Question

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

$\mathbf{V}\left(\mathbf{r}\mathbf{,}\mathbf{\theta }\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_0}\left[\frac{\mathbf{q}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{racos\theta }}}\mathbf{-}\frac{\mathbf{q}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(\frac{\mathbf{ra}}{\mathbf{R}}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{racos\theta }}}\right]$

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

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

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

c) Calculate the energy of this configuration.

Short answer

a) It is proved that the equation of the potential is 0 for $\mathit{r}\mathbf{=}\mathit{R}$.

b) The induced charge on the surface is ${\mathit{q}}^{\mathbf{1}}$.

c) The energy of the configuration is $\frac{{\mathbf{q}}^{\mathbf{2}}\mathbf{R}}{\mathbf{8}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)}}$.

Step-by-step solution

Prove the formula for the law of cosine as follows:(a) 

Consider the figure for the given condition as shown in figure below:

From the figure write the equation as:

Consider the equations as:

$$\begin{gathered} r=\sqrt{r^2+a^2-2ra\cos \theta } \\ {r}_1=\sqrt{r^2+b^2-2rb\cos \theta } \end{gathered}$$

Since, $b=\frac{R^2}{a}$, $q^1=\frac{-R}{a}q$and $\frac{q^1}{r_1}=\frac{-R}{a}\frac{q}{r_1}$.

Write the equation as:

$\frac{q^1}{r_1}=\frac{-R}{a}\frac{q}{\sqrt{r^2+b^2-2rb\cos \theta }}$

Consider the equation for the potential for the condition as follows:

$V\left(r\right)=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{r}+\frac{q^1}{r_1}\right)$

Substitute the values and rewrite the equation for the potential as:

$\begin{array}{rcl}V\left(r\right)& =& \frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{\sqrt{r^2+a^2-2ra\cos \theta }}+\frac{-R}{a}\frac{q}{\sqrt{r^2+{\left(\frac{R^2}{a}\right)}^2-2r\left(\frac{R^2}{a}\right)\cos \theta }}\right)\\ & =& \frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{\sqrt{r^2+a^2-2ra\cos \theta }}-\frac{1}{\sqrt{r^2+a^2-2ra\cos \theta }}\right)\\ & =& 0\\ & & \end{array}$

Therefore, it is proved that the equation of the potential is 0 for $R=r$.

Solve for the induced surface charge on the sphere. (b)

Consider the formula for the induced surface charge density as:

$\sigma \left(\theta \right)=-{\epsilon }_0\frac{\partial V}{\partial n}$

Since, $\frac{\partial V}{\partial n}=\frac{\partial V}{\partial r}$for the value of requal R.

Substitute the values and solve as:

$\begin{array}{rcl}\sigma \left(\theta \right)& =& -{\epsilon }_0\frac{\partial }{\partial n}\left(\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{\sqrt{r^2+a^2-2ra\cos \theta }}-\frac{1}{\sqrt{R^2+{\left(\frac{ar}{R}\right)}^2-2ra\cos \theta }}\right)\right)\\ & =& -{\epsilon }_0\frac{\partial }{\partial n}{\left(\frac{q}{4\pi {\epsilon }_0}{\left(r^2+a^2-2ra\cos \theta \right)}^{\frac{3}{2}}\left(2r-2a\cos \theta \right)+\frac{1}{2}{\left(R^2+{\left(\frac{ar}{R}\right)}^2-2ra\cos \theta \right)}^{\frac{3}{2}}\left({\left(\frac{a}{R}\right)}^{2}2r-2a\cos \theta \right)\right)}_{r=R}\\ & =& \frac{q}{4\pi {\epsilon }_0}{\left(R^2+a^2-2Ra\cos \theta \right)}^{\frac{-3}{2}}\left(R^2-a^2\right)\\ & & \end{array}$

Consider the expression for the induced surface charge as:

$q_{ind}=\int \sigma da$

Substitute the values and solve as:

$\begin{array}{rcl}{q}_{ind}& =& {\int }_0^{2\pi }{\int }_0^{\pi }\frac{q}{4\pi R}{\left(R^2+a^2-2Ra\cos \theta \right)}^{\frac{-3}{2}}\left(R^2-a^2\right)R^2\sin \theta d\theta d\varphi \\ & =& \frac{q}{4\pi R}\left(R^2-a^2\right)2\pi R^2{\left(-\frac{1}{Ra}{\left(R^2+a^2-2Ra\cos \theta \right)}^{\frac{-1}{2}}\right)}_0^{\pi }\\ & =& \frac{q}{2a}\left(a^2-R^2\right)\left(\frac{1}{\sqrt{R^2+a^2+2Ra}}-\frac{1}{\sqrt{R^2+a^2-2Ra}}\right)\\ & & \end{array}$

Consider for $a>R$rewrite the equation as:

$\begin{array}{rcl}{q}_{ind}& =& \frac{q}{2a}\left(a^2-R^2\right)\left(\frac{1}{a+R}-\frac{1}{a-R}\right)\\ & =& -\frac{q}{a}R\\ & =& q^1\\ & & \end{array}$

Therefore, the induced charge on the surface is q¹.

Solve for the energy of the configuration as: (c)

Consider the formula for the force of image charge q¹on the charge q is obtained as:

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

Substitute the values and solve for the force as:

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

Solve for the work done as follows:

$\begin{array}{rcl}W& =& {\int }_{\infty }^a\frac{1}{4\pi {\epsilon }_0}\frac{q^{2}Ra}{{\left(a^2-R^2\right)}^2}da\\ & =& -\frac{q^{2}R}{4\pi {\epsilon }_0}{\left(\frac{-1}{2\left(a^2-R^2\right)}\right)}_{\infty }^a\\ & =& \frac{q^{2}R}{8\pi {\epsilon }_0\left(a^2-R^2\right)}\\ & & \end{array}$

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