Q34P - Introduction to Electrodynamics

Question

Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:

(a)Use Eq. 2.43. You found the potential in Prob. 2.21.

(b)Use Eq. 2.45. Don't forget to integrate over all space.

(c)Use Eq. 2.44. Take a spherical volume of radius  a . What happens as $a\to \infty$?

Step-by-Step Solution

Answer

(a)The total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{4}\frac{q^2}{R}\end{array}\right)$.

(b)The total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{4}\frac{q^2}{R}\end{array}\right)$.

(c)The total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{4}\frac{q^2}{R}\end{array}\right)$.

The contribution due to the surface integral becomes small as the value$a\to \infty$approaches zero.

Step 1: Determine the expression for the energy stored.

Write the expression for the amount of energy stored in a uniformly charged sphere of radius R and charge q.

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\mathit{p}\mathit{V}\mathbf{ }\mathit{d}\mathit{\tau }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here,$\sigma$is the volume charge density, W  is the stored energy V and is the potential.

Step 2: Solve for part (a) determine the charge density.

(a)

 

The charge per unit volume of the sphere is defined as its volume charge density. It can be expressed in the following way:

$p=\frac{q}{\mathrm{Volume}}$

Here, q is the charge of the solid sphere.

 

The cube of the radius of the volume determines the volume of the sphere.

$Volume=\frac{4}{3}{\mathrm{\pi R}}^3$

Here, R is the radius of the solid sphere.

Now, substitute $\frac{4}{3}{\mathrm{\pi R}}^3$for Volume of the sphere in the equation $p=\frac{q}{Volume}$ and Solving for the P.

$$\begin{gathered} p=\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ =\frac{3q}{4{\mathrm{\pi R}}^3} \end{gathered}$$

Therefore, the charge density is $p=\frac{3q}{4{\mathrm{\pi R}}^3}$.

Step 3: Determine stored energy.

(b)

 

Using the result of problem 2.21, a charged sphere of radius has the following potential:

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

Write the expression for change in the volume of the sphere is,

$$\begin{gathered} d\tau ={\int }_o^{\infty }{\int }_o^x{\int }_o^{2x}rdrd\theta d\varnothing \\ =4{\mathrm{\pi r}}^2\mathrm{dr} \end{gathered}$$

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{R}\left(\begin{array}{c}\frac{3}{2}-\frac{r^2}{2R^2}\end{array}\right)$ for V , $\frac{3q}{4{\mathrm{\pi R}}^2}$for P and$4{\mathrm{\pi r}}^2\mathrm{dr}$for$d\tau$in equation$W=\frac{1}{2}\int pV d\tau$ and solving for W.

$$\begin{gathered} W=\frac{3q^2}{16{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^6}{\int }_0^R(3R^2-r^2)r^{2}dr \\ =\frac{3q^2}{16{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^6}{\left(\begin{array}{c}3R^2\left(\begin{array}{c}\frac{r^3}{3}\end{array}\right)-\frac{r^5}{5}\end{array}\right)}_0^R \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{3}{5}\frac{q^2}{R}\right) \end{gathered}$$

Thus, the total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{3}{5}\frac{q^2}{R}\right)$

Step 4: Determine the stored energy in the uniformly charged sphere.

(c)

Using the equation 2.45, determine the amount of energy contained in an evenly charged solid sphere.

Write the expression for the energy stored in the sphere is,

$W=\frac{{\epsilon }_0}{2}\int E^2 d\tau$

Here, E is the electric field intensity and is the permittivity of the free space.

 

The electric field of a conducting sphere is expressed as follows:$E=\left\{\begin{array}{l}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qr}{R^3},r<R(\mathrm{inside} \mathrm{the} \mathrm{sphere})\\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^3},r>R(out\mathrm{side} \mathrm{the} \mathrm{sphere})\end{array}\right.$

Now, Substitute the electric field values inside and outsides the sphere and$4{\mathrm{\pi r}}^2\mathrm{dr}$for$d\tau$in the equation$W=\frac{{\epsilon }_0}{2}\int E^2 d\tau$and soling for W.$$\begin{gathered} W=\frac{{\epsilon }_0}{2}{\int }_0^R{\left(\begin{array}{c}\frac{qr}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}\end{array}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr}+\frac{{\mathrm{\epsilon }}_0}{2}{\int }_0^{\infty }{\left(\begin{array}{c}\frac{\mathrm{qr}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}\end{array}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr} \\ =\left(\begin{array}{c}\begin{array}{c}\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^6}\end{array}{\int }_0^{\mathrm{R}}{\mathrm{r}}^2\mathrm{dr} +\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}{\int }_0^{\mathrm{R}}{\mathrm{r}}^{-2}\mathrm{dr}\end{array}\right) \end{gathered}$$

Simplifying the above equation,

$$\begin{gathered} W=\left(\begin{array}{c}\frac{q^2}{8{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^6}\end{array}\left(\begin{array}{c}\frac{R^5}{5}\end{array}\right)+\frac{q^2}{8{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{R}\end{array}\right)\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q^2}{2}\left(\frac{\begin{array}{c}1\end{array}}{5R}+\frac{1}{R}\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{5}\frac{q^2}{R}\end{array}\right) \end{gathered}$$

Thus, the total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{5}\frac{q^2}{R}\end{array}\right).$

Using the equation 2.44, determine the amount of energy stored in an evenly charged solid sphere.

                            $w=\frac{{\epsilon }_0}{2}\left[{\int }_v{E}^{2}d\tau +{\oint }_{s}VE.da\right]$

Here, W is the large enough to enclose all the charge, E is the electric field, V is the Voltage and da is the small area.

 

The electric field of a conducting sphere is expressed as follows:

$E=\left\{\begin{array}{l}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qr}{R^3},r<R(\mathrm{inside} \mathrm{the} \mathrm{sphere})\\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2},r>R(out\mathrm{side} \mathrm{the} \mathrm{sphere})\end{array}\right.$

Let's use a radius of sphere a > R.

 

Substitute the expression for electric field for (r < R and r > R) in the equation$W=\frac{{\epsilon }_0}{2}\left[{\int }_V{E}^{2}d\tau +{\oint }_{s}VE . da\right]$ and simplifying.

Solve as further,

$$\begin{gathered} W=\frac{q^2}{40{\mathrm{\pi \epsilon }}_0\mathrm{R}}+\frac{q^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{R}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{3}{4}\frac{q^2}{R}\end{array}\right) \end{gathered}$$

Hence, the total energy stored in the uniformly charged sphere is $W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{3}{5}\begin{array}{c}\frac{q^2}{R}\end{array}\right)$

The contribution due to the surface integral becomes small as the value$a\to \infty$approaches zero.