In Ex. 3.8 we determined the electric field outside a spherical conductor

(radiusR)placed in a uniform external field ${\mathbf{E}}_{\mathbf{0}}$. Solve the problem now using

the method of images, and check that your answer agrees with Eq. 3.76. [Hint:Use

Ex. 3.2, but put another charge, -q,diametrically opposite q.Let$\mathbf{a}\mathbf{\to }\mathbf{\infty }$, with$\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{2}\mathbf{q}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}{\mathbf{E}}_{\mathbf{0}}$held constant.]

The radius of the sphere is R.

The uniform external field is$E_0$.

The external field ${\mathrm{E}}_0$is replaced by a charge q.

role="math" localid="1657522471977" ${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{2}}}$

Here, a is the area.

Consider a charge q at a distance a long x axis from the origin.

An image charge -q is then placed at x=-a .

Let the induced charge on the sphere be ${\mathrm{q}}^{\text{'}}\left(=-\frac{\mathrm{R}}{\mathrm{a}}\mathrm{q}\right)\mathrm{at}\mathrm{x}=\mathrm{b}\left(=\frac{{\mathrm{R}}^2}{\mathrm{a}}\right)\mathrm{where}\mathrm{b}<\mathrm{R}.$

The corresponding image charge $-{\mathrm{q}}^{\text{'}}$is considered at x=-b.

The potential at r is then,

$V=\frac{1}{4\pi {\epsilon }_0}\left[q\left(\frac{1}{r_1}-\frac{1}{r^2}\right)+q^{\text{'}}\left(\frac{1}{r^3}-\frac{1}{r^4}\right)\right]...\left(1\right)$

Here

$$\begin{gathered} {\mathrm{r}}_1=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{ra}\cos \mathrm{\theta }} \\ {\mathrm{r}}_2=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2+2\mathrm{ra}\cos \mathrm{\theta }} \\ {\mathrm{r}}_3=\sqrt{{\mathrm{r}}^2+{\mathrm{b}}^2-2\mathrm{rb}\cos \mathrm{\theta }} \\ {\mathrm{r}}_4=\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2+2\mathrm{rb}\cos \mathrm{\theta }} \end{gathered}$$

localid="1657523731983" $$\begin{gathered} Fora\gg rthefollowingsimplificationfollows, \\ \frac{1}{r^1}-\frac{1}{r^2}=\frac{1}{\sqrt{r^2+a^2-2ra\cos \theta }}-\frac{1}{\sqrt{r^2+a^2+2ra\cos \theta }} \\ \approx \frac{1+{\displaystyle \frac{r}{a}}\cos \theta }{a}-\frac{1-{\displaystyle \frac{r}{a}}\cos \theta }{a} \\ =\frac{2r}{a^2}\cos \theta \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly}\mathrm{for}\mathrm{r}\gg \mathrm{b}\mathrm{the}\mathrm{second}\mathrm{term}\mathrm{simplifies}\mathrm{as}\mathrm{follows}, \\ \frac{1}{{\mathrm{r}}^3}-\frac{1}{{\mathrm{r}}_4}\approx \frac{2\mathrm{b}}{{\mathrm{r}}^2}\mathrm{cos\theta } \\ =\frac{2{\mathrm{R}}^2}{{\mathrm{ar}}^2}\mathrm{cos\theta } \end{gathered}$$

$$\begin{gathered} \mathrm{Substitutions}\mathrm{of}\mathrm{equation}\mathrm{results}\mathrm{in}, \\ \mathrm{V}=-{\mathrm{E}}_0\left(\mathrm{r}-\frac{{\mathrm{R}}^3}{{\mathrm{r}}^2}\right)\cos \mathrm{\theta } \\ \mathrm{This}\mathrm{is}\mathrm{exactly}\mathrm{the}\mathrm{result}\mathrm{mentioned}\mathrm{in}\mathrm{Eq}.3.76. \end{gathered}$$