91Ó°ÊÓ

Write the expression for the potential at the point on the sphere at a distance rfrom the charge q.

$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{r}}$ …… (1)

Here, ${\mathit{Î\mu }}_{\mathbf{0}}$ is the permittivity for the free space, V is the potential, qis the charge, r is the distance.

At the center of the sphere the magnitude of electric field which is z distance away from the charge q.

Write the expression for charge q.

${\mathit{E}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$ …… (2)

Here, ${\mathit{Î\mu }}_{\mathbf{0}}$ is the permittivity for the free space, E is the electric field.

a)

The figure of sphere is shown below.

Here, R is the radius with its center at origin, a charge qis placed on the z-axis at a distance z from the center. An area element is taken at distance r from the charge.

Write the formula for average electric field.

$\begin{array}{rcl}\stackrel{\mathbf{→}}{{\mathbf{E}}_{\mathbf{a}\mathbf{v}\mathbf{g}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{R}}^{\mathbf{2}}}\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{E}}\mathit{d}\mathit{a}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{R}}^{\mathbf{2}}}\mathbf{âˆ\circledR }\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}\right)\mathbf{cos}\mathit{Î\pm }\mathit{d}\mathit{a}\\ & & \end{array}$ …… (3)

Refer the above figure, apply the parallelogram law of vector addition to write the expression for r.

$$\begin{gathered} {\mathit{r}}^{\mathbf{2}}\mathbf{=}{\mathit{z}}^{\mathbf{2}}\mathbf{+}{\mathit{R}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{z}\mathit{R}\mathit{c}\mathit{o}\mathit{s}\mathit{θ} \\ \mathit{r}\mathbf{=}\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\mathbf{R}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{R}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}} \end{gathered}$$ …… (4)

By the law of cosines,

$$\begin{gathered} {\mathit{R}}^{\mathbf{2}}{\mathit{z}}^{\mathbf{2}}\mathbf{+}{\mathit{r}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{z}\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{Î\pm } \\ \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{Î\pm }\mathbf{=}\frac{{\mathbf{Z}}^{\mathbf{2}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{\mathbf{2}\mathbf{z}\mathbf{r}} \end{gathered}$$ …… (5)

Substitute $\mathit{r}\mathbf{=}\sqrt{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\mathbf{R}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{R}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}}$in equation (5)

$$\begin{gathered} \mathbf{cos}\mathrm{Î\pm }\mathbf{=}\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\left(\sqrt{z^2+R^2-2\mathrm{zR}\cos \mathrm{θ}}\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{\mathbf{2}\mathbf{z}\left(\sqrt{z^2+R^2-2\mathrm{zR}\cos \mathrm{θ}}\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{R}\mathbf{}\mathbf{cos}\mathbf{θ}}{\mathbf{2}\mathbf{z}\left(\sqrt{z^2+R^2-2zR\cos \mathrm{θ}}\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{θ}}{\left(\sqrt{z^2+R^2-2zR\cos \mathrm{θ}}\right)} \end{gathered}$$

Rewrite the equation,

$$\begin{gathered} \frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{Î\pm }}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}}{\left(\sqrt{z^2+R^2-2zRcos\mathrm{θ}}\right)}\left(\frac{1}{r^2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}}{\left(\sqrt{z^2+R^2-2zRcos\mathrm{θ}}\right)}\left(\frac{1}{{\left(\sqrt{z^2+R^2-2zRcos\mathrm{θ}}\right)}^2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}}{{\left(z^2+R^2-2zRcos\mathrm{θ}\right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}} \end{gathered}$$

Now put the value of $\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{Î\pm }}{{\mathbf{r}}^{\mathbf{2}}}$in equation (3)

localid="1658314561163" $\begin{array}{rcl}{\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{R}}^{\mathbf{2}}}\mathbf{âˆ\circledR }\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}\right)\left(\frac{z-R\cos \mathrm{θ}}{{\left(z^2+R^2-2zR\cos \mathrm{θ}\right)}^{\frac{3}{2}}}\right)\mathit{d}\mathit{a}\left(-\hat{z}\right)\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{R}}^{\mathbf{2}}}\mathbf{âˆ\circledR }\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}\right)\left(\frac{z-R\cos \mathrm{θ}}{{\left(z^2+R^2-2zR\cos \mathrm{θ}\right)}^{\frac{3}{2}}}\right)\left(R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{16{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\right)\left(2\mathrm{Ï€}\right)\underset{\mathbf{0}}{\overset{\mathbf{Ï€}}{\mathbf{∫}}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{cos}\mathbf{θ}}{{\left(z^2+R^2-2zR\cos \mathrm{θ}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathbf{sin}\mathit{θ}\mathit{d}\mathit{θ}\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\underset{\mathbf{0}}{\overset{\mathbf{Ï€}}{\mathbf{∫}}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{cos}\mathbf{θ}}{{\left(z^2+R^2-2zR\cos \mathrm{θ}\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathbf{sin}\mathit{θ}\mathit{d}\mathit{θ}\left(-\hat{z}\right)\\ & & \end{array}$

Now let’s consider that,

$\begin{array}{c}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}\mathbf{=}\mathbf{u}\\ \mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{d}\mathbf{θ}\mathbf{=}\mathbf{d}\mathbf{u}\\ \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{d}\mathbf{s}\mathbf{=}\mathbf{-}\mathbf{d}\mathbf{u}\end{array}$

Then the equation of electric field is modified as,

localid="1658314627313" $\begin{array}{rcl}{\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \mathbf{-}\left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\underset{\mathbf{1}}{\overset{\mathbf{-}\mathbf{1}}{\mathbf{∫}}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{u}}{{\left(z^2+R^2-2zRu\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathit{d}\mathit{u}\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\underset{\mathbf{-}\mathbf{1}}{\overset{\mathbf{1}}{\mathbf{∫}}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{u}}{{\left(z^2+R^2-2zRu\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathit{d}\mathit{u}\left(-\hat{z}\right)\\ & & \end{array}$

Solving the above integral we get,

${\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{1}{R}\left(\frac{1}{\left|z-R\right|}-\frac{1}{\left|z+R\right|}\right)-\frac{1}{2z^{2}R}\left(\left|z-R\right|-\left(z+R\right)+\left(R^2+z^2\right)\left(\frac{1}{z-R}-\frac{1}{z+R}\right)\right)\right)\left(-\hat{z}\right)$

Now, $\mathit{z}\mathbf{>}\mathbf{>}\mathit{R}$so that the terms becomes,

Therefore, the average electric field $\mathbf{\left(}\frac{\mathbf{q}}{\mathbf{8}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{(}\mathbf{}\widehat{\mathbf{-}}\mathbf{z}\mathbf{)}$over a spherical surface, due to charges outside the sphere, is the same as the field at the center.

b)

Write the expression for the electric filed in the region $\mathit{z}\mathbf{<}\mathit{R}$is modified as,

$\begin{array}{rcl}{\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{1}{R}\left(\frac{1}{z-R}-\frac{1}{z+R}\right)-\frac{1}{2z^{2}R}\left(z-R-\left(z+R\right)+\left(R^2+z^2\right)\left(\frac{1}{z-R}-\frac{1}{z+R}\right)\right)\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(\frac{1}{R}\left(\frac{2z}{R^2-z^2}\right)-\frac{1}{2z^{2}R}\left(-2z+\left(R^2+z^2\right)\left(\frac{2z}{R^2-z^2}\right)\right)\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\left(0\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \mathbf{0}\\ & & \end{array}$

Hence, the average due to charges inside the sphere is ${\stackrel{→}{E}}_{avg}=0$.