Chapter 5, Problem 5.20P

Consider the basic image problem of a point charge q at z=d, suspended over an infinite conducting plane at z=0. Eq. (10) in chapter 2 to find E and D everywhere at the conductor surface as functions of cylindrical radius p. (b) Use your from part a to find the charge density, and the total induced charge on the conductor.

To determine

(a)

The expression for E and D as a function of cylindrical radius $\rho$.

Answer to Problem 5.20P

The required expressions are:

   $\begin{array}{l}\text{<mover accent="true"><mn>E</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\epsilon }_0{\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\\ \text{<mover accent="true"><mn>D</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\end{array}$

Explanation of Solution

Given Information:

The point charge q is at $z=d$ . The infinite conducting plane is at $z=0$.

Calculation:

Let the point charge and conducting plane are in cylindrical coordinate system. Consider a point $P\left(\rho ,0,0\right)$ on the plane, $z=0$ . The charge is at $Q\left(0,0,d\right)$.

$\begin{array}{l}\text{r}=\rho {\text{a}}_{\rho }\\ \text{<msup><mn>r</mn><mo>′</mo></msup>}=d{\text{a}}_z\\ \Rightarrow \text{r}-\text{<msup><mn>r</mn><mo>′</mo></msup>}=\rho {\text{a}}_{\rho }-d{\text{a}}_z\\ \left|\text{r}-\text{<msup><mn>r</mn><mo>′</mo></msup>}\right|={\left({\rho }^2+d^2\right)}^{1/2}\end{array}$

So, the electric filed intensity:

$\begin{array}{l}\text{<mover accent="true"><mn>E</mn><mo>→</mo></mover>}=\frac{Q\left(\text{r}-\text{<msup><mn>r</mn> <mo>′</mo></msup>}\right)}{4\pi {\epsilon }_0{\left|\text{r}-\text{<msup><mn>r</mn><mo>′</mo> </msup>}\right|}^3}\\ =\frac{q}{4\pi {\epsilon }_0{\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\end{array}$

The displacement field:

$\begin{array}{l}\text{<mover accent="true"><mn>D</mn><mo>→</mo></mover>}=\frac{Q\left(\text{r}-\text{<msup><mn>r</mn> <mo>′</mo></msup>}\right)}{4\pi {\left|\text{r}-\text{<msup><mn>r</mn><mo>′</mo> </msup>}\right|}^3}\\ =\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\end{array}$

Conclusion:

The required expressions are:

   $\begin{array}{l}\text{<mover accent="true"><mn>E</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\epsilon }_0{\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\\ \text{<mover accent="true"><mn>D</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\end{array}$

To determine

(b)

The charge density and total induced charge on the conductor.

Answer to Problem 5.20P

The charge density is ${\rho }_S=-\frac{qd}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}{\text{C/m}}^2$ , and total induced charge is $Q=-\frac{q}{2}\text{C}$.

Explanation of Solution

Given Information:

The point charge q is at $z=d$ . The infinite conducting plane is at $z=0$.

   $\begin{array}{l}\text{<mover accent="true"><mn>E</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\epsilon }_0{\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\\ \text{<mover accent="true"><mn>D</mn><mo>→</mo></mover>}=\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\end{array}$

Calculation:

The conducting plate is xy plane.

So, the unit normal vector:

   $\begin{array}{c}n={\text{a}}_z\\ D_N={\text{<mover accent="true"><mn>D</mn><mo>→</mo></mover>}\cdot n|}_S\\ =\left(\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\right)\cdot {\text{a}}_z\\ =-\frac{qd}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\end{array}$

So, the charge density:

   $\begin{array}{l}{\rho }_S=D_N\\ \Rightarrow {\rho }_S=-\frac{qd}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}{\text{C/m}}^2\end{array}$

The induced charge,

   $\begin{array}{c}Q={\displaystyle \underset{S}{\oint }D\cdot d\text{S}}\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{q}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}\left(\rho {\text{a}}_{\rho }-d{\text{a}}_z\right)\right)\cdot {\text{a}}_z\rho d\rho }d\varphi }\\ =-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{qd\rho }{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}d\rho }d\varphi }\\ =-\frac{qd}{4\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\left[-\frac{1}{\sqrt{{\rho }^2+d^2}}\right]}_0^{\infty }d\varphi }\\ =-\frac{qd}{4\pi }\left(\frac{1}{d}\right){\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }\\ =-\frac{2\pi q}{4\pi }\\ =-\frac{q}{2}\text{C}\end{array}$

Conclusion:

The charge density is ${\rho }_S=-\frac{qd}{4\pi {\left({\rho }^2+d^2\right)}^{3/2}}{\text{C/m}}^2$ , and total induced charge is $Q=-\frac{q}{2}\text{C}$.