Q35P - Introduction to Electrodynamics

Question

Here is a fourth way of computing the energy of a uniformly charged

solid sphere: Assemble it like a snowball, layer by layer, each time bringing in an infinitesimal charge $\mathit{d}\mathit{q}$from far away and smearing it uniformly over the surface, thereby increasing the radius. How much work$\mathit{d}\mathit{W}$does it take to build up the radius by an amount$\mathit{d}\mathit{r}$? Integrate this to find the work necessary to create the entire sphere of radius R and total charge q .

Step-by-Step Solution

Answer

The work done is$W=\frac{3q^2}{4{\mathrm{\pi \epsilon }}_0\left(5\mathrm{R}\right)}$

Step 1: Define function

Write the expression for the potential due to sphere or radius r on the surface is,

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

Here,$\overline{q}$is the total charge on sphere and r is the radius.

Step 2: Determine the expression for the change in the work.

Write the expression for work done,

$\mathit{d}\mathit{W}\mathbf{=}\mathit{d}\overline{\mathbf{q}}\mathit{V}$

Here,$d\overline{q}$is the charge brought far away from the sphere and V is the potential due to sphere r radius r on the surface.

Substitute$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\overline{q}}{r}$or V in above equation.

$dW=d\overline{q}\left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\overline{q}}{r}\end{array}\right)$

The total charge on a sphere of radius r is calculated as follows:

$\overline{q}=\frac{4}{3}{\mathrm{\pi r}}^{3}p$

Here, p is the volume charge density of sphere

 

The sphere's volume charge density is given by,

$p=\left(\begin{array}{c}\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\end{array}\right)$

Here, R is the radius of the sphere with charge .

Substitute$\left(\begin{array}{c}\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\end{array}\right)$for P in the equation.

 $$\begin{gathered} \overline{q}=\frac{4}{3}{\mathrm{\pi r}}^3\left(\begin{array}{c}\frac{\mathrm{q}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\end{array}\right) \\ =\mathrm{q}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3} \end{gathered}$$

Differentiate the above equation,

 

Substitute$\frac{3q}{R^3}{r}^{2}dr$for dr and$q\frac{r^3}{R^3}$for$\overline{q}$in the equation $dW=d\overline{q}\left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\overline{q}}{r}\end{array}\right)$

$$\begin{gathered} dW=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{q{r}^3}{R^3}\end{array}\right)\frac{1}{r}\left(\begin{array}{c}\frac{3q}{R^3}\end{array}{r}^{2}dr\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3q}{R^6}\end{array}{r}^{4}dr\right) \end{gathered}$$

Step 3: Determine expression for the work done.

The amount of effort required generating the whole sphere with radius and total charge is

$$\begin{gathered} W={\int }_0^R\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{3q^2}{R^6}{r}^{4}dr \\ =\frac{3q^2}{4{\mathrm{\pi \epsilon }}_0{R}^6}{\left(\begin{array}{c}\frac{r^5}{5}-0\end{array}\right)}_0^R \\ =\frac{3q^2}{4{\mathrm{\pi \epsilon }}_0{R}^6}\left(\begin{array}{c}\frac{R^5}{5}\end{array}\right) \\ =\frac{3q^2}{4{\mathrm{\pi \epsilon }}_0\left(5R\right)} \end{gathered}$$

Therefore, work done is$W=\frac{3q^2}{4{\mathrm{\pi \epsilon }}_0\left(5R\right)}.$