V_O= & {} \int _P^O \mathbf {E}\cdot d\mathbf {l}=\int _P^{P_s} \mathbf {E}_\mathrm{ext}\cdot d\mathbf {l}+\int _{P_s}^O \mathbf {E}_\mathrm{in}\cdot d\mathbf {l}\\ {}= & {} \int _r^{h/2} \frac{\rho h}{2\varepsilon _0 }\, dr + \int _{h/2}^0 \frac{\rho r}{\varepsilon _0}\,dr=\left. \frac{\rho h}{2\varepsilon _0}\, r \right| _r^{h/2}+\left. \frac{\rho }{2\varepsilon _0}\, r^2 \right| _{h/2}^0\\ {}= & {} \frac{\rho }{2\varepsilon _0}\left( \frac{h^2}{2}- hr-\frac{h^2}{4}\right) =\frac{\rho h}{2\varepsilon _0}\left( \frac{h}{4}- r\right) .\end{aligned}$$

It can be observed that calculated potentials for point P and for point $P'$ are negative, which coincides with the fact that field $\mathbf {E}$ has the direction of decreasing potentials.

Fig. 2.13

Calculation of the potential of a charged plate

(b) If the plate is a conductor, charge is distributed over the lower and upper surfaces, since interior charge in conductors is null. Charge density $\sigma$ on the surfaces must be homogeneous, because if not, field inside the plate wouldn’t be null, as it has to be a balanced conductor. The field at internal points of the plate is therefore null. To calculate the field at external points, symmetry reasonings are the same as the ones in section (a). The Gaussian surface is the same as before, but now the charge is only in the intersection of the Gaussian surface with the plate surfaces (Fig. 2.14). The flux through the surface is obtained as in (2.45),

$$ \Phi _E=E2S. $$

If Gauss’ law is applied the result is

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0} =\frac{\sigma 2S}{\varepsilon _0}. $$

If both expressions are equated, the result is

$$ E_\mathrm{ext}=\frac{\sigma }{\varepsilon _0}, $$

So the field at any external point is constant. If we express it with vectors

$$ \displaystyle \mathbf {E}= \left\{ \begin{array}{ccc}0&{}\quad &{} r\le h/2 \quad (|z|\le h/2),\\ \\ \displaystyle \frac{\sigma }{\varepsilon _0}\,\frac{\mathbf r}{r}=\frac{\sigma }{\varepsilon _0}\,\text{ sgn }(z)\mathbf {u}_z&{}\quad &{}r\ge h/2 \quad (|z|\ge h/2), \end{array}\right. $$

Fig. 2.14

Gaussian surfaces and field vectors to apply Gauss’ law to a conducting plate

To calculate the potential, since the field is null inside the plate, internal points have the same potential (zero) as the ones on the centre of the plate. To calculate the potential at an external point P, it is enough to circulate $\mathbf {E}$ from P to the plate’s surface, since circulation from the surface to the centre of the plate is null. If the circulation path indicated in Fig. 2.13 is followed, the result is

$$ V_P=V_P-V_O=\int _P^O \mathbf {E}\cdot d\mathbf {l}=\int _P^{P_s} \mathbf {E}_\mathrm{ext}\cdot d\mathbf {l}=\int _r^{h/2} \frac{\sigma }{\varepsilon _0}\, dr =\left. \frac{\sigma }{\varepsilon _0}\, r \right| _r^{h/2}=\frac{\sigma }{\varepsilon _0}\left( \frac{h}{2}- r\right) . $$

If the plate has no thickness, the problem is the same as the previous one with the only difference that there aren’t two charged surfaces with density $\sigma$ but only one; so if the same calculus is repeated,

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0} =\frac{\sigma S}{\varepsilon _0}. $$

And if we equate with (2.45) the result is

$$ E=\frac{\sigma }{2\varepsilon _0}. $$

The resulting potential is

$$ V_P=V_P-V_O=\int _P^O \mathbf {E}\cdot d\mathbf {l}=\int _r^0 \frac{\sigma }{2\varepsilon _0}\, dr =-\frac{\sigma }{2\varepsilon _0}r. $$

2.8 :

We have a wire AB with length l and its line charge density is $\lambda _1=\lambda (1+kx)$, where x is the distance of a point of the wire to the central point M of segment ${ {OA}}$ (Fig. 2.15), and $\lambda$ and k are two known constants. Perpendicularly to this wire at a distance a of its extreme A, an infinite wire with line charge density $\lambda _2=\lambda$ is placed. Determine the electric field at point M.

Fig. 2.15

Figure of Problem 2.8

Solution

Fig. 2.16

Field produced by the finite wire

To solve the problem, the superposition principle for electric fields is applied: fields at point M is the addition of the fields produced by both wires at that point. To calculate the field of wire AB we consider Fig. 2.16, where distance from any element of charge $dq'=\lambda _1 dx$ to point M is expressed by variable x. Field $d\mathbf {E}_1$ produced at point M by the differential element of charge $dq'$ is given by (2.15):

$$\begin{aligned}d\mathbf {E}_1= & {} \frac{dq'}{4\pi \varepsilon _0 d^2}\mathbf {u}=\frac{1}{4\pi \varepsilon _0}\frac{\lambda _1 dx}{x^2}(-\mathbf {u}_x)=\\ {}= & {} \frac{1}{4\pi \varepsilon _0}\frac{\lambda (1+kx)dx}{x^2}(-\mathbf {u}_x), \end{aligned}$$

To calculate total field due to wire AB, the superposition principle (2.16) is applied:

$$\begin{aligned}\mathbf {E}_1= & {} \int _{a/2}^{l+a/2} \frac{\lambda }{4\pi \varepsilon _0}\frac{(1+kx)dx}{x^2}(-\mathbf {u}_x)= \frac{\lambda }{4\pi \varepsilon _0}\left( \int _{a/2}^{l+a/2}\frac{1}{x^2}dx+\int _{a/2}^{l+a/2}\frac{k}{x}dx\right) (-\mathbf {u}_x)=\\ {}= & {} \frac{\lambda }{4\pi \varepsilon _0}\left. \left( \frac{-1}{x}+k\ln x\right) \right| _{a/2}^{l+a/2}(-\mathbf {u}_x)=\frac{\lambda }{4\pi \varepsilon _0}\left[ \frac{2}{a}-\frac{2}{a+2l}+k\ln \frac{l+a/2}{a/2}\right] (-\mathbf {u}_x)=\\= & {} \frac{\lambda }{2\pi \varepsilon _0}\left[ -\frac{1}{a}+\frac{1}{a+2l}-\frac{k}{2}\ln \frac{a+2l}{a}\right] \mathbf {u}_x.\end{aligned}$$

It should be observed that the limits of the integral are the ends of the charged wire.

To calculate the field produced by the infinite wire with density $\lambda _2$ the problem’s 2.3 result is applied,

$$ \mathbf {E}_2=\frac{\lambda _2}{2\pi \varepsilon _0 r}\mathbf {u}_\rho =\frac{\lambda }{2\pi \varepsilon _0 a/2}\mathbf {u}_x=\frac{\lambda }{\pi \varepsilon _0 a}\mathbf {u}_x, $$

where distance from the wire to the point field is a / 2 and the radial unit vector is, in this case, $\mathbf {u}_x$.

The electric field at point M, applying superposition principle, is

$$ \mathbf {E}=\mathbf {E}_1+\mathbf {E}_2=\frac{\lambda }{2\pi \varepsilon _0}\left( \frac{1}{a}+\frac{1}{a+2l}-\frac{k}{2}\ln \frac{a+2l}{a}\right) \mathbf {u}_x. $$

2.9 :

Two straight conductors, parallel and infinite, with respective density charge $\lambda _1=\lambda$ and $\lambda _2=-2\lambda$ are separated by a distance d. Calculate the potential difference between points A and B in Fig. 2.17.

Fig. 2.17

Figure of Problem 2.9

Solution

To calculate the potential difference between A and B, $V_A-V_B$, it is necessary to know the electrostatic field at every point in a line between these points. For this, the superposition principle is applied, and the resulting field at each point is the addition of the fields produced by each wire independently. From Problem 2.3 it is known that electric field produced by an infinite line is perpendicular to this line, and with the same magnitude at every point of a cylindrical surface whose axis is the line. The electric field at P (Fig. 2.18) is given by (2.36), which for the line of density $\lambda _1$ is

$$ \mathbf {E}_1=\frac{\lambda _1}{2\pi \varepsilon _0 r_1}\frac{\mathbf {r}_1}{r_1}=\frac{\lambda }{2\pi \varepsilon _0 r_1}\frac{\mathbf {r}_1}{r_1}. $$

The field produced by the line of density $\lambda _2$ is

$$ \mathbf {E}_2=\frac{\lambda _2}{2\pi \varepsilon _0 r_2}\frac{\mathbf {r}_2}{r_2}=\frac{-2\lambda }{2\pi \varepsilon _0 r_2}\frac{\mathbf {r}_2}{r_2}=\frac{\lambda }{\pi \varepsilon _0 (d-r_1)}\frac{\mathbf {r}_1}{r_1}. $$

Adding both fields, the total electric field at any point P is obtained

$$ \mathbf {E}=\mathbf {E}_1+\mathbf {E}_2=\frac{\lambda }{2\pi \varepsilon _0}\left( \frac{1}{r_1}+\frac{2}{d-r_1}\right) \frac{\mathbf {r}_1}{r_1}. $$

To calculate the potential difference between A and B, a circulation of $\mathbf {E}$ from A to B must be done. A line from A to $B'$ has been taken (Fig. 2.19), where $\mathbf {E}$ and $d\mathbf r$ are parallel, and then from $B'$ to B where they are perpendicular, and circulation null.¹⁹ Applying (2.22),

$$\begin{aligned}V_A-V_B= & {} \int _{r_A}^{r_B} \mathbf E\cdot d\mathbf r= \\ {}= & {} \int _{d/4}^{3d/4} \frac{\lambda }{2\pi \varepsilon _0}\left( \frac{1}{r}+\frac{2}{d-r}\right) dr= \\ {}= & {} \frac{\lambda }{2\pi \varepsilon _0}\left( \left. \ln r\right| _{d/4}^{3d/4}-\left. 2\ln (d-r)\right| _{d/4}^{3d/4}\right) =\frac{\lambda }{2\pi \varepsilon _0}3\ln 3.\end{aligned}$$

2.10 :

We have an isolated spherical conductor whose radius is $R_1=4\,$cm, and whose potential is $9000\,$V referring to ground. After, it is surrounded with a concentric spherical conducting layer, with inner radius $R_2=8\,$cm and exterior one $R_3=10\,$cm, isolated and with null total charge. Determine charges and potentials on the inner conductor, as well as the conducting layer, for the following cases: (a) Inner conductor and conducting layer isolated. (b) If the conducting layer is connected to ground. (c) If the layer is once again isolated and the conductor is connected to ground by a conducting wire that goes through a small hole in the layer.

Fig. 2.18

Fields produced by the two infinite wires

Fig. 2.19

Scheme to calculate the potential difference between A and B

Solution

Fig. 2.20

Isolated conducting sphere and isolated conducting layer

(a) Note the charge distribution is not known. The isolated spherical conductor will have certain charge $q_1$, since its potential is not zero. If the expression of potential for a spherical conductor is applied ((2.43) of Problem 2.5), it results

$$ V=\frac{q_1}{4\pi \varepsilon _0 R_1}. $$

From where conductor’s charge $q_1$ is obtained,

$$ q_1=4\pi \varepsilon _0 R_1 V=40\,\mathrm {nC}. $$

If the conductor is surrounded by the conducting layer, Fig. 2.20, charges for both conductors are reorganized, until the balance is reached, all properties in Sect. 2.8 are verified. Since conductor A is isolated, charge $q_1$ remains, and this charge can only be on its outsider surface. Due to the spherical symmetry of the figure, it will be uniformly distributed over the surface, and thus the electric field inside the conductor is null. On the conducting layer B, charges are distributed so that the field inside it is null. If Gauss’ law (2.28) is applied to a Gaussian surface $\partial V$ totally inside the conductor (discontinuous line in the figure),

$$ \Phi _E=\oint _{\partial V} \mathbf {E}\cdot d\mathbf {S} =0=\frac{q_\mathrm{in}}{\varepsilon _0} \Rightarrow q_\mathrm{in}=0=q_1+q_\mathrm{B,in}, $$

where $q_\mathrm{B, in}$ is the charge of the inner surface of the conducting layer B. It should be observed that the flux is null because the field at every point of $\partial V$ is zero. Therefore

$$ q_\mathrm{B,in}=-q_1. $$

If the principle of conservation of charge is applied to conductor B, the external surface charge of B is

$$ q_B=0=-q_1+q_\mathrm{B,ex}\Rightarrow q_\mathrm{B,ex}=+q_1. $$

The charge distribution is the one shown in Fig. 2.20.

Potential at any point is derived from the superposition of potentials created by each of the charge distributions. For every distribution, (2.43) obtained in Problem 2.5 is applied. For conductor A, the distance r from any interior point P to the centre is lower (or equal) than the distance from the centre to the charge distributions ($r\le R_1, r<R_2, r<R_3$), so it results

$$ V_A=\frac{1}{4\pi \varepsilon _0 }\left( \frac{q_1}{R_1}-\frac{q_1}{R_2}+\frac{q_1}{R_3}\right) =9\cdot 10^9 \cdot 40 \cdot 10^{-9} \left( \frac{1}{0.04}-\frac{1}{0.08}+\frac{1}{0.1}\right) =8100\,\mathrm{V}. $$

For conductor B, the distance $r'$ from any interior point $P'$ to the centre results in $r'>R_1, r'\ge R_2, r'\le R_3$, so if (2.43) is applied,

$$ V_B=\frac{1}{4\pi \varepsilon _0 }\left( \frac{q_1}{r'}-\frac{q_1}{r'}+\frac{q_1}{R_3}\right) =9\cdot 10^9 \cdot \frac{40 \cdot 10^{-9}}{0.1}=3600\,\mathrm{V}. $$

Fig. 2.21

Isolated conducting sphere and grounded conducting layer

(b) Connecting the conducting layer to ground, Fig. 2.21, we equalize the potentials of conductor B and the ground. Ground potential is taken as a reference ($0\,$V) and, therefore,

$$ V_B=0. $$

Since there is no potential difference between B and ground, there cannot exist any electric field between them (due to (2.18), $\mathbf {E}=-\nabla V$). The field is null in B, as well as outside the conductor B. From (2.29) ($\nabla \cdot \mathbf {E}=\rho /\varepsilon _0$), it is obtained that on the outside surface of conductor B there cannot be charges. The other charge distributions remain the same, as seen by reapplying the reasoning from section (a). It can be observed how grounded conductor B does not have a null charge (it is not an isolated system), but it remains negatively charged. If (2.43) is applied, the result for conductor A is,

$$ V_A=\frac{1}{4\pi \varepsilon _0 }\left( \frac{q_1}{R_1}-\frac{q_1}{R_2}\right) =9\cdot 10^9 \cdot 40 \cdot 10^{-9} \left( \frac{1}{0.04}-\frac{1}{0.08}\right) =4500\,\mathrm{V}. $$

A device like this, a grounded conductor which surrounds another one, is the base of electrostatic shields and it is called a Faraday cage. Even though charge or potential inside it is changed, the electric field and potential outside it is always zero. Also, any external electric field would affect neither the electric field nor the potential of the conductors.

(c) If conducting layer B is disconnected from ground, its charge, $-q_1=-40\,$nC, remains and distributes between the inner and outsider surface of B,

$$ -q_1=q_2'+q_3', $$

as seen in Fig. 2.22.

Fig. 2.22

Grounded conducting sphere and isolated conducting layer

Conductor A does not keep its charge anymore, since it is grounded. New charge is called $q_1'$. Applying the reasoning from section (a), the charge on the inner surface of conductor B will be

$$ q_2'=-q_1'. $$

It is also known that conductor A potential is zero, since it is grounded. If (2.43) is applied to conductor A the result is

$$ V_A=0=\frac{1}{4\pi \varepsilon _0 }\left( \frac{q_1'}{R_1}+\frac{q_2'}{R_2}+\frac{q_3'}{R_3}\right) \Rightarrow \frac{q_1'}{0.04}+\frac{q_2'}{0.08}+\frac{q_3'}{0.1}=0. $$

With these three equations the new values of the charges are obtained,

$$ q_1'=\frac{4}{9}\,q_1=17.8\,\text {nC},\quad q_2'=-\frac{4}{9}\,q_1=-17.8\,\text {nC},\quad q_3'=-\frac{5}{9}\,q_1=-22.2\,\text {nC}. $$

If (2.43) is applied we obtain conductor B potential,

$$ V_B=\frac{1}{4\pi \varepsilon _0 }\left( \frac{q_1'}{r'}+\frac{q_2'}{r'}+\frac{q_3'}{R_3}\right) =9\cdot 10^9 \cdot \frac{-22.2 \cdot 10^{-9}}{0.1}=-2000\,\mathrm{V}. $$

2.11 :

Consider two coaxial conductor cylindrical surfaces, A and B, with infinite length, whose radii are a and b. Outer conductor B is grounded and potential of inner conductor A is $V_a$. The space between both conductors and outside of conductor B is a vacuum. (a) Calculate surface charge densities on both conductors. (b) If P is a point between A and B, and $P'$ is outside of conductor B, calculate potential difference $V_P-V_P'$.

Solution

Fig. 2.23

Set of two coaxial hollow cylindrical conductors, the exterior one connected to ground

(a) Figure 2.23 shows both cylinders. Firstly, since charges can freely move inside conductors, we must study how charges distribute inside both conductors. As it was seen in Sect. 2.8, charges distribute on conductor surfaces, so the field inside them is zero. Due to the symmetry, charge has to be distributed uniformly on the surface. The entire charge of conductor A is distributed on its external surface.²⁰ Let’s suppose that $q_a$ is the charge on conductor A for a finite length L. If Gauss’ law for a coaxial cylindrical surface $\partial V$ totally inside conductor B, with length L is applied, the result is, following the same reasoning as Problem 2.10,

$$ \Phi _E=\oint _{\partial V} \mathbf {E}\cdot d\mathbf {S} =0=\frac{q_\mathrm{in}}{\varepsilon _0} \Rightarrow q_\mathrm{in}=0=q_a+q_\mathrm{B,in}, $$

where $q_\mathrm{B,in}$ is the charge on inner surface of conductor B. It should be observed that the flux is null since the field on every point of the considered Gaussian surface is zero. Therefore

$$ q_\mathrm{B,in}=-q_a. $$

As conductor B is grounded, its potential is zero. The electric field outside B is zero and charge on the outside surface of the cylinder is also zero, as it was previously explained on Problem 2.10.

To obtain charge densities, we find the electric field’s expression at any point P between both conductors. Due to the cylindrical symmetry, the problem can be solved by applying Gauss’ law (2.28) in a similar way as it was done in Problem 2.6. To do it, we take as a cylindrical Gaussian surface $\partial V$, with any length L, with the same axis as the cylinder, and with radius r so that the surfaces passes through point P where we want to calculate the field (discontinuous line in Fig. 2.23). The flux through the bases of the Gaussian surface is zero, since $d\mathbf {S}$ and $\mathbf {E}$ are perpendicular at any point of the bases. So it only remains to calculate the flux through the lateral surface, where $d\mathbf {S}$ and $\mathbf {E}$ are parallel:

$$ \Phi _E=\oint _{\partial V}\mathbf {E}\cdot d\mathbf {S} =\int _{S_\mathrm{lat}} E\,dS=E\int _{S_\mathrm{lat}}dS=E2\pi rL, $$

where $S_\mathrm{lat}$ is the cylinder lateral surface. On the other side, if we apply Gauss’ law, we obtain

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{\sigma _a 2\pi a L}{\varepsilon _0}, $$

where $\sigma _a$ is charge density of cylinder A. It should be observed that $q_\mathrm{in}$ is the charge inside the Gaussian surface, and that there’s only charge on the lateral surface of cylinder A, whose radius is a. If both expressions are equalized, electric field is obtained,

$$ E=\frac{\sigma _a a}{\varepsilon _0 r}\Rightarrow \mathbf {E}=\frac{\sigma _a a}{\varepsilon _0 r}\mathbf {u}_\rho , $$

where $\mathbf {u}_\rho$ is the radial unit vector for cylindrical coordinates. As additional information, the potential of both conductors is known. The potential difference $V_A-V_B$ between them is $V_a-0=V_a$, since B is grounded. If we circulate the electric field by following a line perpendicular to the cylinder axis, so that $\mathbf {E}$ and $d\mathbf {l}$ are parallel, the result is

$$ V_A-V_B=V_a=\int _a^b \mathbf {E}\cdot d\mathbf {l}= \int _a^b \frac{\sigma _a a}{\varepsilon _0 r} dr= \frac{\sigma _a a}{\varepsilon _0} \ln \frac{b}{a}, $$

from which charge density $\sigma _a$ of conductor A can be calculated,

$$ \sigma _a=\frac{V_a \varepsilon _0}{a\ln \frac{b}{a}}. $$

On a piece of cylinder with length L, A’s charge will be $q_a=\sigma _a 2\pi a L$. As charge at inner surface of conductor B is $-q_a$, it results that its density $\sigma _b$ can be obtained from

$$ \sigma _a 2\pi a L=-\sigma _b 2\pi b L \Rightarrow \sigma _b=-\frac{a}{b}\sigma _a=-\frac{V_a \varepsilon _0}{b\ln \frac{b}{a}}. $$

The outer surface of conductor B mentioned previously is not charged due to the fact that it is grounded.

(b) To calculate the potential between points P and $P'$ in Fig. 2.23, at a distance r and $r'$ from the axis, we circulate the electric field between both points,²¹

$$ V_P-V_{P'}=V_P=\int _r^{r'} \mathbf {E}\cdot d\mathbf {l}= \int _r^b \frac{\sigma _a a}{\varepsilon _0 r} dr= \frac{\sigma _a a}{\varepsilon _0} \ln \frac{b}{r}=\frac{\ln \frac{b}{r}}{\ln \frac{b}{a}}V_a. $$

It should be observed that $P'$’s potential is null: it has ground potential since there is no electric field outside of conductor B. This is also the reason to use b as the superior limit of the integral, and not $r'$: between b and $r'$, the electric field is zero.

Problems C

2.12 :

Determine the field produced at the coordinate’s origin by a circular-shaped arc wire with radius R in Fig. 2.24, symmetrically placed with respect to the X axis, and charged with positive charge density $\lambda =k|\sin \theta |$, where $\theta$ is the angle with the horizontal of the position vector of any differential element in the wire, and k is a constant.

Fig. 2.24

Figure of Problem 2.12

Solution

The wire in the problem has a symmetric charge with respect to the X axis. If any charge element dq is taken (Fig. 2.25), field $d\mathbf {E}$ produced by this element of charge has the same magnitude, and forms the same angle $\theta$ with X axis as the field produced by its symmetric $dq'$ in reference to this axis. Components parallel to the Y axis of these fields cancel; components along the X axis will be added. The total field produced by a wire has, therefore, a null vertical component, while a horizontal component is twice the one produced by the piece of wire placed in the first quadrant.

Considering the field produced by element $dq=\lambda dl$, which according to (2.15) is

$$ d\mathbf {E}=\frac{\lambda dl}{4 \pi \varepsilon _{0}|\mathbf {r}-\mathbf {r\,'}|^{2}}\, \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|}=\frac{k|\sin \theta |dl}{4 \pi \varepsilon _{0}R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y), $$

where $\mathbf {r}=0$ is field point O position and $\mathbf {r\,'}=R\cos \theta \mathbf {u}_x+R\sin \theta \mathbf {u}_y$ is the source point dq position.

The total field produced by the wire is

$$\begin{aligned}\mathbf {E}= & {} \int d\mathbf {E}=\int _L \frac{k|\sin \theta |dl}{4 \pi \varepsilon _{0}R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y)= \\ {}= & {} \int _{L+} \frac{k\sin \theta dl}{4 \pi \varepsilon _{0}R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y)+\int _{L-} \frac{k(-\sin \theta )dl}{4 \pi \varepsilon _0 R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y), \end{aligned}$$

where $L+$ indicates the piece of wire in the first quadrant and $L-$ indicates the piece of wire in the fourth quadrant. Length element dl, arch of circle, can be expressed as a function of angle $\theta ,\, dl=R d\theta$, and the integral results in²²

$$\begin{aligned}\mathbf {E}&=\int _0^{\frac{\pi }{6}} \frac{k\sin \theta R d\theta }{4 \pi \varepsilon _0 R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y)+\int _{-\frac{\pi }{6}}^0 \frac{k(-\sin \theta )R d\theta }{4 \pi \varepsilon _0 R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y)=\\&=-\frac{k}{4\pi \varepsilon _0 R}\left[ \left( \int _0^{\frac{\pi }{6}} \sin \theta \cos \theta d\theta -\int _{-\frac{\pi }{6}}^0 \sin \theta \cos \theta d\theta \right) \mathbf {u}_x\right. \\&\qquad \qquad \quad \left. +\left( \int _0^{\frac{\pi }{6}} \sin ^2\theta d\theta -\int _{-\frac{\pi }{6}}^0 \sin ^2\theta d\theta \right) \mathbf {u}_y\right] =\\&=-\frac{k}{4\pi \varepsilon _0 R}\left[ \left( \left. \frac{\sin ^2\theta }{2}\right| _0^{\frac{\pi }{6}}-\left. \frac{\sin ^2\theta }{2}\right| _{-\frac{\pi }{6}}^0\right) \mathbf {u}_x\right. \\&\qquad \qquad \quad \left. +\left( \left. \left( \frac{2\theta -\sin 2\theta }{4}\right) \right| _0^{\frac{\pi }{6}}-\left. \left( \frac{2\theta -\sin 2\theta }{4}\right) \right| _{-\frac{\pi }{6}}^0\right) \mathbf {u}_y\right] \\&=-\frac{k}{4\pi \varepsilon _0 R}\left[ \left( \frac{1}{8}-0-0+\frac{1}{8}\right) \mathbf {u}_x+\frac{1}{4}\left( \pi /3-\sin (\pi /3)-\pi /3-\sin (-\pi /3)\right) \mathbf {u}_y\right] \\&=-\frac{k}{16\pi \varepsilon _0 R}\mathbf {u}_x. \end{aligned}$$

Fig. 2.25

Field produced at coordinate’s origin by the circular-shaped wire

The same results can be reached bearing in mind the previous symmetry considerations,

$$\begin{aligned}\mathbf {E}&=\int _L \frac{k|\sin \theta |dl}{4 \pi \varepsilon _{0}R^2}(-\cos \theta \mathbf {u}_x-\sin \theta \mathbf {u}_y)=2\int _{L+} \frac{k\sin \theta dl}{4 \pi \varepsilon _{0}R^2}(-\cos \theta \mathbf {u}_x)\\&=-2\int _0^{\frac{\pi }{6}} \frac{k\sin \theta \cos \theta R d\theta }{4 \pi \varepsilon _0 R^2}\mathbf {u}_x=\\&=-\frac{k}{2\pi \varepsilon _0 R}\int _0^{\frac{\pi }{6}}\sin \theta \cos \theta d\theta \mathbf {u}_x=-\frac{k}{2\pi \varepsilon _0 R}\left. \frac{\sin ^2\theta }{2}\right| _0^{\frac{\pi }{6}}\mathbf {u}_x\\&=-\frac{k}{2\pi \varepsilon _0 R}\left( \frac{1}{8}-0\right) \mathbf {u}_x=-\frac{k}{16\pi \varepsilon _0 R}\mathbf {u}_x. \end{aligned}$$

2.13 :

The spherical crown in Fig. 2.26 (sectioned by plane XY) is shown, whose centre is at point $C(1\,$m, 0, 0), with inner radius $R_i=20\,$cm and exterior one $R_e=50\,$cm, has a non homogeneous charge density $\rho =k/r$ with $k=2\,\upmu$C/m$^2$ and r the distance to the crown’s centre measured in meters. The wire in the figure, in XY plane, is infinitely long, makes $45^{\circ }$ with X axis, and has a charge per unit of length $\lambda =30\,$nC/m. Determine the electric field produced at point $A(1\,$m, $-1\,$m, 0).

Fig. 2.26

Figure of Problem 2.13

Solution

To calculate the field at point A, the superposition principle is applied, adding the fields produced by the spherical crown and the wire at that point. The field produced by each of the distributions can be obtained by applying Gauss’ law (2.28).

For the spherical crown, the same procedure as Problem 2.5 is applied: we take a spherical Gaussian surface, concentric with the charged crown, that passes over point A on the field to be calculated (Fig. 2.27). Field $\mathbf {E}_\rho$ is radial and to determine its magnitude, the flux through this Gaussian surface is calculated

$$ \Phi _{E_\rho }=\oint _{\partial V}\mathbf {E}_\rho \cdot d\mathbf {S}=E_\rho 4\pi r^2, $$

If Gauss’ theorem is applied,

$$ \Phi _{E_\rho }=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _{V_\mathrm{in}} \rho dV=\frac{1}{\varepsilon _0}\int _{V_\mathrm{in}} \frac{k}{r} 4\pi r^2dr=\frac{4\pi k}{\varepsilon _0}\int _{R_i}^{R_e} rdr = \frac{4\pi k}{\varepsilon _0}\frac{R_e^2-R_i^2}{2}, $$

since the volume element in a sphere can be written as $dV=4\pi r^2dr$. Equating both flux calculations and taking the data, results are

$$ E_\rho =\frac{0.105 k}{\varepsilon _0}=23729\,\text {N/C}. $$

And using vector notation, if Fig. 2.27 is observed:

$$ \mathbf {E}_\rho =-23729\,\mathbf {u}_y\,\text {N/C}. $$

Fig. 2.27

Fields produced by the charged distributions and Gaussian surfaces

For the wire, the procedure of Problem 2.4 is applied: a cylindrical Gaussian surface is drawn in Fig. 2.27, whose axis is the wire, with any length L and passing over point A. Field $\mathbf {E}_\lambda$ is perpendicular to the wire, and to obtain its magnitude, the flux through the surface is calculated,

$$ \Phi _{E_\lambda }=\oint _{\partial V}\mathbf {E}_\lambda \cdot d\mathbf {S} =\int _{S_\mathrm{lat}} E_\lambda \,dS=\left. E_\lambda 2\pi rL\right| _{r=\sqrt{2}\,\text {m}}, $$

since the distance from the wire to A, the radius of the cylindrical surface, is $\sqrt{2}\,$ meters long. And applying Gauss’ theorem,

$$ \Phi _{E_\lambda }=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{\lambda L}{\varepsilon _0}. $$

Equating the two calculations of the flux, results are

$$ E_\lambda =\frac{\lambda }{2\pi \varepsilon _0 r}=\frac{30\cdot 10^{-9}}{2\pi \varepsilon _0 \sqrt{2}}=381.5\,\text {N/C}. $$

And using vector notation

$$ \mathbf {E}_\lambda =381.5\,\frac{\mathbf {u}_x-\mathbf {u}_y}{\sqrt{2}}=270\,(\mathbf {u}_x-\mathbf {u}_y)\,\text {N/C}. $$

The total field is obtained by adding both fields (vectorially):

$$ \mathbf {E}=\mathbf {E}_\rho +\mathbf {E}_\lambda =(270\,\mathbf {u}_x-23999\,\mathbf {u}_y)\,\text {N/C}. $$

2.14 :

The space region defined by equation $0<z<2$ (with Cartesian coordinates in meters) has a charge density $\rho =k|z-1|$ with $k=8\,\upmu$C/m$^4$. (a) Determine the electric field value for the points in the region defined by the sphere with its centre at the coordinates’ origin and radius $2\,$m. (b) Determine the divergence value of the electric field at the previous points. (c) Determine the electric field flux through the previous sphere surface.

Solution

(a) Figure 2.28 represents the region of the problem. Charge distribution is symmetric in reference to the center plane $z=1$, so Gauss’ law (2.28) can be easily applied. A similar procedure as the one used for the infinite plate on Problem 2.7 is followed. Now the points at the field to be calculated are a few specific points defined by the sphere indicated in the statement and light grey coloured in the figure. Points in the higher hemisphere are all the inside points of the charged plate, while points in the lower hemisphere are outside the plate. If Gauss’ law is applied as in Problem 2.7, with cylindrical Gaussian surfaces indicated in Fig. 2.28 (which are the ones used before)

Fig. 2.28

Figure of Problem 2.14

$$ \Phi _E=\oint _{\partial V }\mathbf {E}\cdot d\mathbf {S}=\int _{B_\mathrm{upp}}EdS+\int _{B_\mathrm{low}}EdS=E2S. $$

To calculate the charge inside the Gaussian surface, it should be taken into account that the charge density is variable. As the charge is symmetric in reference to the medium plane, we will determine the higher half charge, where $|z-1|=z-1$, and we will double the resulting charge. For points outside of the charged region, as point P, it results

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _{V_\mathrm{in}}\rho dV =\frac{1}{\varepsilon _0}2\int _1^2 k(z-1)Sdz=\frac{2kS}{\varepsilon _0}\left. \frac{(z-1)^2}{2}\right| _1^2=\frac{kS}{\varepsilon _0}. $$

To calculate the integral we have taken a volume element $dV=Sdz$ (in dark grey in the figure), that represents a differential cylinder whose area is S and its height is dz and it is located at any distance from the medium plane. If we compare both flux expressions, the results are

$$ E_\mathrm{ext}=\frac{k}{2\varepsilon _0}=\frac{8\cdot 10^{-6}}{2\varepsilon _0}=4.52\cdot 10^5 \,\text {NC}^{-1}, $$

If Gauss’ law is applied to an interior point $P'$, it is obtained

$$ \Phi _E=\frac{q_\mathrm{in}}{\varepsilon _0} =\frac{1}{\varepsilon _0}2\int _1^z k(z-1)Sdz=\frac{2kS}{\varepsilon _0}\left. \frac{(z-1)^2}{2}\right| _1^z=\frac{kS}{\varepsilon _0} (z-1)^2. $$

where z is the coordinate in meters of point $P'$. Comparing both flux expressions, results are

$$ E_\mathrm{in}=\frac{k}{2\varepsilon _0}(z-1)^2=4.52\cdot 10^5\,(z-1)^2\,\text {NC}^{-1}. $$

Vectorially, at points over the center plane ($z=1$) the sense of field vector is $\mathbf {u}_z$ and for the ones below it is $-\mathbf {u}_z$.

Every point in the lower hemisphere ($z<0$) is exterior to the charged region, and since they are below the center plane, the sense of the vector field $\mathbf {E}$ is the negative Z-axis, as can be observed in Fig. 2.28. Points in the higher hemisphere are all inside the plate. The vector field has the orientation parallel to the Z-axis in the positive direction at points above the center plane, and it has the opposite orientation at points below the medium plane. If we express the result using vector notation and Cartesian coordinates, it results, for the points in the sphere $x^2+y^2+z^2\le 4$:

$$ \displaystyle \mathbf {E}= \left\{ \begin{array}{ccc}-4.52\cdot 10^5\,\mathbf {u}_z\,\text {NC}^{-1}&{}\quad &{} z<0 ,\\ &{}&{}\\ -4.52\cdot 10^5\,(z-1)^2\,\mathbf {u}_z\,\text {NC}^{-1}&{}\quad &{} 0\le z\le 1,\\ &{}&{}\\ 4.52\cdot 10^5\,(z-1)^2\,\mathbf {u}_z\,\text {NC}^{-1}&{}\quad &{} 1\le z\le 2, \end{array}\right. $$

with z expressed in meters.

(b) From (2.29) the result is

$$ \nabla \cdot \mathbf {E}(\mathbf r)=\frac{\rho (\mathbf r)}{\varepsilon _0}=\left\{ \begin{array}{ccc}0&{}\quad &{} x^2+y^2+z^2\le 4 \text{ and } z<0 \,,\\ &{}&{}\\ \displaystyle \frac{k|z-1|}{\varepsilon _0}=9.04\cdot 10^5|z-1|\,\text{ NC }^{-1}\text{ m }^{-1}&{}\quad &{} x^2+y^2+z^2\le 4 \text{ and } z\ge 0. \end{array}\right. $$

It can be observed that the lower hemisphere points are outside of the charged zone ($\rho =0$) and the electric field’s divergence at these points is null.

(c) If Gauss’ law (2.28) is applied to the surface defined by the sphere it results

$$ \Phi _E=\oint _{\partial V} \mathbf {E}\cdot d\mathbf {S} =\frac{q_\mathrm{in}}{\varepsilon _0}=\frac{1}{\varepsilon _0}\int _{V_\mathrm{sph}} \rho dV=\frac{1}{\varepsilon _0}\int _{V_\mathrm{upsph}} k|z-1| dV, $$

where $V_\mathrm{upsph}$ is the upper hemisphere’s volume. To calculate the integral we use spherical coordinates. It can be observed that r varies from 0 to 2 (sphere’s radius), coordinate $\phi$ goes from 0 to $2\pi$ because the whole circumference is described, and coordinate $\theta$ goes from 0 to $\pi /2$, since only the upper hemisphere is charged:

$$ \Phi _E=\frac{1}{\varepsilon _0}\int _{V_\mathrm{uppsph}} k|z-1| dV=\frac{1}{\varepsilon _0}\int _0^R\int _0^{\pi /2}\int _0^{2\pi }k|r\cos \theta -1| r^2\sin \theta dr d\theta d\phi , $$

where we have taken into account that $z=r\cos \theta$. It should be observed that function $|z-1|=|r\cos \theta -1|$ has a different value depending on z, whether $z>1$ or $z<1$. Where $z=1$ and the sphere’s radius $R=2$, results are $\cos \theta =1/2\rightarrow \theta =\pi /3$. Using two integrals to substitute the absolute value expression and solving them, results are

$$\begin{aligned}\Phi _E&=\frac{k}{\varepsilon _0}\left( \int _0^2\int _0^{\pi /3}\int _0^{2\pi }(r\cos \theta -1) r^2\sin \theta dr d\theta d\phi \right. \\&\qquad \,\,-\left. \int _0^2\int _{\pi /3}^{\pi /2}\int _0^{2\pi }(r\cos \theta -1)r^2\sin \theta dr d\theta d\phi \right) =\\&=\frac{2\pi k}{\varepsilon _0}\left( \int _0^2\int _0^{\pi /3}(r\cos \theta -1) r^2\sin \theta dr d\theta -\int _0^2\int _{\pi /3}^{\pi /2}(r\cos \theta -1) r^2\sin \theta dr d\theta \right) =\\&=\frac{2\pi k}{\varepsilon _0}\left( \int _0^{\pi /3}\left( 4\cos \theta \sin \theta -\frac{8}{3}\sin \theta \right) d\theta - \int _{\pi /3}^{\pi /2}\left( 4\cos \theta \sin \theta -\frac{8}{3}\sin \theta \right) d\theta \right) =\\&=\frac{2\pi k}{\varepsilon _0}\left( 4\frac{\sin ^2\theta }{2}\left. \right| _0^{\pi /3}+\frac{8}{3}\left. \cos \theta \right| _0^{\pi /3}- 4\frac{\sin ^2\theta }{2}\left. \right| _{\pi /3}^{\pi /2}-\frac{8}{3}\left. \cos \theta \right| _{\pi /3}^{\pi /2}\right) \\&=\frac{2\pi k}{\varepsilon _0}\left( \frac{3}{2}-\frac{4}{3}-\frac{1}{2}+\frac{4}{3}\right) . \end{aligned}$$

The flux obtained is

$$ \Phi _E=\frac{2\pi k}{\varepsilon _0}=5.68\cdot 10^6\text{ Nm }^2\text{ C }^{-1}. $$

2.15 :

Calculate the electric field produced at point P in Fig. 2.29 by the cylinder with volumetric uniform density $\rho$, whose height is H, inner radius $R_i$ and exterior one $R_e$.

Solution

The problem can be solved by direct integration using cylindrical coordinates. Any charge element dq (Fig. 2.30), at $\mathbf {r\,'}=r\cos \phi \mathbf {u}_x+r\sin \phi \mathbf {u}_y+z\mathbf {u}_z$ is expressed by using cylindrical coordinates as follows

$$ dq=\rho dV=\rho r dr d\phi dz. $$

The field produced by this element at point P, located at $\mathbf r=h\mathbf {u}_z$, is

$$\begin{aligned}d\mathbf {E}= & {} \frac{dq}{4 \pi \varepsilon _0}\; \frac{\mathbf {r}-\mathbf {r\,'}}{|\mathbf {r}-\mathbf {r\,'}|^3}=\\ {}= & {} \frac{\rho }{4 \pi \varepsilon _0}\;\frac{-r\cos \phi \mathbf {u}_x-r\sin \phi \mathbf {u}_y+(h-z)\mathbf {u}_z}{[r^2+(h-z)^2]^{3/2}}r dr d\phi dz. \end{aligned}$$

Fig. 2.29

Figure of Problem 2.15

Fig. 2.30

Charge differentials for a finite cylinder

If the superposition principle is applied

$$ \mathbf {E}= \int _{R_i}^{R_e}\!\!\int _0^{2\pi }\!\!\int _0^H\frac{\rho }{4 \pi \varepsilon _0} \frac{-r^2\cos \phi \mathbf {u}_x-r^2\sin \phi \mathbf {u}_y+(h-z)r\mathbf {u}_z}{(r^2+(h-z)^2)^{3/2}}dr d\phi dz. $$

The sine integral and the cosine integral between 0 and $2\pi$ are null, and therefore, components in $\mathbf {u}_x$ and $\mathbf {u}_y$ are null. This can be deduced from the figure’s symmetry: the field can only have a component in $\mathbf {u}_z$ as each dq has its symmetric $dq'$ in a horizontal plane whose $d\mathbf {E\,'}$ makes the same angle with the vertical, and it gives opposite components in $\mathbf {u}_x$ and $\mathbf {u}_y$, and equal components in $\mathbf {u}_z$. The integral to solve is

$$\begin{aligned}\mathbf {E}\,\!\!= & {} \frac{\rho }{4 \pi \varepsilon _0}\int _{R_i}^{R_e}\!\!\int _0^{2\pi }\!\!\int _0^H \frac{(h-z)r }{(r^2+(h-z)^2)^{3/2}}dr d\phi dz\mathbf {u}_z\\ {}= & {} \frac{\rho }{2\varepsilon _0}\int _{R_i}^{R_e}\!\!\int _0^H \frac{(h-z)r }{(r^2+(h-z)^2)^{3/2}}dr dz\mathbf {u}_z, \end{aligned}$$

because $\displaystyle \int _0^{2\pi } d\phi =2\pi$. Solving in r, bearing in mind that the derivative of the function $\displaystyle [r^2+(h-z)^2]$ is 2r, the result is

$$\begin{aligned}\mathbf {E}\,\!\!= & {} \frac{\rho }{2\varepsilon _0}\int _0^H \left. \frac{-(h-z) }{(r^2+(h-z)^2)^{1/2}}\right| _{R_i}^{R_e}dz\mathbf {u}_z=\\ {}= & {} -\frac{\rho }{2\varepsilon _0}\int _0^H\left( \frac{(h-z) }{\sqrt{R_e^2+(h-z)^2}}-\frac{(h-z)}{\sqrt{R_i^2+(h-z)^2}}\right) dz\mathbf {u}_z. \end{aligned}$$

Hence,

$$\begin{aligned}\mathbf {E}\,\!\!= & {} \frac{\rho }{2\varepsilon _0}\left( \left. \sqrt{R_e^2+(h-z)^2}\right| _0^H-\left. \sqrt{R_i^2+(h-z)^2}\right| _0^H \right) \mathbf {u}_z\\ {}= & {} \frac{\rho }{2\varepsilon _0}\left( \sqrt{R_e^2+(h-H)^2}-\sqrt{R_i^2+(h-H)^2}-\sqrt{R_e^2+h^2}+\sqrt{R_i^2+h^2}\right) \mathbf {u}_z. \end{aligned}$$

2.16 :

Considere an infinite plate with thickness $h=1\,$m in Fig. 2.31, with uniform charge density $\rho =20\,\upmu$C/m$^3$, except inside a spherical cavity with charge density three times that of the plate, with the centre the medium plane, and diameter h. Determine the electric field produced at point P(1, 2, 2), with the coordinates measured in meters referring to the coordinate axis located at the centre of the sphere, as shown in Fig. 2.31.

Fig. 2.31

Infinite plate with a spherical cavity and higher density

Solution

To calculate the electric field produced by the charge distribution, it can be observed that the superposition principle can be applied, and therefore it can be calculated as the addition of the fields produced by the two distributions with high symmetry: an infinite plate with uniform charge density $\rho$ and a uniformly charged sphere with density $2\rho$. In this way the spherical area would have a charge distribution, by adding of the previous ones, of $3\rho$ indicated on the problem’s statement. Fields produced by these distributions (Fig. 2.32) can be obtained from exercises previously worked in this book.

Fig. 2.32

Superposition principie for fields produced by the plate and the sphere

To calculate the field $\mathbf {E}_p$ produced by the whole plate, without the cavity, we apply the obtained results in Problem 2.7, bearing in mind that point P is outside the plate. According to (2.46), the electric field results

$$ \mathbf {E}_p=\frac{\rho h}{2\varepsilon _0 }\,\text{ sgn }(z)\mathbf {u}_z=\frac{20 \cdot 10^{-6} \cdot 1}{2\cdot 8.85\cdot 10^{-12}}\mathbf {u}_z=1.130 \cdot 10^6\mathbf {u}_z\,\text {N/C}. $$

To calculate the field produced by the sphere $\mathbf {E}_s$, we apply the result of Problem 2.5, and bear in mind that the sphere is solid (2.40), with density $2\rho$ and radius h / 2 and that point P is outside the sphere. The distance r from the sphere’s centre to point P is calculated from the vector’s position at point P:

$$ \mathbf {r}=\mathbf {u}_x+2\mathbf {u}_y+2\mathbf {u}_z,\quad r=3, \quad \mathbf {u}_r=\frac{\mathbf {u}_x+2\mathbf {u}_y+2\mathbf {u}_z}{3}. $$

The field results

$$\begin{aligned} \mathbf {E}_s&=\frac{2\rho \displaystyle \left( \frac{h}{2}\right) ^3}{3\varepsilon _0 r^2}\mathbf {u}_r=\frac{\rho h^3}{12\varepsilon _0 \displaystyle r^2}\mathbf {u}_r= \frac{20 \cdot 10^{-6} \cdot 1^3}{12\cdot 8.85\cdot 10^{-12}\cdot 3^2}\frac{\mathbf {u}_x+2\mathbf {u}_y+2\mathbf {u}_z}{3}\\&=6975(\mathbf {u}_x+2\mathbf {u}_y+2\mathbf {u}_z)\,\text {N/C}. \end{aligned}$$

The field $\mathbf {E}$ produced by the plate with the cavity is the addition of the fields $\mathbf {E}_p$ and $\mathbf {E}_s$,

$$ \mathbf {E}=\mathbf {E}_p+\mathbf {E}_s=(6975\mathbf {u}_x+13950\mathbf {u}_y+1.144\cdot 10^6\mathbf {u}_z)\,\text {N/C}. $$

2.17 :

Consider the bent wire in Fig. 2.33, where coordinates are expressed in meters, with uniform charge density $\lambda =8\,\upmu$C/m. The horizontal piece is very long (semi-infinite) and the other one, $L=2\,$m long, making an angle of 60${}^{\circ }$ with respect to the horizontal axis. Determine the electric field and the electric potential at point P(2, 1).

Fig. 2.33

Charged wire of Problem 2.17

Solution

To solve the problem, the superposition principle can be applied if the problem is considered as the addition of the fields and potentials produced by a semi-infinite horizontal wire and a finite one (60${}^{\circ }$) in respect to the horizontal axis. For the case of the semi-infinite wire, we consider Fig. 2.34 and (2.37) of Problem 2.4,

$$ \mathbf {E}_1=\frac{\lambda }{4\pi \varepsilon _0 y}\left( (\sin \varphi _f-\sin \varphi _i)\mathbf {u}_x+(\cos \varphi _i-\cos \varphi _f)\mathbf {u}_y\right) . $$

Fig. 2.34

Field produced by the semi-infinite wire

Angle $\varphi _i$ can be obtained from its tangent: $\tan \varphi _i=y/x=1/2\Rightarrow \varphi _i=30^\circ$. Angle $\varphi _f$ is 180${}^\circ$, since the wire is infinite along the $+X$ direction. Therefore,

$$\begin{aligned} \mathbf {E}_1&=\frac{8\cdot 10^{-6}}{4\pi \varepsilon _0\cdot 1}\left( (\sin 180^{\circ }-\sin 30^{\circ })\mathbf {u}_x+(\cos 30^{\circ }-\cos 180^{\circ })\mathbf {u}_y\right) \\&=71.9\cdot 10^3\left( -0.5\mathbf {u}_x+1.87\mathbf {u}_y\right) , \end{aligned}$$

$$ \mathbf {E}_1=(-36\mathbf {u}_x+134.2\mathbf {u}_y)\cdot 10^3 \text {N/C}. $$

Considering the finite wire of length $L=2\,$m (Fig. 2.35) and applying (2.37) of Problem 2.4 again, the result is

$$ \mathbf {E}_2=\frac{\lambda }{4\pi \varepsilon _0 h}\left( (\sin \varphi _f-\sin \varphi _i)\mathbf {u}_x+(\cos \varphi _i-\cos \varphi _f)\mathbf {u}_y\right) . $$

Fig. 2.35

Field produced by the finite wire

Let’s determine the coordinates of the wire’s extreme $P'(x',y')$:

$$ x'=L\cos 60^{\circ }=1,\quad y'=L\sin 60^{\circ }=\sqrt{3}, $$

so

$$ \mathbf {PP'}=(1-2)\mathbf {u}_x+(\sqrt{3}-1)\mathbf {u}_y=-\mathbf {u}_x+0.732\mathbf {u}_y, $$

with $|PP'|=1.239.$

The vector magnitude $\mathbf r=\mathbf {OP}$ is $|{\textit{OP}}|=\sqrt{5}$, and angle $\theta$ is obtained from

$$ \tan \theta =1/2\rightarrow \theta =26.6^{\circ }. $$

Angle $\varphi _i$ is, therefore,

$$ \varphi _i=60^{\circ }-\theta =33.4^{\circ }. $$

Distance h from point P to the wire is obtained from

$$ h=|OP|\sin \varphi _i=1.232. $$

The value of $\varphi _f$ is obtained from

$$ \sin \varphi _f=\frac{h}{|PP'|}=0.994\rightarrow \varphi _f=83.9^{\circ }. $$

If we substitute in $\mathbf {E}_2$ expression

$$\begin{aligned} \mathbf {E}_2= & {} \frac{8\cdot 10^{-6}}{4\pi \varepsilon _0\cdot 1.232}\left( (\sin 83.9^{\circ }-\sin 33.4^{\circ })\mathbf {u}_x+(\cos 33.4^{\circ }-\cos 83.9^{\circ })\mathbf {u}_y\right) =\\ {}= & {} 58.44\cdot 10^3\left( 0.44\mathbf {u}_x+0.73\mathbf {u}_y\right) =(25.9\mathbf {u}_x+42.6\mathbf {u}_y)\cdot 10^3 \text {N/C}.\end{aligned}$$

The total field at point P, if the superposition principle is applied, is

$$ \mathbf {E}=\mathbf {E}_1+\mathbf {E}_2=\left( -10.1\mathbf {u}_x+176.8\mathbf {u}_y\right) \cdot 10^3 \text {N/C}. $$

2.18 :

A thin, flat plate, which has the shape of a regular n-sided polygon, inscribed in a circle of radius a in plane XY, is considered. This plate is maintained at a fixed potential V, while the rest of the plane XY is held at zero potential. Apply Biot–Savart-like law in electrostatics to calculate the electric field produced by this plate at a point P located over the perpendicular to the plate by its centre.

Solution

Figure 2.36 shows the schematic view of the regular n-sided polygon plate, located in the XY-plane, with its centre at the origin, and kept at a potential V ($n=6$ in Fig. 2.36). In this plane ($z=0$), the potential is set to zero in the region outside the plate. To solve the problem we apply the (2.35),

$$ \mathbf {E}(\mathbf r)=\frac{V}{2\pi }\oint _C \frac{(\mathbf r-\mathbf r')\times d\mathbf {l}'}{|\mathbf r-\mathbf r'|^3}. $$

$\mathbf r$ refers to the field point and $\mathbf r'$ locates the source point. $d\mathbf {l}'$ is an element of length of the integration path C. As discussed in Sect. 2.9, we just need to calculate the contributions coming from the boundary contour C.

Fig. 2.36

Schematic view of the regular n-sided polygon plate

Looking at Fig. 2.36, it can be seen that electric field at point P can be calculated as the superposition of the field due to n triangles obtained by joining the centre of circle O to the vertices of the polygon: the sense of integration along their common sides is opposite for two adjacent triangles and adding all the contributions from these triangles, we obtain the integration around the contour C. Therefore, the only non-zero contributions come from the n sides that define the contour C. By symmetry, when the n fields are added vectorially, only the components located along Z-axis remains; the XY-plane components add to zero.

Then, considering the point P at Z-axis in the Fig. 2.36 and a point $P'$ and $d\mathbf {l}'$ at the upper straight side,

$$ \mathbf r=z\mathbf {u}_z, \quad \mathbf r'=x'\mathbf {u}_x+a\cos \frac{\pi }{n}\mathbf {u}_y, $$

$$ \mathbf r-\mathbf r'=-x'\mathbf {u}_x-a\cos \frac{\pi }{n}\mathbf {u}_y+z\mathbf {u}_z\quad |\mathbf r-\mathbf r'|=(x'^2+a^2\cos ^2(\pi /n)+z^2)^{1/2}, $$

$$ d\mathbf {l}'=dx'\mathbf {u}_x\,\quad (\mathbf r-\mathbf r')\times d\mathbf {l}'=zdx'\mathbf {u}_y+a\cos \frac{\pi }{n}dx'\mathbf {u}_z. $$

Calling $b^2=a^2\cos ^2(\pi /n)+z^2$ and applying the (2.35) to calculate the contribution $E_{iz}$ along $\mathbf {u}_z$,

$$\begin{aligned} E_{iz}= & {} \frac{V}{2\pi }\int _{-a\sin (\pi /n)}^{a\sin (\pi /n)} \frac{a\cos (\pi /n)dx'}{\left( x'^2+b^2\right) ^{3/2}}=\frac{V a\cos (\pi /n)}{2\pi b^2} \left. \frac{x'}{\sqrt{x'^2+b^2}}\right| _{-a\sin (\pi /n)}^{a\sin (\pi /n)} =\\ {}= & {} \frac{V a\cos (\pi /n)}{2\pi b^2}\frac{2a\sin (\pi /n)}{\sqrt{a^2\sin ^2(\pi /n)+b^2}}=\frac{V }{2\pi }\frac{a^2\sin (2\pi /n)}{[a^2\cos ^2(\pi /n)+z^2]\sqrt{a^2+z^2}}. \end{aligned}$$

Adding the n contributions from the n sides that define the contour C, the total electric field at P is obtained,

$$ \mathbf {E}=nE_{iz}=\frac{V }{2\pi }\frac{na^2\sin (2\pi /n)}{[a^2\cos ^2(\pi /n)+z^2]\sqrt{a^2+z^2}}\mathbf {u}_z. $$