91Ó°ÊÓ

Consider the diagram for the condition.

Here, qrepresents the charge on the inner shell, -q represents the charge on the outer shell, represents the configuration's inner radius, arepresents the configuration's outer radius, band r represents the Gaussian surface's radius and It is considered as (a < r < b).

Consider the electric field at point rbetween the two concentric spherical shells of radii aand bis expressed as follows using Gauss' law

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï„}\mathbf{Ï„}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, qis the charge enclosed in Gaussian surface

(a)

Equation 2.45 is used to calculate the energy for this configuration.

The energy of this configuration can be calculated using equation 2.45 as follows:

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

Substitute $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2}$for Eand$4{\mathrm{Ï€°ù}}^2\mathrm{dr}$for$d\mathrm{Ï„}$in above equation.

Integrate the above equation with limits aand b.

$$\begin{gathered} W=\frac{{\mathrm{Î\mu }}_0}{2}{\left(\begin{array}{c}\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\right)}^2{∫}_a^b{\left(\begin{array}{c}\frac{1}{r^2}\end{array}\right)}^{2}4{\mathrm{Ï€°ù}}^2\mathrm{dr} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}{∫}_{\mathrm{a}}^{\mathrm{b}}{\mathrm{r}}^{-2}\mathrm{dr} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}{\left(\begin{array}{c}-\frac{1}{\mathrm{r}}\end{array}\right)}_{\mathrm{a}}^{\mathrm{b}} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}\left(\begin{array}{c}\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\end{array}\right) \end{gathered}$$

Thus, the energy of the given configuration is$W=\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0}\left(\begin{array}{c}\frac{1}{a}-\frac{1}{b}\end{array}\right).$

(b)

The energy of this configuration can be calculated using equation 2.47 as follows:

$$\begin{gathered} W=\frac{{\mathrm{Î\mu }}_0}{2}∫E^{2}d\mathrm{Ï„} \\ =\frac{{\mathrm{Î\mu }}_0}{2}∫{(E_1+E_2)}^{2}d\mathrm{Ï„} \\ =\frac{{\mathrm{Î\mu }}_0}{2}∫(E_1+E_2+2E_1{E}_2)d\mathrm{Ï„} \\ =W_1+W_2+{\mathrm{Î\mu }}_0∫E_1{E}_{2}d\mathrm{Ï„} \end{gathered}$$

Here,$W_1$isthe energy of the spherical shell of radius a ,$W_2$is the energy of the spherical shell of radius b, $E_1$is the electric field due to spherical shell of radius aand $E_2$Electric field due to spherical shell of radius b.

The electric field due to a spherical shell of radius afor (r > a ) is expressed as follows according to Gauss's law:

$E_1=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2}$

The following is the energy of a spherical shell of radius a:

$W_1=\frac{{\mathrm{Î\mu }}_0}{2}{∫}_a^{\mathrm{Î\pm }}{E_1}^{2}d\mathrm{Ï„}$

Substitute$\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2}$for $E_1$and $4{\mathrm{Ï€°ù}}^2\mathrm{dr}$for$d\mathrm{Ï„}$in above equation and integrate.$$\begin{gathered} W_1=\frac{{\mathrm{Î\mu }}_0}{2}{∫}_a^{∞}{\left(\begin{array}{c}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2}\end{array}\right)}^2\left(4{\mathrm{Ï€°ù}}^2\mathrm{dr}\right) \\ =\frac{{\mathrm{Î\mu }}_0}{2}{\left(\begin{array}{c}\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\right)}^2{∫}_{\mathrm{a}}^{∞}{\left(\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^{2}4{\mathrm{Ï€°ù}}^2\mathrm{dr} \\ =\frac{\mathrm{q}}{8{\mathrm{Ï€Î\mu }}_0}{\left(-\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}_{\mathrm{a}}^{∞} \\ =\frac{1}{8{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}} \end{gathered}$$

The electric field due to a spherical shell of radius afor (r > b ) is expressed as follows according to Gauss's law:

$E_2=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{-q}{r^2}$

The following is the energy of a spherical shell of radius b:

$W_2=\frac{{\mathrm{Î\mu }}_0}{2}{∫}_b^{∞}{E_2}^{2}d\mathrm{Ï„}$

Substitute $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{-q}{r^2}$for $E_2$and$4{\mathrm{Ï€°ù}}^2\mathrm{dr}$for$d\mathrm{Ï„}$in above equation and integrate the equation.

$$\begin{gathered} W_2=\frac{{\mathrm{Î\mu }}_0}{2}{∫}_b^{∞}{\left(\begin{array}{c}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\frac{q}{r^2}\right)}^2\left(4{\mathrm{Ï€°ù}}^2\mathrm{dr}\right) \\ =\frac{{\mathrm{Î\mu }}_0}{2}{\left(\begin{array}{c}\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\right)}^2{∫}_{\mathrm{b}}^{∞}{\left(\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^{2}4{\mathrm{Ï€°ù}}^2\mathrm{dr} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}{\left(\begin{array}{c}-\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^2 \\ =\frac{1}{8{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}^2}{\mathrm{b}} \end{gathered}$$

The product of two fields is now calculated as follows:

$E_1-E_2=-{\left(\begin{array}{c}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\right)}^2\frac{q^2}{r^4}$

Integrate the above equation,

$$\begin{gathered} {∫}_b^{∞}{E}_1-E_{2}d\mathrm{Ï„}={∫}_b^{∞}-{\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)}^2\frac{q^2}{r^4}d\mathrm{Ï„} \\ =-{\left(\begin{array}{c}\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\end{array}\right)}^2\left(4\mathrm{Ï€}\right){∫}_b^{∞}\frac{1}{r4}\left(r^2\right)dr \\ =-{\left(\begin{array}{c}\frac{q}{4{{\mathrm{Ï€Î\mu }}_0}^2}\end{array}\right)}^2{∫}_b^{∞}\left(r^{-2}\right)dr \\ =-{\left(\begin{array}{c}\frac{q}{4{{\mathrm{Ï€Î\mu }}_0}^2}\end{array}\right)}^2{\left(-\frac{1}{r}\right)}_b^{∞} \\ =-\frac{q^2}{4{{\mathrm{Ï€Î\mu }}_0}^{2}b} \end{gathered}$$

Equation 2.47 is used to calculate the energy for this configuration:

Substitute the$\frac{1}{8{\mathrm{Ï€Î\mu }}_0}\frac{q^2}{a}$for$W_1$and$\frac{1}{8{\mathrm{Ï€Î\mu }}_0}\frac{q^2}{b}$for$W_2$and$-\frac{q^2}{4{{\mathrm{Ï€Î\mu }}_0}^2\mathrm{b}}$for${∫}_b^{∞}{E}_1.E_{2}d\mathrm{Ï„}$in equation.

$$\begin{gathered} W=W_1+W_2+{\mathrm{Î\mu }}_0{∫}_b^{∞}{E}_1.E_{2}d\mathrm{Ï„}. \\ W=\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0}\left[\frac{1}{a}+\frac{1}{b}-\frac{2}{b}\right] \\ =\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0}\left[\frac{1}{a}-\frac{1}{b}\right] \end{gathered}$$

Therefore, the energy of the given configuration is$\boxed{W=\frac{q^2}{8{\mathrm{Ï€Î\mu }}_0}\left[\frac{1}{b}-\frac{1}{b}\right]}$