91Ó°ÊÓ

Here, the configuration is asymptotically symmetric,

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

$V\left(cos\mathrm{θ}\right)=\underset{i=1}{\overset{∞}{∑}}\left(A_l{r}^l+\frac{B_l}{r^{l+1}}\right)P_l\left(cos\mathrm{θ}\right)$ …… (1)

a)

For $r>b$

In equation (1), $A_l=0$for all because $V→0$at infinity.

Thus,$V\left(r,\mathrm{θ}\right)=\underset{l=1}{\overset{∞}{∑}}\frac{B_l}{r^{l+1}}{P}_l\left(\cos \mathrm{θ}\right)$

For $a<r<b$

$V\left(r,\mathrm{θ}\right)=∑\left(C_l{r}^l+\frac{D_l}{r^{l+1}}\right)P_l\left(\cos \mathrm{θ}\right)$

For $r<a$,$V\left(r,\mathrm{θ}\right)=V_0$

Using boundary conditions,

V is continuous at a …… (a)

V is continuous at b …… (b)

At b , $\mathrm{Δ}\left(\frac{∂V}{∂r}\right)=\frac{-1}{{\mathrm{Î\mu }}_0}\mathrm{σ}\left(\mathrm{θ}\right)$ …… (c)

Here, $\mathrm{σ}\left(\mathrm{θ}\right)$is the surface charge density.

Now, substitute $k\cos \mathrm{θ}$for $\mathrm{σ}\left(\mathrm{θ}\right)$in equation (c)

$\mathrm{Δ}\left(\frac{∂V}{∂r}\right)=\frac{-k\cos \mathrm{θ}}{{\mathrm{Î\mu }}_0}$

Now, using boundary condition (b), V is continuous at b ,

$$\begin{gathered} \underset{l=1}{\overset{∞}{∑}}\frac{B_l}{b^{l+1}}{P}_l\left(\cos \mathrm{θ}\right)=\underset{l=1}{\overset{∞}{∑}}\left(C_l{b}^l+\frac{D_l}{b^{l+1}}\right)P_l\left(\cos \mathrm{θ}\right) \\ {C}_l{b}^l+\frac{D_l}{b^{l+1}}=\frac{B_l}{b^{l+1}} \end{gathered}$$

$B_l=C_l{b}^{2l+1}+D_l$ …… (2)

Using boundary condition (a), V is continuous at

$\underset{l=1}{\overset{∞}{∑}}\left[C_{|}{a}^{|}+\frac{D_{|}}{a^{|+1}}\right]P_{|}\left(\cos \mathrm{θ}\right)=v_0$ …… (3)

If $l=0$, then

$\begin{array}{rcl}{V}_0& =& C_0{a}^0+\frac{D_0}{a^{0+1}}\\ & =& C_0+\frac{D_0}{a}\\ D_0& =& a{V}_0-a{C}_0\\ & & \end{array}$ …… (4)

Substitute$l=0$in equation (2),

$B_0=b{C}_0+D_0$ ……. (5)

Substitute (4) in (5),

$B_0=\left(b-a\right)C_0+a{V}_0$ …… (6)

If $l≠0$,

$C_l{a}^l+\frac{D_l}{a^{l+1}}=0$

Substitute equation (7) in equation (2),

$B_l=C_l\left(b^{2l+1}-a^{2l+1}\right)$ …… (8)

From the boundary condition (c)

$\left[\underset{l=1}{\overset{∞}{∑}}{B}_l\left(-\left(l+1\right)\right)\frac{1}{b^{l+2}}{P}_1\left(\cos \mathrm{θ}\right)+\underset{l=1}{\overset{∞}{∑}}\left(C_{l}l{b}^{l-1}+D_l\frac{-\left(l+1\right)}{b^{l+2}}\right)P_l\left(\cos \mathrm{θ}\right)\right]=\frac{-k\cos \mathrm{θ}}{{\mathrm{Î\mu }}_0}$

If$l≠1$, then

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

$\left(l+1\right)\left(B_l-D_l\right)=-l{b}^{2l+1}{C}_l$ ....(9)

If $l=1$, then

$$\begin{gathered} \frac{2B_1}{b^2}+\left(C_1+D_1\left(\frac{-2}{b^2}\right)\right)=\frac{k}{{\mathrm{Î\mu }}_0} \\ {C}_1+\frac{2}{b_3}\left(B_1-D_1\right)=\frac{k}{{\mathrm{Î\mu }}_0} \end{gathered}$$ ……. (10)

From equation (7), (8), (9)

For ,$l≠0or1$

$$\begin{gathered} \left(l+1\right)\left[\left(b^{2l+1}-a^{2l+1}\right)C_l+a^{2l+1}{C}_l\right]+l{b}^{2l+1}{C}_l=0 \\ \left(l+1\right)b^{2l+1}{C}_l+l{b}^{2l+1}{C}_l=0 \\ \left(2l+1\right)C_l=0 \\ {C}_l=0 \end{gathered}$$

Therefore, $B_l=C_l=D_l=0$, for $l>1$.

For , $l=1$

$\begin{array}{rcl}{C}_1+\frac{2}{b^3}\left[\left(b^3-a^3\right)C_1+a^3{C}_1\right]& =& \frac{k}{{\mathrm{Î\mu }}_0}\\ 3C_1& =& \frac{k}{{\mathrm{Î\mu }}_0}\\ C_1& =& \frac{k}{3{\mathrm{Î\mu }}_0}\end{array}$

Thus,$C_1=\frac{k}{3{\mathrm{Î\mu }}_0}$

From equation (7),

$D_1=-a^3{C}_1$

Substitute$C_1=\frac{k}{3{\mathrm{Î\mu }}_0}$in above equation,

$D_1=\frac{-a^{3}k}{3{\mathrm{Î\mu }}_0}$

As,$B_l=\left(b^3-a^3\right)C_1$

Substitute$C_1=\frac{k}{3{\mathrm{Î\mu }}_0}$ in above equation,

$B_1=\left(b^3-a^3\right)\frac{k}{3{\mathrm{Î\mu }}_0}$

Now, from equation (9), for $l=0$

$$\begin{gathered} B_0-D_0=0 \\ {B}_0=D_0 \end{gathered}$$

Now, equating equation (4) and (5)

$$\begin{gathered} \left(b-a\right)C_0+a{V}_0=a{V}_0-a{C}_0 \\ \left(b-a\right)C_0=-a{C}_0 \end{gathered}$$

Thus,$C_0=0$

Therefore,$D_0=B_0=a{V}_0$

Hence, for ,$râ\copyright ¾b$

$V\left(r,\mathrm{θ}\right)=\frac{a{V}_0}{r}+\frac{\left(b^3-a^3\right)k}{3r^2{\mathrm{Î\mu }}_0}$

For ,$aâ\copyright ½râ\copyright ½b$

$V\left(r,\mathrm{θ}\right)=\frac{a{V}_0}{r}+\frac{K}{3{\mathrm{Î\mu }}_0}\left(r-\frac{a^3}{r^2}\right)\cos \mathrm{θ}$

b)

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

$$\begin{gathered} {\mathrm{σ}}_i\left(\mathrm{θ}\right)=-{\mathrm{Î\mu }}_0{\left.\frac{∂V}{∂r}\right|}_{r=a} \\ {\mathrm{σ}}_i\left(\mathrm{θ}\right)=-{\mathrm{Î\mu }}_0\left[\frac{-a{V}_0}{r^2}+\frac{K}{3{\mathrm{Î\mu }}_0}\left(1+\frac{2a^3}{r^3}\right)\cos \mathrm{θ}\right] \end{gathered}$$

Solve as further,

$$\begin{gathered} {\mathrm{σ}}_i\left(\mathrm{θ}\right)=-{\mathrm{Î\mu }}_0\left[\frac{-a{V}_0}{a^2}+\frac{K}{3{\mathrm{Î\mu }}_0}\left(1+\frac{2a^2}{r^3}\right)\cos \mathrm{θ}\right] \\ {\mathrm{σ}}_i\left(\mathrm{θ}\right)=-{\mathrm{Î\mu }}_0\left[\frac{-V_0}{a}+\frac{K}{{\mathrm{Î\mu }}_0}\cos \mathrm{θ}\right] \\ {\mathrm{σ}}_i\left(\mathrm{θ}\right)=\frac{+{\mathrm{Î\mu }}_0{V}_0}{a}-k\cos \mathrm{θ} \end{gathered}$$

Thus, induced charge density is ${\mathrm{σ}}_i\left(\mathrm{θ}\right)=\frac{+{\mathrm{Î\mu }}_0{V}_0}{a}-k\cos \mathrm{θ}$.

c)

Write the expression for total charge on system.

$\begin{array}{rcl}q& =& ∫{\mathrm{σ}}_{i}da\\ & =& \frac{V_0{\mathrm{Î\mu }}_0}{a}4\mathrm{Ï€}{a}^2\\ & =& 4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}a{V}_0\\ & & \end{array}$

At large V ,

$\begin{array}{rcl}V& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r}\\ & =& \frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}a{V}_0}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}\\ & =& \frac{a{V}_0}{r}\\ & & \end{array}$

Therefore, the total charge on system is $4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}a{V}_0$.