Q28P - Introduction to Electrodynamics

Question

Use Eq. 2.29 to calculate the potential inside a uniformly charged

solid sphere of radius R and total charge q . Compare your answer to Pro b. 2.21.

Step-by-Step Solution

Answer

The potential inside a uniformly charged solid sphere is$\frac{q}{8\pi {\epsilon }_{0}R}\left(3-\frac{z^2}{R^2}\right).$

Step 1: Define the potential at the point P.

Consider the sphere for the given condition.

                                 

Write the formula for volume charge density of the sphere,

$\mathit{p}=\frac{q}{V}$

Here, q Charge on the solid sphere, V  is the volume of the sphere.

 

Write the potential to the infinitesimal element at the point P.

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{d\tau }}{\mathbf{r}}$

Here, r is the distance between point P and element and q is the constant charge density.

Step 2: Determine potential at point P inside solid sphere

$$\begin{gathered} V=\frac{1}{2{\epsilon }_0}\left(\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\right)\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{\left(8{\mathrm{\pi R}}^3\right){\mathrm{\epsilon }}_0}\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{3R^2}\right) \\ =\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{potential} \mathrm{inside} \mathrm{a} \mathrm{uniformly} \mathrm{charged} \mathrm{solid} \mathrm{sphere} \mathrm{is} \frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right). \end{gathered}$$Consider the spherical co-ordinates; an infinitesimal volume element is described as,

$\mathit{d}\mathit{\tau }\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{\phi }$

At point P the potential to the infinitesimal element is,

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{d}\mathbf{\tau }}{\mathbf{r}}$

Substitute $r^2 \sin \theta drd\theta d\phi \mathrm{for} d\tau \mathrm{and} \sqrt{r^2+z^2-2rz \cos \mathit{\theta }} \mathrm{for} r$for and for in above equation.

$V=\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\int }_{\varphi -0}^{2\mathrm{\pi }}d\varphi {\int }_{r=0}^R{r}^{2}dr{\int }_{r=0}^{\mathrm{\pi }}\frac{\sin \theta d\theta }{\sqrt{r^2+z^2-2rz \cos \theta }}$

Consider the expression.

$r^2+z^2-2rz \cos \mathit{\theta }=t$

Differentiate the above equation,

$$\begin{gathered} 2rz \sin \mathbf{ }\mathit{\theta }d\mathit{\theta } =dt \\ \sin \mathit{\theta }d\mathit{\theta } =\frac{dt}{2rz} \end{gathered}$$

If $\mathit{\theta }=0$, then solve as,

$$\begin{gathered} t=r^2+z^2-2rz\cos \left(0\right) \\ =r^2+z^2-2rz \\ ={(r-z)}^2 \end{gathered}$$

If $\theta =\mathrm{\pi }$, then solve as,

$$\begin{gathered} t=r^2+z^2-2rz \cos \left(\mathrm{\pi }\right) \\ ={\mathrm{r}}^2+{\mathrm{z}}^2+2\mathrm{rz} \\ ={(\mathrm{r}+\mathrm{z})}^2 \end{gathered}$$

Substitute $r^2+z^2-2rz \cos \mathit{\theta }\mathbf{ }\mathit{f}\mathit{o}\mathit{r} t \mathrm{and} \sin \mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{ }\mathrm{for} \frac{dt}{2rz}$values for and for into the equation.

$$\begin{gathered} {\int }_0^x\frac{\sin \theta }{\sqrt{r^2+z^2-2 rz \cos \theta }}d\theta =\frac{1}{2 rz}{\int }_{{(r-z)}^2}^{{(r+z)}^2}\frac{1}{\sqrt{t}}dt \\ =\frac{2}{2rz}{\left(\sqrt{t}\right)}_{{(r-z)}^2}^{{(r+z)}^2} \\ =\frac{1}{rz}(r+z-\left|r-z\right|) \end{gathered}$$

If r < z, then $\left|r-z\right|=-(r-z)$then, solve as,

$$\begin{gathered} {\int }_0^x\frac{\sin \mathbf{ }\mathit{\theta }}{\sqrt{r^2+z^2-2 rz \cos \mathit{\theta }}}d\mathit{\theta }\mathbf{ }\mathbf{=}\frac{1}{rz}(r+z+(r-z\left)\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } =\frac{2}{z} \\ \mathrm{If} \mathrm{r} > \mathrm{z} , \mathrm{then} \left|\mathrm{r}-\mathrm{z}\right|=(\mathrm{r}-\mathrm{z}) \mathrm{solve} \mathrm{as}, \\ {\int }_0^{\mathrm{x}}\frac{\sin \mathbf{ }\mathbf{\theta }}{\sqrt{{\mathrm{r}}^2+{\mathrm{z}}^2-2 \mathrm{rz} \cos \mathbf{\theta }}}\mathrm{d}\mathbf{\theta }\mathbf{ }\mathbf{=}\frac{1}{\mathrm{rz}}(\mathrm{r}+\mathrm{z}-(\mathrm{r}-\mathrm{z}\left)\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } =\frac{2}{\mathrm{z}} \end{gathered}$$

Thus, potential at point P inside solid sphere is,

$$\begin{gathered} V=\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\int }_{\varphi -0}^{2x}d\varphi {\int }_{r-0}^{r-R}{r}^{2}dr{\int }_{\theta -0}^x\frac{\sin \theta d\theta }{\sqrt{r^2+z^2-2rz cos \theta }} \\ =\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\left(\varphi \right)}_0^{2x}\left({\int }_{r-0}^{r-r}{r}^{2}dr\left(\frac{2}{z}\right)+{\int }_{r-z}^{r-R}{r}^{2}dr\left(\frac{2}{z}\right)\right) \\ =\frac{p}{4{\mathrm{\pi \epsilon }}_0}(2\mathrm{\pi }-0) \left(\left(\frac{2}{z}\right){\left(\frac{{\mathrm{r}}^3}{3}\right)}_0^{\mathrm{z}}+2{\left(\frac{r^2}{2}\right)}_z^R\right) \\ =\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_0} \left(\left(\frac{2}{\mathrm{z}}\right)\left(\frac{{\mathrm{z}}^3}{3}-0\right)+2\left(\frac{{\mathrm{R}}^2}{2}-\frac{{\mathrm{r}}^2}{2}\right)\right) \\ \mathrm{Solving} \mathrm{further}, \\ \mathrm{V}=\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left(\frac{2{\mathrm{z}}^2}{3}+({\mathrm{R}}^2-{\mathrm{z}}^2)\right) \\ =\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left({\mathrm{R}}^2-\frac{{\mathrm{z}}^2}{3}\right) \\ \mathrm{Substitute} \frac{\mathrm{q}}{{\displaystyle \frac{4}{3}}\mathbf{\pi }{\mathrm{R}}^3}\mathrm{for} \mathit{p} \mathrm{in} \mathrm{the} \mathrm{equation}. \end{gathered}$$

$$\begin{gathered} V=\frac{1}{2{\epsilon }_0}\left(\frac{q}{{\displaystyle \frac{4}{3}{\mathrm{\pi R}}^3}}\right)\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{\left(8{\mathrm{\pi R}}^3\right){\mathrm{\epsilon }}_0}\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{3R^2}\right) \\ =\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{potential} \mathrm{inside} \mathrm{a} \mathrm{uniformly} \mathrm{charged} \mathrm{solid} \mathrm{sphere} \mathrm{is} \frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right). \end{gathered}$$