Question

Consider a spherical shell with radius R and surface charge density: The electric field is given by: if r<R E, 0 if r > R 0 (a) Find the energy stored in the field by: (b) Find the energy stored in the field by: Jall space And compare the result with part (a)

Answer 1

We rewrite the \hat{z} in the polar coordinates, and that is
   $\hat{z}=\hat{r}\cos\theta -\hat{\theta}\sin\theta$
And given the electric fields, we have
for r < R,
   $\vec{E}_{in}(\vec{r})=-\frac{\sigma_0}{3\epsilon_0}(\hat{r}\cos\theta-\hat{\theta}\sin\theta)$
And for r > R, we have
  
   $\vec{E}_{out}(\vec{r})=\frac{\sigma_0}{3\epsilon_0}\frac{R^3}{r^3}(3\hat{r}\cos\theta-\hat{r}\cos\theta+\hat{\theta}\sin\theta)$
$\Rightarrow \vec{E}_{out}(\vec{r})=\frac{\sigma_0}{3\epsilon_0}\frac{R^3}{r^3}(2\hat{r}\cos\theta+\hat{\theta}\sin\theta)$
Now the potential defined as
    $V(\vec{r})=-\int_{\infty}^{r}\vec{E}.d\vec{r}$
So,
   $V_{out}(\vec{r})=-\int_{\infty}^{r}\vec{E}_{out}.d\vec{r}$
$\Rightarrow V_{out}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}\int_{\infty}^{r}\frac{1}{r^3}(2\hat{r}\cos\theta+\hat{\theta}\sin\theta).d\vec{r}$
Along the radial paths, we have
    $\Rightarrow V_{out}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}\int_{\infty}^{r}\frac{1}{r^3}(2\hat{r}\cos\theta+\hat{\theta}\sin\theta).\hat{r}d{r}$
$\Rightarrow V_{out}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}2\cos\theta\int_{\infty}^{r}\frac{1}{r^3} d{r}$
$\Rightarrow V_{out}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}2\cos\theta\bigg[-\frac{1}{2r^2}\bigg]_{\infty}^{r}$
$\Rightarrow V_{out}(\vec{r})=\frac{\sigma_0R^3}{3\epsilon_0}\frac{\cos\theta}{r^2}$
And
    $V_{in}(\vec{r})=-\int_{\infty}^{r}\vec{E}.d\vec{r}=-\int_{\infty}^{R}\vec{E}_{out}.d\vec{r}-\int_{R}^{r}\vec{E}_{in}.d\vec{r}$
$\Rightarrow V_{in}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}\int_{\infty}^{R}\frac{1}{r^3}(2\hat{r}\cos\theta+\hat{\theta}\sin\theta).d\vec{r}-\int_{R}^r \left (-\frac{\sigma_0}{3\epsilon_0} \right )(\hat{r}\cos\theta-\hat{\theta}\sin\theta).d\vec{r}$
Along the radial paths, we have
    $\Rightarrow V_{in}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}\int_{\infty}^{R}\frac{1}{r^3}(2\hat{r}\cos\theta+\hat{\theta}\sin\theta).\hat{r}d{r}+\frac{\sigma_0}{3\epsilon_0}\int_{R}^r (\hat{r}\cos\theta-\hat{\theta}\sin\theta).\hat{r}d{r}$
$\Rightarrow V_{in}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}2\cos\theta \int_{\infty}^{R}\frac{1}{r^3}d{r}+\frac{\sigma_0}{3\epsilon_0}\cos\theta \int_{R}^r d{r}$
$\Rightarrow V_{in}(\vec{r})=-\frac{\sigma_0R^3}{3\epsilon_0}2\cos\theta \bigg[-\frac{1}{2r^2}\bigg]_{\infty}^{R}+\frac{\sigma_0}{3\epsilon_0}\cos\theta \big[r\big]_{R}^r$
$\Rightarrow V_{in}(\vec{r})=\frac{\sigma_0R^3}{3\epsilon_0} \frac{\cos\theta}{R^2}+\frac{\sigma_0}{3\epsilon_0}\cos\theta (r-R)$
$\Rightarrow V_{in}(\vec{r})=\frac{\sigma_0R}{3\epsilon_0} {\cos\theta}+\frac{\sigma_0}{3\epsilon_0}\cos\theta (r-R)$
$\Rightarrow V_{in}(\vec{r})=\frac{\sigma_0}{3\epsilon_0} r{\cos\theta}$
a)
  By calculating the surface integral at r = R, we have the expression
    $U_a=\frac{1}{2}\oint_S \sigma Vda'$
$\Rightarrow U_a=\frac{1}{2}\int \sigma_0\cos\theta \frac{\sigma_0}{3\epsilon_0}R \cos\theta ~R^2 \sin\theta d\theta ~d\phi$
$\Rightarrow U_a=\frac{\sigma_0^2}{6\epsilon_0}R^3\int_0^\pi \cos^2\theta \sin\theta d\theta ~\int_0^{2\pi}d\phi$
$\Rightarrow U_a=\frac{\sigma_0^2}{6\epsilon_0}R^3~2\pi~ \frac{2}{3}$
$\Rightarrow U_a=\frac{2\pi \sigma_0^2}{9\epsilon_0}R^3$
b)
Using the volume integral over the all space, we have
      $U_{b}=\frac{\epsilon_0}{2}\int_{all~space}E^2 ~d^3 r$
$\Rightarrow U_{b}=\frac{\epsilon_0}{2}\int_{r<R}E_{in}^2 ~d^3 r+\frac{\epsilon_0}{2}\int_{r >R}E_{out}^2 ~d^3 r$
$\Rightarrow U_{b}=\frac{\epsilon_0}{2}\int_{r<R}\left ( \frac{\sigma_0}{3\epsilon_0} \right )^2 ~d^3 r+\frac{\epsilon_0}{2}\int_{r >R} \left ( \frac{\sigma_0R^3}{3\epsilon_0} \right )^2\frac{1}{r^6}(4\cos^2\theta +\sin^2\theta)~d^3 r$
$\Rightarrow U_{b}=\frac{\epsilon_0}{2}\left ( \frac{\sigma_0}{3\epsilon_0} \right )^2 \frac{4\pi R^3}{3}+\frac{\epsilon_0}{2}\left ( \frac{\sigma_0R^3}{3\epsilon_0} \right )^2 \int_{R}^{\infty} \frac{r^2~dr}{r^6}\int_0^\pi (4\cos^2\theta +\sin^2\theta)\sin\theta ~d\theta ~\int_0^{2\pi}~d\phi$
$\Rightarrow U_{b}=\frac{2\pi\sigma^2_0R^3}{27\epsilon_0}+\frac{\epsilon_0}{2}\left ( \frac{\sigma_0R^3}{3\epsilon_0} \right )^2 2\pi\bigg[-\frac{1}{3r^3}\bigg]_{R}^{\infty} \int_0^\pi (3\cos^2\theta +1)\sin\theta ~d\theta$
$\Rightarrow U_{b}=\frac{2\pi\sigma^2_0R^3}{27\epsilon_0}+\frac{\pi \sigma^2_0R^6}{9\epsilon_0} \frac{1}{3R^3} (2 +2)$
$\Rightarrow U_{b}=\frac{2\pi\sigma^2_0R^3}{27\epsilon_0}+\frac{4\pi \sigma^2_0R^3}{27\epsilon_0}$
$\Rightarrow U_{b}=\frac{6\pi\sigma^2_0R^3}{27\epsilon_0}$
$\Rightarrow U_{b}=\frac{2\pi\sigma^2_0R^3}{9\epsilon_0}$
And we have seen that
   $U_a =U_{b}=\frac{2\pi\sigma^2_0R^3}{9\epsilon_0}$