(a) Using the law of cosines, show that Eq. 3.17 can be written as follows:

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{[}\frac{\mathbf{q}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{racos}\mathbf{}\mathbf{\theta }}}\mathbf{-}\frac{\mathbf{q}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{ra}\mathbf{/}\mathbf{R}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{racos}\mathbf{}\mathbf{\theta }}}\mathbf{]}$

Where$\mathbf{r}$and $\mathbf{\theta }$are the usual spherical polar coordinates, with the $\mathbf{z}$axis along the

line through $\mathbf{q}$. In this form, it is obvious that$\mathbf{V}\mathbf{=}\mathbf{0}$on the sphere, localid="1657372270600" $\mathbf{r}\mathbf{=}\mathbf{R}$.

(a) Find the induced surface charge on the sphere, as a function of $\mathbf{\theta }$. Integrate this to get the total induced charge . (What should it be?)

(b) Calculate the energy of this configuration.

Write the expression for potential using law of cosines.

$\mathbf{\text{V(r,θ)=}}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{q}{\sqrt{r^2+a^2-2racos\theta }}-\frac{q}{\sqrt{R^2+(ra/R{)}^2-2racos\theta }}\right]$ …… (1)

Here, role="math" localid="1657372493931" $\mathrm{r}$and $\mathrm{\theta }$are the spherical polar coordinates, role="math" localid="1657372543438" $\mathrm{q}$is the charge, ${\mathrm{\epsilon }}_0$is the permittivity for the free space and $\mathrm{R}$is the radius of the sphere.

a) From the figure given below we can write the equation as,

$\mathrm{\sigma }\left(\mathrm{\theta }\right)=-{\mathrm{\epsilon }}_0\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}{\left(\begin{array}{l}-\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}(2\mathrm{r}-2\mathrm{acos}\mathrm{\theta })+\\ \frac{1}{2}{\left({\mathrm{R}}^2+{\left(\frac{\mathrm{ra}}{\mathrm{R}}\right)}^2-2\mathrm{racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{{\mathrm{R}}^2}2\mathrm{r}-2\mathrm{acos}\mathrm{\theta }\right)\end{array}\right)}_{\mathrm{r}=\mathrm{R}}$…… (2)

Write the equation for $\mathrm{r}$

$\mathrm{r}=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}$

${\mathrm{r}}_1=\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}\mathrm{\theta }}$

As we know that,

$\mathrm{b}=\frac{{\mathrm{R}}^2}{\mathrm{a}}$

${\mathrm{q}}^1=\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}$ ……. (3)

$\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}=\frac{-\mathrm{R}}{\mathrm{a}}\frac{\mathrm{q}}{{\mathrm{r}}_1}$

Substitute $\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}\mathrm{\theta }}$for ${\mathrm{r}}_1$in equation (3).

$\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}=\frac{-\mathrm{R}}{\mathrm{a}}\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}\mathrm{\theta }}}$ …… (4)

Now, write the expression for the potential for this configuration.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{q}}{\mathrm{r}}+\frac{{\mathrm{q}}^1}{{\mathrm{r}}_1}\right)$……. (5)

Substitute $\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}$for $\mathrm{r},\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}\mathrm{\theta }}$for${\mathrm{r}}_1$and$\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}$for${\mathrm{q}}^1$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}+\frac{{\mathrm{a}}_{\mathrm{q}}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}\mathrm{\theta }}}\right)$ ……(6)

Substitute $\frac{{\mathrm{R}}^2}{\mathrm{a}}$for $\mathrm{b}$.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}+\frac{\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2-2\mathrm{r}\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos \mathrm{\theta }}}\right)$

$\mathrm{V}\left(\mathrm{r}\right)-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\left.\frac{\mathrm{a}}{\mathrm{R}}\sqrt{{\mathrm{r}}^2+{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2-2\mathrm{r}\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos \mathrm{\theta }}\right)}\right.$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^{2}2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\left.\sqrt{{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\mathrm{r}}^2+{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)}^2-2\mathrm{r}{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^2\left(\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\cos \mathrm{\theta }}\right)}\right.$

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\sqrt{{\left.\frac{\mathrm{a}}{\mathrm{R}}\right)}^2{\mathrm{r}}^2+(\mathrm{R}{)}^2-2\mathrm{racos}\mathrm{\theta }}}\right)$

Rearranging the statement,

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\frac{\mathrm{ar}}{\mathrm{R}}\right)}^2-2\mathrm{racos}\mathrm{\theta }}}\right)$

The potential is zero, when $\mathrm{r}=\mathrm{R}$.

$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}0}}+\frac{\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}}{\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rbcos}0}}\right)$ …… (7)

$=0$

Hence, the potential is zero, when $\mathrm{r}=\mathrm{R}$.

b)

Write the expression for the induced surface charge density.

$\mathrm{\sigma }\left(\mathrm{\theta }\right)={\mathrm{\epsilon }}_0\frac{\mathrm{\partial }\mathrm{V}}{\widehat{\mathrm{o}}\mathrm{n}}$ ……. (8)

But $\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{n}}=\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{r}}$at $\mathrm{r}=\mathrm{R}$

Substitute $\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\begin{array}{c}\mathrm{ar}\\ \mathrm{R}\end{array}\right)}^2-2\mathrm{racos}\mathrm{\theta }}}\right)$ for $\mathrm{V}$,

$\mathrm{\sigma }\left(\mathrm{\theta }\right)=-{\mathrm{\epsilon }}_0\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{n}}\left(\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }}}-\frac{1}{\sqrt{{\mathrm{R}}^2+{\left(\frac{\mathrm{ar}}{\mathrm{R}}\right)}^2-2\mathrm{racos}\mathrm{\theta }}}\right)\right)$

$=-{\mathrm{\epsilon }}_0\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}{\left(\begin{array}{l}-\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}(2\mathrm{r}-2\mathrm{acos}\mathrm{\theta })\\ +\frac{1}{2}{\left({\mathrm{R}}^2+{\left(\frac{\mathrm{ra}}{\mathrm{R}}\right)}^2-2\mathrm{racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{{\mathrm{R}}^2}2\mathrm{r}-2\mathrm{acos}\mathrm{\theta }\right)\end{array}\right)}_{\mathrm{r}-\mathrm{R}}$

$=\frac{-\mathrm{q}}{4\mathrm{\pi }}\left(\begin{array}{l}-{\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{\frac{3}{2}}(\mathrm{R}-\mathrm{acos}\mathrm{\theta })\\ +{\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{\frac{3}{2}}\left(\frac{{\mathrm{a}}^2}{\mathrm{R}}-\mathrm{acos}\mathrm{\theta }\right)\end{array}\right)$

Write induced surface charge formula.

${\mathrm{q}}_{\mathrm{irvi}}=\int \mathrm{\sigma da}$ …… (9)

Substitute $\frac{\mathrm{q}}{4\mathrm{\pi R}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right)$for $\mathrm{\sigma }$in equation (9).

${\mathrm{q}}_{\mathrm{in}}={\int }_0^{2\mathrm{\pi }} {\int }_0^{\mathrm{\pi }} \frac{\mathrm{q}}{4\mathrm{\pi R}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right){\mathrm{R}}^2\sin \mathrm{\theta d\theta d\varphi }$

$=\frac{\mathrm{q}}{4\mathrm{\pi R}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right)2{\mathrm{\pi R}}^2{\int }_0^2 {\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{-\frac{3}{2}}\sin \mathrm{\theta d\theta }$

$=\frac{\mathrm{q}}{4\mathrm{\pi R}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right)2{\mathrm{\pi R}}^2{\left(\frac{-1}{\mathrm{Ra}}{\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\mathrm{\theta }\right)}^{-\frac{1}{2}}\right)}_0^{\mathrm{\pi }}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right)(-1)\left({\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\left(\mathrm{\pi }\right)-\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Racos}\left(0\right)\right)\right)}^{-\frac{1}{2}}\right)$

Further solving,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{R}}^2-{\mathrm{a}}^2\right)(-1){\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Ra}(-1)-{\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Ra}\left(1\right)\right)}^{\frac{-1}{2}}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2-{\mathrm{R}}^2\right){\left(\left({\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}\right)-\left({\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Ra}\right)\right)}^{\frac{-1}{2}}$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)\left(\frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}}}-\frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Ra}}}\right)$ …… (10)

For$\mathrm{a}>\mathrm{R}$ ,

$\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^{2}2\mathrm{Ra}}=\sqrt{(\mathrm{arR}{)}^2}$

$=\mathrm{a}-\mathrm{R}$

Substitute $\mathrm{a}-\mathrm{R}$for $\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2-2\mathrm{Ra}}$and $\mathrm{a}+\mathrm{R}$for $\sqrt{{\mathrm{R}}^2+{\mathrm{a}}^2+2\mathrm{Ra}}$ in equation (10).

Therefore,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)\left(\frac{1}{\mathrm{a}+\mathrm{R}}-\frac{1}{\mathrm{a}-\mathrm{R}}\right)$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[(\mathrm{a}+\mathrm{R})(\mathrm{a}-\mathrm{R})\left(\frac{1}{\mathrm{a}+\mathrm{R}}-\frac{1}{\mathrm{a}-\mathrm{R}}\right)\right]$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[\left(\frac{(\mathrm{a}+\mathrm{R})(\mathrm{a}-\mathrm{R})}{\mathrm{a}-\mathrm{R}}-\frac{(\mathrm{a}+\mathrm{R})(\mathrm{a}-\mathrm{R})}{\mathrm{a}-\mathrm{R}}\right)\right]$

$=\frac{\mathrm{q}}{2\mathrm{a}}\left[(\mathrm{a}-\mathrm{R})-(\mathrm{a}+\mathrm{R}{)}_{-}^{-}\right.$

Solve as further,

${\mathrm{q}}_{\mathrm{in}}=\frac{\mathrm{q}}{2\mathrm{a}}(-2\mathrm{R})$

$=\frac{\mathrm{q}}{\mathrm{a}}\mathrm{R}$

$={\mathrm{q}}^1$

Hence, the induced surface charge is ${\mathrm{q}}^1$.

c)

Write the expression for the force of image charge ${\mathrm{q}}^1$on $\mathrm{q}$.

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{qq}}^1}{(\mathrm{a}-\mathrm{b}{)}^2}$…… (11)

Substitute $\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}$for ${\mathrm{q}}^1$and $\frac{{\mathrm{R}}^2}{\mathrm{a}}$for $\mathrm{b}$in equation (11).

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}\left(\frac{-\mathrm{R}}{\mathrm{a}}\mathrm{q}\right)}{{\left(\begin{array}{c}\mathrm{a}-{\mathrm{R}}^2\\ \mathrm{a}\end{array}\right)}^2}$

$=\frac{1}{4{\mathrm{\pi z}}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}^2}$

Thus, the force of image charge ${\mathrm{q}}^1$on $\mathrm{q}$is $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}^2}$.

Write the expression for work done.

$\mathrm{W}={\int }_{\mathrm{a}}^{\mathrm{A}} \mathrm{Fda}$ …… (12)

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}^2}$for $\mathrm{F}$in equation (12)

$\mathrm{W}={\int }_{\mathrm{a}}^{\mathrm{a}} -\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2\mathrm{Ra}}{{\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}^2}\mathrm{da}$

$=-\frac{{\mathrm{q}}^2\mathrm{R}}{4{\mathrm{\pi \epsilon }}_0}{\left(\frac{-1}{2\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}\right)}^{\mathrm{a}}$

$=\frac{{\mathrm{a}}^2\mathrm{R}}{8{\mathrm{\pi \epsilon }}_0\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}$

Therefore, the energy of this configuration is$\frac{{\mathrm{a}}^2\mathrm{R}}{8{\mathrm{\pi \epsilon }}_0\left({\mathrm{a}}^2-{\mathrm{R}}^2\right)}$