Chapter 24, Problem 50P

(a)

To determine

ToCalculate: The electric field and the energy density as a function of $r$ ,where $r$ is the distance from the center of the sphere, for $0\le r<\infty$ .

Answer to Problem 50P

  $u=\frac{k^2{\in }_0{Q}^2}{2r^4}{R}_1\le r\le R_2$

  $E_r=0$

Explanation of Solution

Given information :

Radius of inner radius = $R_1$

Charge of inner radius = $+Q$

Radius of outer radius = $R_2$

Charge of outer radius = $-Q$

Formula used :

From Gauss’s law:

  $E_r\left(4\pi r^2\right)=\frac{Q_{inside}}{{\in }_0}$

Where, $E_r$ is the electric field, $Q_{inside}$ is the enclosed charge, r is the radius and ${\in }_0$ is the absolute permittivity.

Energy density is given by:

  $u=\frac{1}{2}{\in }_0{E}_{}^2$

Calculation:

Use Gauss’s law to find the electric field for spherical surfaces of radii $r<R_1,R_1<r<R_2$ , and $r>R_2$

  

For $r<R_1$ :

  $\begin{array}{c}{E}_r\left(4\pi r^2\right)=\frac{Q_{inside}}{{\in }_0}=0\\ Q_{inside}=0\\ E_r=0\end{array}$

Because, $E=0$ for $r<R_1$

  $u=0$

For $R_1<r<R_2$ :

  $E_r\left(4\pi r^2\right)=\frac{Q_{inside}}{{\in }_0}=\frac{Q}{{\in }_0}$

  $\begin{array}{l}{E}_{r_2}=\frac{Q}{4\pi {\in }_0{r}^2}\\ E_{r_2}=\frac{kQ}{r^2}{R}_1\le r\le R_2\end{array}$

  $\begin{array}{l}u=\frac{1}{2}{\in }_0{E}_{R_1<r<R_2}^2\\ u=\frac{1}{2}{\in }_0{\left(\frac{kQ}{r^2}\right)}^2\\ u=\frac{k^2{\in }_0{Q}^2}{2r^4}{R}_1\le r\le R_2\end{array}$

For $r>R_2$ :

  $E_r\left(4\pi r^2\right)=\frac{Q_{inside}}{{\in }_0}=0$

  $Q_{inside}=0$

  $E_r=0$

Because $E=0$ for $r>R_2$

  $u_r=0$

Conclusion:

The energy density is:

  $u=\frac{k^2{\in }_0{Q}^2}{2r^4}{R}_1\le r\le R_2$

The electric field is:

  $E_r=0$

(b)

To determine

ToCalculate: The energy associated with the electrostatic field in a spherical shell between the conductors that has a radius $r$ , a thickness $dr$ , and a volume $4\pi r^{2}dr$ .

Answer to Problem 50P

  $dU=\frac{k{Q}^2}{2r^2}dr$

Explanation of Solution

Given information:

Radius of inner radius = $R_1$

Charge of inner radius = $+Q$

Radius of outer radius = $R_2$

Charge of outer radius = $-Q$

The energy density is:

  $u=\frac{k^2{\in }_0{Q}^2}{2r^4}{R}_1\le r\le R_2$

Formula Used:

The energy in the electrostatic field in a spherical shell of radius $r$ , thickness $dr$ , and volume $4\pi r^{2}dr$ between the conductors is:

  $dU=4\pi r^{2}u\left(r\right)dr$

Calculation:

  $\begin{array}{l}dU=4\pi r^{2}u\left(r\right)dr\\ dU=4\pi r^2\left(\frac{k^2{\in }_0{Q}^2}{2r^4}\right)dr\end{array}$

  $dU=\frac{k{Q}^2}{2r^2}dr$

Conclusion:

The energy associated with the electrostatic field is,

  $dU=\frac{k{Q}^2}{2r^2}dr$

(c)

To determine

ToFind: The total energy and compare the result with the result obtained using $U=\frac{1}{2}QV$ .

Answer to Problem 50P

  $U=\frac{1}{2}{Q}^2\left(\frac{\left(R_2-R_1\right)}{4\pi {\in }_0{R}_1{R}_2}\right)$

Explanation of Solution

Given information:

Radius of inner radius = $R_1$

Charge of inner radius = $+Q$

Radius of outer radius = $R_2$

Charge of outer radius = $-Q$

Formula used:

Energy density:

  $U=\frac{1}{2}QV$

Where, Q is the charge and V is the voltage.

The energy associated with the electrostatic field is $dU=\frac{k{Q}^2}{2r^2}dr$ .

Calculation:

Integrate $dU$ from $r=R_1$ to $R_2$ .

  $\begin{array}{l}U=\frac{k{Q}^2}{2}{\displaystyle \underset{R_1}{\overset{R_2}{\int }}\frac{dr}{r^2}}\\ U=\frac{k{Q}^2\left(R_2-R_1\right)}{2R_1{R}_2}\\ U=\frac{1}{2}{Q}^2\left(\frac{\left(R_2-R_1\right)}{4\pi {\in }_0{R}_1{R}_2}\right)\end{array}$

Conclusion:

Theintegration of the given expression is:

  $U=\frac{1}{2}{Q}^2\left(\frac{\left(R_2-R_1\right)}{4\pi {\in }_0{R}_1{R}_2}\right)$