Question

The potential at the surface of a sphere (radius R) is given by

${\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{k}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{\theta }$,

Where k is a constant. Find the potential inside and outside the sphere, as well as the surface charge density $\mathit{\sigma }\left(\theta \right)$on the sphere. (Assume there's no charge inside or outside the sphere.)

Short answer

Answer

The electric potential inside the sphere is $\frac{k}{5}\left(\frac{r}{R}\right)\cos \theta \left[4{\left(\frac{r}{R}\right)}^2\left(5{\cos }^2\theta -3\right)-3\right]$.

The potential outside the sphere is $\frac{k}{5}{\left(\frac{r}{R}\right)}^2\cos \theta \left[4{\left(\frac{r}{R}\right)}^2\left(5{\cos }^3\theta -3\right)-3\right]$.

The surface charge density on the sphere is $\frac{{\epsilon }_{0}k}{5R}\cos \theta \left[140{\cos }^2\theta -93\right]$.

Step-by-step solution

Given data

Write the expression for the potential at the surface of a sphere.

${\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{k}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{\theta }$ …… (1)

Here, k is constant.

Write the trigonometry formula for $\cos 3\theta$.

$V_0=k4{\cos }^3\theta -3\cos \theta$ …… (2)

Now, substitute $4{\cos }^3\theta -3\cos \theta$for $\cos 3\theta$in equation (1).

$V_0=k4{\cos }^3\theta -3\cos \theta$ …… (3)

Write the general expression for the Legendre polynomial.

${\mathit{P}}_{\mathbf{n}}\left(x\right)\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{n}}}{\mathbf{d}{\mathbf{x}}^{\mathbf{n}}}{\left(x^2-1\right)}^{\raisebox{1ex}{$\mathbf{n}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}$ …… (4)

Thus, the value of $P_1\left(\cos \theta \right)$and $P_3\left(\cos \theta \right)$is,

$$\begin{gathered} P_1\left(\cos \theta \right)=\frac{d^1}{d{x}^1}{\left({\cos }^2\theta -1\right)}^{\frac{1}{2}} \\ =\frac{d^1}{d{x}^1}\left(\sin \theta \right) \\ =\cos \theta \\ {P}_3\left(\cos \theta \right)=\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right) \end{gathered}$$

Determine electric potential inside the sphere

Write the expression potential of the sphere of the sphere of the radius R in terms of Legendre polynomials.

$$\begin{gathered} V_0\left(\theta \right)=k\left[4{\cos }^3\theta -3\cos \theta \right] \\ =k\left[\alpha P_3\left(\cos \theta \right)+\beta P_1\left(\cos \theta \right)\right] \end{gathered}$$ …… (5)

Here, $\alpha$and $\beta$are the Legendre polynomial constants.

Substitute $\cos \theta$for $P_1\left(\cos \theta \right)$and $\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right)$for $P_3\left(\cos \theta \right)$in equation (5).

$\begin{array}{rcl}k\left[4{\cos }^3\theta -3\cos \theta \right]& =& k\left[\alpha P_3\left(\cos \theta \right)+\beta P_1\left(\cos \theta \right)\right]\\ 4{\cos }^3\theta -3\cos \theta & =& k\left(\alpha \left[\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right)\right]+\beta \cos \theta \right)\\ 4{\cos }^3\theta -3\cos \theta & =& \frac{5\alpha }{2}{\cos }^3\theta +\left(\beta -\frac{3a}{2}\cos \theta \right)\\ & & \end{array}$

Compare the above equation on both sides, then the value of constant $\alpha$is ,

$\begin{array}{rcl}\frac{5\alpha }{2}& =& 4\\ \alpha & =& \frac{8}{5}\\ & & \end{array}$

Calculate the value of $\beta$.

$\begin{array}{rcl}\beta -\frac{3\alpha }{2}& =& -3\\ \beta -\frac{3}{2}\left(\frac{8}{5}\right)& =& -3\\ \beta -\frac{24}{10}& =& -3\\ \beta & =& -3+\frac{24}{10}\\ & & \end{array}$

Simplifying further,

$$\begin{gathered} \beta =\frac{-30+24}{10} \\ \beta =\frac{-6}{10} \\ \beta =\frac{-3}{5} \end{gathered}$$

Substitute $\frac{8}{5}$ for $\alpha$and $\frac{-3}{5}$for $\beta$in equation (5).

$\begin{array}{rcl}{V}_0\left(\theta \right)& =& k\left[\frac{8}{5}{P}_3\left(\cos \theta \right)-\frac{3}{5}{P}_1\left(\cos \theta \right)\right]\\ & =& \frac{k}{5}\left[8P_3\left(\cos \theta \right)-3P_1\left(\cos \theta \right)\right]\\ & & \end{array}$ …… (6)

Write the expression the potential inside the sphere.

$V\left(r,\theta \right)=\sum _{l=0}^a{A}_l{r}^l{P}_l\left(\cos \theta \right)$ ……. (7)

Multiply the above equation with $P_1\left(\cos \theta \right)\left(\sin \theta \right)$on both sides.

$\begin{array}{rcl}{P}_1\left(\cos \theta \right)\left(\sin \theta \right)V\left(r,\theta \right)& =& \sum _{l=0}^a{A}_l{r}^l{P}_l\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)\\ \int V\left(r,\theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta & =& A_l{r}^l\int \sum _{l=0}^a{P}_l\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)\\ A_l{r}^l\left(\frac{2}{2l+1}\right)& =& \underset{0}{\overset{a}{\int }}{V}_0\left(\theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta \\ & & \end{array}$ …… (8)

Use the condition,

$$\begin{gathered} \underset{0}{\overset{a}{\int }}{V}_0\left(\theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta =\left\{\frac{2}{2l+1}ifl=1 \\ 0ifl=0\right. \end{gathered}$$ in equation (8).

$A_l=\frac{2l+1}{2r^l}\underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta$

Find the value of A_l.

$\begin{array}{rcl}{A}_l& =& \frac{2l+1}{2R^l}\underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta \\ & =& \frac{2l+1}{2R^l}\left[\underset{0}{\overset{\pi }{\int }}\frac{k}{5}\left\{8P_3\left(\cos \theta \right)-3P_1\left(\cos \theta \right)\right\}{P}_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta \right]\\ & =& \frac{2l+1}{2R^l}\frac{k}{5}\left[8\underset{0}{\overset{\pi }{\int }}{P}_3\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta -3\underset{0}{\overset{\pi }{\int }}{P}_1\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta \right]\\ & & \end{array}$ …… (9)

Use the Legendre polynomial and the integral equation (9).

$\begin{array}{rcl}\underset{-1}{\overset{+1}{\int }}{P}_n\left(x\right)P_m\left(x\right)dx& =& {\delta }_{mn}\frac{2}{2n+1}\\ & =& 0\left(form\ne n\right)\\ & =& \frac{2}{2n+1}\left(form\ne n\right)\\ & & \end{array}$

Use the above equation.

$\begin{array}{rcl}{A}_l& =& \frac{2l+1}{2R^l}\frac{k}{5}\left[8\underset{0}{\overset{\pi }{\int }}{P}_3\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta -3\underset{0}{\overset{\pi }{\int }}{P}_1\left(\cos \theta \right)P_1\left(\cos \theta \right)\left(\sin \theta \right)d\theta \right]\\ A_l& =& \frac{k}{5}\frac{2l+1}{2R^l}\left\{8\left(\frac{2l+1}{2R^l}\right){\delta }_{13}-3\left(\frac{2l+1}{2R^l}\right){\delta }_{11}\right\}\\ & =& \frac{k}{5R^l}\left(8{\delta }_{13}-3{\delta }_{11}\right)\\ & =& \frac{-3k}{5R}ifl=1\\ & =& \frac{8k}{5R^3}ifl=3\\ & & \end{array}$

Thus, write the potential inside the sphere

$\begin{array}{rcl}V\left(r,\theta \right)& =& \sum _{l=0}^a{A}_l{r}^l{P}_l\cos \theta \\ & =& \frac{-3k}{5R}r{P}_1\left(\cos \theta \right)+\frac{8k}{5R^3}{r}^3{P}_3\left(\cos \theta \right)\\ & & \end{array}$ …… (10)

Substitute $\cos \theta$for $P_1\left(\cos \theta \right)$and $\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right)$for $P_3$in equation (10).

$\begin{array}{rcl}V\left(r,\theta \right)& =& \frac{k}{5}\left[-3\left(\frac{r}{R}\right)\cos \theta +8{\left(\frac{r}{R}\right)}^3\frac{\left(5{\cos }^3\theta -3\cos \theta \right)}{2}\right]\\ & =& \frac{k}{5}\left(\frac{r}{R}\right)\cos \theta \left[4{\left(\frac{r}{R}\right)}^2\left(5{\cos }^2\theta -3\right)-3\right]\\ & & \end{array}$

Thus, the electric potential inside the sphere is $\frac{k}{5}\left(\frac{r}{R}\right)\cos \theta \left[4{\left(\frac{r}{R}\right)}^2\left(5{\cos }^2\theta -3\right)-3\right]$.

Determine electric potential outside the sphere

Write the potential outside the sphere $\left(r\ge R\right)$.

$V\left(r,\theta \right)=\sum _{I=0}^a\frac{B_I}{r^{I+1}}{P}_1\left(\cos \theta \right)$ …… (11)

The value of $B_I$is given as,

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

If $I=1$, then value of $B_I$,

$B_I=A_I{R}^3$

Substitute $-\frac{3k}{5R}$for $A_I$in above equation.

$$\begin{gathered} B_I=\frac{-3k}{5R}{R}^3 \\ =\frac{-3k{R}^2}{5} \end{gathered}$$

If $I=3$then the value of $B_2$

$B_3=A_3{R}^7$

Substitute $\frac{8k}{5R^3}$for $A_3$.

$$\begin{gathered} B_3=\frac{8k}{5R^3}{R}^7 \\ =\frac{8k}{5R^4} \end{gathered}$$

Write the expression for potential outside the sphere.

$$\begin{gathered} V\left(r,\theta \right)=\sum _{I=0}^a\frac{B_I}{r^{I-1}}{P}_1\left(\cos \theta \right) \\ =-\left(\frac{3k{R}^2}{5}\right)\frac{1}{r^2}{P}_1\left(\cos \theta \right)+\left(\frac{8k{R}^4}{5}\right)\frac{1}{r^4}{P}_3\left(\cos \theta \right) \end{gathered}$$ …… (12)

Substitute $\cos \theta$for $P_1\left(\cos \theta \right)$and $\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right)$for $P_3$in equation (12).

$$\begin{gathered} V\left(r,\theta \right)=\frac{k}{5r}\left[-3{\left(\frac{R}{r}\right)}^4\left(\frac{5{\cos }^3\theta -3\cos \theta }{2}\right)\right] \\ =\frac{k}{5}{\left(\frac{R}{r}\right)}^2\cos \theta \left[4{\left(\frac{R}{r}\right)}^2\left(5{\cos }^3\theta -3\right)-3\right] \end{gathered}$$

Hence, the potential outside the sphere $=\frac{k}{5}{\left(\frac{R}{r}\right)}^2\cos \theta \left[4{\left(\frac{R}{r}\right)}^2\left(5{\cos }^3\theta -3\right)-3\right]$.

Determine Surface charge density

Using the equation 3.83, write the expression surface charge density on the sphere

$\begin{array}{rcl}\sigma \left(\theta \right)& =& {\epsilon }_0\sum _{l=0}^a\left(2l+1\right)A_l{R}^{l-1}{P}_l\left(\cos \theta \right)\\ & =& {\epsilon }_0\left(3A_l{P}_1\left(\cos \theta \right)+7A_3{R}^2{P}_3\right)\\ & & \end{array}$ …… (13)

Substitute $\frac{-3k}{5R}$for A_l, $\cos \theta$for $P_1\left(\cos \theta \right)$, $\frac{8k}{5R^3}$for $A_2$and $\frac{1}{2}\left(5{\cos }^3\theta -3\cos \theta \right)$ for P₃in equation (13).

$\begin{array}{rcl}\sigma \left(\theta \right)& =& {\epsilon }_0\left[3\left(\frac{-3k}{5R}\right)\cos \theta +7\left(\frac{8k}{5R^3}\right)R^2\frac{\left(5{\cos }^3\theta -3\cos \theta \right)}{2}\right]\\ & =& \frac{{\epsilon }_{0}k}{5R}\left[-9\cos \theta +\frac{56}{2}\left(5{\cos }^3\theta -3\cos \theta \right)\right]\\ & =& \frac{{\epsilon }_{0}k}{5R}\cos \theta \left[140{\cos }^2\theta -93\right]\\ & & \end{array}$

Hence, the surface charge density on the sphere is $\frac{{\epsilon }_{0}k}{5R}\cos \theta \left[140{\cos }^2\theta -93\right]$.