Question

(a) Show that the average electric field over a spherical surface, due to charges outside the sphere, is the same as the field at the center.

(b) What is the average due to charges inside the sphere?

Short answer

Answer

a) The average electric field $\left(\frac{q}{4\pi {\epsilon }_0{z}^2}\right)\left(\widehat{-}z\right)$over a spherical surface, due to charges outside the sphere, is the same as the field at the center.

b) The average due to charges inside the sphere is $\stackrel{\boxed{\mathbf{∆}\mathbf{∆}}}{{\mathbf{E}}_{\mathbf{a}\mathbf{v}\mathbf{g}}}\mathbf{=}\mathbf{0}$.

Step-by-step solution

Define function

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

$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}$ …… (1)

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$ 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.

${\mathit{E}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$ …… (2)

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

Determine the average electric field over a spherical surface

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.

$\begin{array}{rcl}\overrightarrow{{\mathbf{E}}_{\mathbf{a}\mathbf{v}\mathbf{g}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{R}}^{\mathbf{2}}}\mathbf{\oint }\overrightarrow{\mathbf{E}}\mathit{d}\mathit{a}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{R}}^{\mathbf{2}}}\mathbf{\oint }\left(\frac{q}{4\pi {\epsilon }_0{r}^2}\right)\mathbf{cos}\mathit{\alpha }\mathit{d}\mathit{a}\\ & & \end{array}$ …… (3)

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

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

By the law of cosines,

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

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

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

Rewrite the equation,

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

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

localid="1658314561163" $\begin{array}{rcl}{\overrightarrow{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{R}}^{\mathbf{2}}}\mathbf{\oint }\left(\frac{q}{4\pi {\epsilon }_0{r}^2}\right)\left(\frac{z-R\cos \theta }{{\left(z^2+R^2-2zR\cos \theta \right)}^{\frac{3}{2}}}\right)\mathit{d}\mathit{a}\left(-\hat{z}\right)\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{R}}^{\mathbf{2}}}\mathbf{\oint }\left(\frac{q}{4\pi {\epsilon }_0{r}^2}\right)\left(\frac{z-R\cos \theta }{{\left(z^2+R^2-2zR\cos \theta \right)}^{\frac{3}{2}}}\right)\left(R^2\sin \theta d\theta d\varphi \right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{16{\pi }^2{\epsilon }_0}\right)\left(2\pi \right)\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{cos}\mathbf{\theta }}{{\left(z^2+R^2-2zR\cos \theta \right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta }\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\pi {\epsilon }_0}\right)\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{cos}\mathbf{\theta }}{{\left(z^2+R^2-2zR\cos \theta \right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta }\left(-\hat{z}\right)\\ & & \end{array}$

Now let’s consider that,

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

Then the equation of electric field is modified as,

localid="1658314627313" $\begin{array}{rcl}{\overrightarrow{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \mathbf{-}\left(\frac{q}{8\pi {\epsilon }_0}\right)\underset{\mathbf{1}}{\overset{\mathbf{-}\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{u}}{{\left(z^2+R^2-2zRu\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathit{d}\mathit{u}\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\pi {\epsilon }_0}\right)\underset{\mathbf{-}\mathbf{1}}{\overset{\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{z}\mathbf{-}\mathbf{R}\mathbf{u}}{{\left(z^2+R^2-2zRu\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathit{d}\mathit{u}\left(-\hat{z}\right)\\ & & \end{array}$

Solving the above integral we get,

${\overrightarrow{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\left(\frac{q}{8\pi {\epsilon }_0}\right)\left(\frac{1}{R}\left(\frac{1}{\left|z-R\right|}-\frac{1}{\left|z+R\right|}\right)-\frac{1}{2z^{2}R}\left(\left|z-R\right|-\left(z+R\right)+\left(R^2+z^2\right)\left(\frac{1}{z-R}-\frac{1}{z+R}\right)\right)\right)\left(-\hat{z}\right)$

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

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

Determine the average due to charges inside the sphere

b)

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

$\begin{array}{rcl}{\overrightarrow{\mathbf{E}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \left(\frac{q}{8\pi {\epsilon }_0}\right)\left(\frac{1}{R}\left(\frac{1}{z-R}-\frac{1}{z+R}\right)-\frac{1}{2z^{2}R}\left(z-R-\left(z+R\right)+\left(R^2+z^2\right)\left(\frac{1}{z-R}-\frac{1}{z+R}\right)\right)\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\pi {\epsilon }_0}\right)\left(\frac{1}{R}\left(\frac{2z}{R^2-z^2}\right)-\frac{1}{2z^{2}R}\left(-2z+\left(R^2+z^2\right)\left(\frac{2z}{R^2-z^2}\right)\right)\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \left(\frac{q}{8\pi {\epsilon }_0}\right)\left(0\right)\left(-\hat{z}\right)\\ & \mathbf{=}& \mathbf{0}\\ & & \end{array}$

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