91Ó°ÊÓ

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

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

Here,$\mathrm{σ}$is the volume charge density, W is the stored energy V and is the potential.

(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, qis the charge of the solid sphere.

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

$Volume=\frac{4}{3}{\mathrm{Ï€¸é}}^3$

Here, Ris the radius of the solid sphere.

Now, substitute $\frac{4}{3}{\mathrm{Ï€¸é}}^3$for Volumeof the sphere in the equation $p=\frac{q}{Volume}$ and Solving for the P.

$$\begin{gathered} p=\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{Ï€¸é}}^3} \\ =\frac{3q}{4{\mathrm{Ï€¸é}}^3} \end{gathered}$$

Therefore, the charge density is $p=\frac{3q}{4{\mathrm{Ï€¸é}}^3}$.

(b)

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

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

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

$$\begin{gathered} d\mathrm{Ï„}={∫}_o^{∞}{∫}_o^x{∫}_o^{2x}rdrd\mathrm{θ}dâˆ\dots \\ =4{\mathrm{Ï€°ù}}^2\mathrm{dr} \end{gathered}$$

Substitute $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{R}\left(\begin{array}{c}\frac{3}{2}-\frac{r^2}{2R^2}\end{array}\right)$ for V , $\frac{3q}{4{\mathrm{Ï€¸é}}^2}$for Pand$4{\mathrm{Ï€°ù}}^2\mathrm{dr}$for$d\mathrm{Ï„}$in equation$W=\frac{1}{2}∫pVd\mathrm{Ï„}$ and solving for W.

$$\begin{gathered} W=\frac{3q^2}{16{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^6}{∫}_0^R(3R^2-r^2)r^{2}dr \\ =\frac{3q^2}{16{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^6}{\left(\begin{array}{c}3R^2\left(\begin{array}{c}\frac{r^3}{3}\end{array}\right)-\frac{r^5}{5}\end{array}\right)}_0^R \\ =\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{3}{5}\frac{q^2}{R}\right) \end{gathered}$$

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

(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,

$W=\frac{{\mathrm{Î\mu }}_0}{2}∫E^{2}d\mathrm{Ï„}$

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

The electric field of a conducting sphere is expressed as follows:localid="1654671404051" $E=\left\{\begin{array}{l}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{qr}{R^3},r<R(\mathrm{inside}\mathrm{the}\mathrm{sphere})\\ \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^3},r>R(out\mathrm{side}\mathrm{the}\mathrm{sphere})\end{array}\right.$

Now, Substitute the electric field values inside and outsides the sphere and$4{\mathrm{Ï€°ù}}^2\mathrm{dr}$for$d\mathrm{Ï„}$in the equation$W=\frac{{\mathrm{Î\mu }}_0}{2}∫E^{2}d\mathrm{Ï„}$and soling for W.localid="1654671414121" $$\begin{gathered} W=\frac{{\mathrm{Î\mu }}_0}{2}{∫}_0^R{\left(\begin{array}{c}\frac{qr}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^3}\end{array}\right)}^{2}4{\mathrm{Ï€°ù}}^2\mathrm{dr}+\frac{{\mathrm{Î\mu }}_0}{2}{∫}_0^{∞}{\left(\begin{array}{c}\frac{\mathrm{qr}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^3}\end{array}\right)}^{2}4{\mathrm{Ï€°ù}}^2\mathrm{dr} \\ =\left(\begin{array}{c}\begin{array}{c}\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^6}\end{array}{∫}_0^{\mathrm{R}}{\mathrm{r}}^2\mathrm{dr}+\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}{∫}_0^{\mathrm{R}}{\mathrm{r}}^{-2}\mathrm{dr}\end{array}\right) \end{gathered}$$

Simplifying the above equation,

localid="1654671482761" $$\begin{gathered} W=\left(\begin{array}{c}\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{R}}^6}\end{array}\left(\begin{array}{c}\frac{R^5}{5}\end{array}\right)+\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0}\left(\begin{array}{c}\frac{1}{R}\end{array}\right)\right) \\ =\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q^2}{2}\left(\frac{\begin{array}{c}1\end{array}}{5R}+\frac{1}{R}\right) \\ =\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\begin{array}{c}\frac{3}{5}\frac{q^2}{R}\end{array}\right) \end{gathered}$$

Thus, the total energy stored in the uniformly charged sphere is localid="1654671492285" $W=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\begin{array}{c}\frac{3}{5}\frac{q^2}{R}\end{array}\right).$

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

localid="1654671499692" $w=\frac{{\mathrm{Î\mu }}_0}{2}\left[{∫}_v{E}^{2}d\mathrm{Ï„}+{âˆ\circledR }_{s}VE.da\right]$

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

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

localid="1654671510116" $E=\left\{\begin{array}{l}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{qr}{R^3},r<R(\mathrm{inside}\mathrm{the}\mathrm{sphere})\\ \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2},r>R(out\mathrm{side}\mathrm{the}\mathrm{sphere})\end{array}\right.$

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

Substitute the expression for electric field for (r < R and r > R) in the equationlocalid="1654671519764" $W=\frac{{\mathrm{Î\mu }}_0}{2}\left[{∫}_V{E}^{2}d\mathrm{Ï„}+{âˆ\circledR }_{s}VE.da\right]$and simplifying.

Solve as further,

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

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

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