Question

A spherical shell of radius R carries a uniform surface charge ${\mathit{a}}_{\mathbf{0}}$on the "northern" hemisphere and a uniform surface charge ${\mathit{a}}_{\mathbf{0}}$on the "southern "hemisphere. Find the potential inside and outside the sphere, calculating the coefficients explicitly up to ${\mathit{A}}_{\mathbf{6}}$and ${\mathit{B}}_{\mathbf{6}}$.

Short answer

Answer

The potential inside the sphere is

$\frac{{\mathbf{\sigma }}_{\mathbf{0}}\mathbf{r}}{\mathbf{2}{\mathbf{r}}_{\mathbf{0}}}\left[P_1\left(-cos\right)+\frac{1}{4}{\left(\frac{r}{R}\right)}^2{P}_3\left(+cos\theta \right)+\frac{1}{8}{\left(\frac{r}{R}\right)}^5{P}_5\left(cos\theta \right)\right]$.

The potential outside the sphere is

$\frac{{\mathbf{\sigma }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{3}}}{\mathbf{2}{\mathbf{r}}_{\mathbf{0}}}\left[P_1\left(cos\right)-\frac{1}{4}P{\left(\frac{R}{r}\right)}^2\left(cos{\theta }_3\right)+\frac{1}{8}{P}_5{\left(\frac{R}{r}\right)}^4\left(cos\theta \right)+.....\right]$.

Step-by-step solution

Define function

Write the expression for the potential inside the sphere.

$\mathit{V}\left(F,\right)\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathit{A}\mathit{r}\left(\frac{B_I}{r^{I+1}}\right)\left(cos\theta \right)\left(r\le R\right)$ …… (1)

Here, $V\left(r,\theta \right)$is the potential and $P_I\left(\cos \theta \right)$is the Legendre polynomial.

Write the expression for $A_I$.

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\mathit{\theta }\mathit{P}\frac{\mathbf{1}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}}{\mathbf{2}{\mathbf{R}}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\left(sin\theta \right){\mathit{d}}_{\mathbf{I}}\mathit{\theta }$ …… (2)

Here, for northern hemisphere varies form 0 to $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and for southern hemisphere varies from $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$to $\pi$.

Now, consider the charge density at northern hemisphere is $S_0$and charge density at southern hemisphere is $-S_0$.

Rewrite the equation (2),

${\mathit{\sigma }}_{\mathbf{I}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left[APa{R}_{0}cos{\theta }_{I-1}\underset{\pi 2}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(sin\theta d_0{\theta }_I\right)-\sigma Pcos\theta \underset{I}{\overset{\pi }{\int }}\left(sin\theta d_0{\theta }_I\right)\right]$ …… (3)

Determine potential inside the sphere

Assume $x=\cos \theta thendx=-\sin \theta$

Substitute $\cos \theta$for x and $-\sin \theta$for dx in equation (3)

$A_I=\frac{1}{2{\epsilon }_0{R}^{I-1}}\left[\underset{1}{\overset{0}{\int }}{\sigma }_0{P}_I\left(x\right)dx-\underset{0}{\overset{-1}{\int }}{\sigma }_0{P}_I\left(x\right)dx\right]$ …… (4)

Now use the following property,

$$\begin{gathered} \underset{-1}{\overset{0}{\int }}{P}_I\left(x\right)dx=\underset{1}{\overset{0}{\int }}{P}_I\left(-x\right)d\left(-x\right) \\ ={\left(-1\right)}^I\underset{-1}{\overset{0}{\int }}{P}_I\left(x\right)dx \end{gathered}$$

Thus, the equation (4) is,

role="math" localid="1655791505378" $A_I=\frac{{\sigma }_0}{2{\epsilon }_0{R}^{I-1}}\left\{\underset{0}{\overset{1}{\int }}{P}_I\left(x\right)dx-\underset{-1}{\overset{1}{\int }}{P}_1\left(x\right)dx\right\}$

Solve as further,

$A_I=\frac{{\sigma }_0}{2{\epsilon }_0{R}^{I-1}}\left(1-{\left(-1\right)}^I\right)\left[\underset{-1}{\overset{0}{\int }}{P}_I\left(x\right)dx\right]$ …… (5)

If $I$is even, then the term $A_I$is zero.

$A_I=0$

This is implies that term $A_0,A_2,A_4,A_8$are equal to zero.

If $I$is odd, then the term $A_I$is

role="math" localid="1655791831465" $A_I=\frac{{\sigma }_0}{{\epsilon }_0{R}^{I-1}}\left[\underset{0}{\overset{1}{\int }}{P}_I\left(x\right)dx\right]$

The value of odd term $A_I$is,

role="math" localid="1655791818630" $A_I=\frac{{\sigma }_0}{{\epsilon }_0}\underset{0}{\overset{1}{\int }}{P}_I\left(x\right)dx$

Substitute x for $P_1\left(x\right)$

$$\begin{gathered} A_I=\frac{{\sigma }_0}{{\epsilon }_0}\underset{0}{\overset{1}{\int }}xdx \\ =\frac{{\sigma }_0}{2{\epsilon }_0}{\left[x^2\right]}_0^1 \\ =\frac{{\sigma }_0}{2{\epsilon }_0} \end{gathered}$$

Write the expression for finding odd value of $A_3$is,

$A_3=\frac{{\sigma }_0}{{\epsilon }_0{R}^2}\underset{0}{\overset{1}{\int }}{P}_3\left(x\right)dx$

Substitute $\left(5x^3-3x\right)$for $P_3$

$A_3=\frac{{\sigma }_0}{{\epsilon }_0{R}^2}\underset{0}{\overset{1}{\int }}\left(5x^3-3x\right)dx$

Solve as further,

$$\begin{gathered} A_3=\frac{{\sigma }_0}{2{\epsilon }_0{R}^2}\left\{{\left[\frac{5x^4}{4}\right]}_0^1-{\left[\frac{3x^2}{2}\right]}_0^1\right\} \\ =\frac{{\sigma }_0}{2{\epsilon }_0{R}^2}\left[\frac{5}{4}-\frac{3}{2}\right] \\ =\frac{{\sigma }_0}{2{\epsilon }_0{R}^2}\left[\frac{-1}{4}\right] \\ =\frac{{\sigma }_0}{8{\epsilon }_0{R}^2} \end{gathered}$$

The value of odd term $A_5$is,

$A_5=\frac{{\sigma }_0}{{\epsilon }_0{R}^4}\underset{0}{\overset{1}{\int }}{P}_5\left(x\right)dx$

Substitute $\left(63x^5-70x^3+15x\right)$for $P_5\left(x\right)$

role="math" localid="1655792486072" $A_5=\frac{{\sigma }_0}{8{\epsilon }_0{R}^4}\underset{0}{\overset{1}{\int }}\left(63x^5-70x^3+15x\right)dx$

Solve as further,

role="math" localid="1655792742416" $$\begin{gathered} A_5=\frac{{\sigma }_0}{8{\epsilon }_0{R}^4}\left\{63{\left[\frac{x^6}{6}\right]}_0^1-70{\left[\frac{x^4}{4}\right]}_0^1+15{\left[\frac{x^2}{2}\right]}_0^1\right\} \\ =\frac{{\sigma }_0}{8{\epsilon }_0{R}^4}\left[\frac{63}{6}-\frac{70}{4}+\frac{15}{2}\right] \\ =\frac{{\sigma }_0}{16{\epsilon }_0{R}^4} \end{gathered}$$

Write the expression for potential inside the sphere.

$V\left(r,\theta \right)=A_1{r}^1{P}_1\left(\cos \theta \right)+A_2{r}^2{P}_2\left(\cos \theta \right)+A_3{r}^3{P}_3\left(\cos \theta \right)+A_5{r}^5{P}_5\left(\cos \theta \right)+...$ …..(6)

Substitute $\frac{{\sigma }_0}{2{\epsilon }_0}$for $A_1$,$\frac{-{\sigma }_0}{8{\epsilon }_0{R}^2}$for $A_3$,$\frac{{\sigma }_0}{16{\epsilon }_0{R}^2}$for $A_5$in equation (6).

$$\begin{gathered} V\left(r,\theta \right)=\frac{{\sigma }_{0}r}{2{\epsilon }_0}{P}_1\left(\cos \theta \right)-\frac{{\sigma }_0}{4{\epsilon }_0{R}^2}{r}^3{P}_3\left(\cos \theta \right)+\frac{{\sigma }_0}{16{\epsilon }_0{R}^4}{r}^5{P}_5\left(\cos \theta \right)+... \\ =\frac{{\sigma }_{0}r}{2{\epsilon }_0}\left[P_1\left(\cos \theta \right)-+\frac{1}{4}{\left(\frac{r}{R}\right)}^2{P}_3\left(\cos \theta \right)+\frac{1}{8}{\left(\frac{r}{R}\right)}^5{P}_5\left(\cos \theta \right)\right] \end{gathered}$$

Thus, the potential inside the sphere is $\frac{{\sigma }_{0}r}{2{\epsilon }_0}\left[P_1\left(\cos \theta \right)-\frac{1}{4}{\left(\frac{r}{R}\right)}^2{P}_3\left(\cos \theta \right)+\frac{1}{8}{\left(\frac{r}{R}\right)}^5{P}_5\left(\cos \theta \right)\right]$.

Determine potential outside the sphere

Write the expression for the potential outside the sphere.

$$\begin{gathered} V\left(r,\theta \right)=\sum _{I=0}^{\infty }\frac{B_I}{r^{I+1}}{P}_I\left(\cos \theta \right) \\ =\frac{B_1}{r^2}{P}_1\left(\cos \theta \right)+\frac{B_3}{r^4}{P}_3\left(\cos \theta \right)+\frac{B_5}{r^6}{P}_5\left(\cos \theta \right)+...... \end{gathered}$$ …… (7)

Here, the value of $B_I$is,

$B_i=A_I{R}^{2I+1}$

If value of $A_I$is zero for even values of $I$then the value of $B_I$is also zero for even value of $I$.

The value of $B_1$is,

$$\begin{gathered} B_1=A_1{R}^3 \\ =\frac{{\sigma }_0}{2{\epsilon }_0}{R}^3 \end{gathered}$$

The value of $B_3$is,

$$\begin{gathered} B_3=A_3{R}^7 \\ =\frac{-\sigma }{8{\epsilon }_0{R}^2}{R}^7 \\ =\frac{-{\sigma }_0}{8{\epsilon }_0}{R}^5 \end{gathered}$$

The value of $R^5$is,

$$\begin{gathered} B_5=A_1{R}^{11} \\ =\frac{{\sigma }_0}{16{\epsilon }_0} \end{gathered}$$

Substitute $\frac{{\sigma }_0}{2{\epsilon }_0}{R}^3$for $B_1,\frac{-{\sigma }_0}{8{\epsilon }_0}{R}^5$, for $B_3,\frac{-{\sigma }_0}{16{\epsilon }_0}$for $B_5$in equation (7).

$$\begin{gathered} V\left(r,\theta \right)=\frac{{\sigma }_0{R}^3}{2{\epsilon }_0{r}^2}{P}_1\left(\cos \theta \right)-\frac{{\sigma }_0{R}^5}{2{\epsilon }_0{r}^4}{P}_3\left(\cos \theta \right)+\frac{{\sigma }_0}{16{\epsilon }_0{r}^6}{R}^7{P}_5\left(\cos \theta \right)+... \\ =\frac{{\sigma }_0{R}^3}{2{\epsilon }_0{r}^2}\left[P_1\left(\cos \theta \right)-\frac{1}{4}{\left(\frac{R}{r}\right)}^2{P}_3\left(\cos \theta \right)+\frac{1}{8}{\left(\frac{R}{r}\right)}^4{P}_5\left(\cos \theta \right)+.....\right] \end{gathered}$$

Therefore, the potential outside the sphere is

$\frac{{\sigma }_0{R}^3}{2{\epsilon }_0{r}^2}\left[P_1\left(\cos \theta \right)-\frac{1}{4}{\left(\frac{R}{r}\right)}^2{P}_3\left(\cos \theta \right)+\frac{1}{8}{\left(\frac{R}{r}\right)}^4{P}_5\left(\cos \theta \right)+.....\right]$.