A conducting sphere of radius a, at potential ${\mathit{V}}_{\mathbf{0}}$, is surrounded by a

thin concentric spherical shell of radius b,over which someone has glued a surface charge

$\mathit{\sigma }\left(\theta \right)\mathbf{=}\mathit{k}\mathbf{}\mathbf{cos}\mathbf{}\mathit{\theta }$,

where k is a constant and $\mathit{\theta }$is the usual spherical coordinate.

a) Find the potential in each region: (i) r>b, and (ii) a<r<b.

b) Find the induced surface charge ${\mathit{\sigma }}_{\mathbf{i}}\left(\theta \right)$on the conductor.

c) What is the total charge of this system? Check that your answer is consistent with the behavior of V at large.

Here, the configuration is asymptotically symmetric,

Write the expression for the electric potential in spherical polar co-ordinates.

$\mathit{V}\left(cos\theta \right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left(A_l{r}^l+\frac{B_l}{r^{l+l}}\right){\mathit{P}}_{\mathbf{l}}\mathbf{}\left(cos\theta \right)$ …… (1)

a)

For r > b

In equation (1), A, =0 for all / because $V\to 0$at infinity.

Thus, $V\left(r,\theta \right)=\underset{}{\sum \left(C_l{r}^l+{\displaystyle \frac{D_l}{r^{l+l}}}\right)P_l\left(\cos \theta \right)}$

$$\begin{gathered} \mathrm{For}\mathrm{r}<\mathrm{a},\mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)={\mathrm{V}}_0 \\ \mathrm{Using}\mathrm{boundary}\mathrm{conditions}, \\ \mathrm{V}\mathrm{is}\mathrm{continuous}\mathrm{at}\mathrm{a}\dots \dots \left(\mathrm{a}\right) \\ \mathrm{V}\mathrm{is}\mathrm{continuous}\mathrm{at}\mathrm{b}\dots \dots \left(\mathrm{b}\right) \\ \mathrm{At}\mathrm{b},∆\left(\frac{\partial \mathrm{V}}{\partial \mathrm{r}}\right)=\frac{-1}{{\mathrm{\epsilon }}_0}\mathrm{\sigma }\left(\mathrm{\theta }\right)\dots \dots \left(\mathrm{c}\right) \\ \mathrm{Here},\mathrm{\sigma }\left(\mathrm{\theta }\right)\mathrm{is}\mathrm{the}\mathrm{surface}\mathrm{charge}\mathrm{density}. \\ \mathrm{Now},\mathrm{substitute}\mathrm{k}\cos \mathrm{\theta }\mathrm{for}\mathrm{\sigma }\left(\mathrm{\theta }\right)\mathrm{in}\mathrm{equation}\left(\mathrm{c}\right) \\ ∆\left(\frac{\partial \mathrm{V}}{\partial \mathrm{r}}\right)=\frac{-\mathrm{k}\cos \mathrm{\theta }}{{\mathrm{\epsilon }}_0} \\ \mathrm{Now},\mathrm{using}\mathrm{boundary}\mathrm{condition}\left(\mathrm{b}\right),\mathrm{V}\mathrm{is}\mathrm{continuous}\mathrm{at}\mathrm{b}, \\ \sum _{l=1}^{\infty }\frac{B_l}{b^{l+1}}{P}_l\left(\cos \theta \right)=\sum _{l=1}^{\infty }\left(C_l{b}^l+\frac{D_l}{b^{l+1}}\right)P_l\left(\cos \theta \right) \\ {C}_l{b}^l+\frac{D_l}{b^{l+1}}=\frac{B_l}{b^{l+1}} \\ {B}_l=C_l{b}^{2l+1}+D_i\dots \dots \left(2\right) \\ \mathrm{Using}\mathrm{boundary}\mathrm{condition}\left(\mathrm{a}\right),\mathrm{V}\mathrm{is}\mathrm{continuous}\mathrm{at}\mathrm{a} \end{gathered}$$

$$\begin{gathered} \sum _{l+1}^{\infty }\left(c_l{a}^l+\frac{D_l}{a^{l+1}}\right)P_l\left(\cos \theta \right)=V_0\dots \dots \left(3\right) \\ Ifl=0,then \\ {V}_0=C_0{a}^0+\frac{D_0}{a^{0+1}} \\ =C_0+\frac{D_0}{a}\dots \dots \left(4\right) \\ {D}_0=a{V}_0-a{C}_0 \\ \mathrm{Substitute}\mathrm{in}\mathrm{equation}\left(2\right), \\ {B}_0=b{C}_0+D_0\dots \dots .\left(5\right) \\ \mathrm{Substitute}\left(4\right)\mathrm{in}\left(5\right), \\ {\mathrm{B}}_0=\left(\mathrm{b}-\mathrm{a}\right){\mathrm{C}}_0+{\mathrm{aV}}_0\dots \dots \left(6\right) \\ \mathrm{If}\mathrm{l}\ne 0, \\ {C}_l{a}^l+\frac{D_l}{a^{l+1}}=0 \\ {D}_l=C_l{a}^{2l+1\dots \dots \left(7\right)} \end{gathered}$$

$$\begin{gathered} Substituteequation\left(7\right)inequation\left(2\right), \\ {B}_l=C_l\left(b^{2l+1}-a^{2l+1}\right)\dots \dots \left(8\right) \\ Fromtheboundarycondition\left(c\right) \\ \left[\sum _{l=1}^{\infty }{B}_l\left(-\left(l+1\right)\right)\frac{1}{b^{l+2}}{P}_1\left(\cos \theta \right)+\sum _{l=1}^{\infty }\left(C_{l}l{b}^{l+1}+D_l\frac{\left(l+1\right)}{b^{l+2}}\right)P_1\left(\cos \theta \right)\right]=\frac{-k\cos \theta }{{\epsilon }_0} \\ Ifl\ne 1,then \end{gathered}$$

$$\begin{gathered} \frac{-\left(l+1\right)}{b^{l+2}}{B}_l-\left(C_{l}l{b}^{l-1}+D_l\frac{-\left(l+1\right)}{b^{l+2}}\right)=0 \\ -\left(l+1\right)B_l-l{C}_{l}l{b}^{2l-1}+\left(l+1\right)D_l=0 \end{gathered}$$

$$\begin{gathered} \left(l+1\right)\left(B_l-D_i\right)=-l{b}^{2l+1}C\dots \dots \left(9\right) \\ Ifl=1,then \\ \frac{2B_1}{b^2}+\left(C_1+D_1\left(\frac{-2}{b^2}\right)\right)=\frac{k}{{\epsilon }_0} \\ {C}_1+\frac{2}{b_3}\left(B_1-D_1\right)=\frac{k}{{\epsilon }_0}\dots \dots .\left(10\right) \\ Fromequation\left(7\right),\left(8\right),\left(9\right) \\ Forl\ne 0or1, \\ \left(l+1\right)\left[\left(b^{2l+1}-a^{2l+1}\right)C_l+a^{2l+1}{C}_l\right]+l{b}^{2l+1}{C}_l=0 \\ \left(l+1\right)l{b}^{2l+1}{C}_l+l{b}^{2l+1}{C}_l=0 \\ \left(2l+1\right)C_l=0 \\ {C}_l=0 \\ Therefore,B_l=C_l=D_l=0,forl>1. \end{gathered}$$

$$\begin{gathered} \mathrm{For}l=1 \\ {C}_1+\frac{2}{b^3}\left[\left(b^3-a^3\right)C_1+a^3{C}_1\right]=\frac{k}{{\epsilon }_0} \\ 3C_1=K/{\epsilon }_0 \\ {C}_1=\frac{k}{3{\epsilon }_0} \\ \mathrm{Thus},{\mathrm{C}}_1=\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0} \\ \mathrm{From}\mathrm{equation}\left(7\right), \\ {\mathrm{D}}_1=-{\mathrm{a}}^3{\mathrm{C}}_1 \\ \mathrm{Substitute}{\mathrm{C}}_1=\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0}\mathrm{in}\mathrm{above}\mathrm{equation} \\ {\mathrm{D}}_1=\frac{-{\mathrm{a}}^3\mathrm{k}}{3{\mathrm{\epsilon }}_0} \end{gathered}$$

$$\begin{gathered} \mathrm{As},{\mathrm{B}}_{\mathrm{l}}=\left({\mathrm{b}}^3-{\mathrm{a}}^3\right){\mathrm{C}}_1 \\ \mathrm{Substitute}{\mathrm{C}}_1=\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0}\mathrm{in}\mathrm{above}\mathrm{equation} \\ {\mathrm{B}}_1=\left({\mathrm{b}}^3-{\mathrm{a}}^3\right)\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0} \\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(9\right),\mathrm{for}l=0 \\ {B}_0-D_0=0 \\ {B}_0=D_0 \\ \mathrm{Now},\mathrm{equating}\mathrm{equation}\left(4\right)\mathrm{and}\left(5\right) \\ \left(\mathrm{b}-\mathrm{a}\right){\mathrm{C}}_0+{\mathrm{aV}}_0={\mathrm{aV}}_0-{\mathrm{aC}}_0 \\ \mathrm{Thus},{\mathrm{C}}_0=0 \\ \mathrm{Therefore},{\mathrm{D}}_0={\mathrm{B}}_0={\mathrm{aV}}_0 \\ \mathrm{Hence},\mathrm{for}\mathrm{r}\ge \mathrm{b}, \\ \mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{{\mathrm{aV}}_0}{\mathrm{r}}+\frac{\left({\mathrm{b}}^3-{\mathrm{a}}^3\right)\mathrm{k}}{3{\mathrm{r}}^2{\mathrm{\epsilon }}_0} \\ \mathrm{For}\mathrm{a}\le \mathrm{r}\le \mathrm{b}, \\ \mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{{\mathrm{aV}}_0}{\mathrm{r}}+\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0}\left(\mathrm{r}-\frac{{\mathrm{a}}^3}{{\mathrm{r}}^2}\right)\cos \mathrm{\theta } \end{gathered}$$

b)

Write the expression for induced surface charge density on the conductor.

$$\begin{gathered} {\sigma }_i\left(\theta \right)=-{\epsilon }_0{\overline{)\frac{\partial V}{\partial t}}}_{r=a} \\ {\sigma }_i\left(\theta \right)=-{\epsilon }_0\left[\frac{-a{V}_0}{r_0}+\frac{K}{3{\epsilon }_0}\left(1+\frac{2a^3}{r^3}\right)\cos \theta \right] \\ \mathrm{Solve}\mathrm{as}\mathrm{further}, \\ {\mathrm{\sigma }}_{\mathrm{i}}\left(\mathrm{\theta }\right)=-{\mathrm{\epsilon }}_0\left[\frac{-{\mathrm{aV}}_0}{{\mathrm{a}}_0}+\frac{\mathrm{K}}{3{\mathrm{\epsilon }}_0}\left(1+\frac{2{\mathrm{a}}^3}{{\mathrm{r}}^3}\right)\cos \mathrm{\theta }\right] \\ {\mathrm{\sigma }}_{\mathrm{i}}\left(\mathrm{\theta }\right)=-{\mathrm{\epsilon }}_0\left[\frac{-{\mathrm{aV}}_0}{{\mathrm{r}}_0}+\frac{\mathrm{K}}{3{\mathrm{\epsilon }}_0}\cos \mathrm{\theta }\right] \\ {\mathrm{\sigma }}_{\mathrm{i}}\left(\mathrm{\theta }\right)=\frac{+{\mathrm{\epsilon }}_0{\mathrm{V}}_0}{\mathrm{a}}-\mathrm{k}\cos \mathrm{\theta } \\ \mathrm{Thus},\mathrm{induced}\mathrm{charge}\mathrm{density}\mathrm{is}{\mathrm{\sigma }}_{\mathrm{i}}\left(\mathrm{\theta }\right)=\frac{+{\mathrm{\epsilon }}_0{\mathrm{V}}_0}{\mathrm{a}}-\mathrm{k}\cos \mathrm{\theta }. \end{gathered}$$

c)

Write the expression for total charge on system.

$$\begin{gathered} q=\int {\sigma }_{i}da \\ =\frac{V_0{\epsilon }_0}{a}4\pi a^2 \\ =4\pi {\epsilon }_{0}a{V}_0 \\ AtlargeV \\ V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r} \\ =\frac{4\pi {\epsilon }_{0}a{V}_0}{4\pi {\epsilon }_{0}r} \\ =\frac{a{V}_0}{r} \end{gathered}$$

Therefore, the total charge on system is $4\pi {\epsilon }_{0}a{V}_0$.