Q36P - Introduction to Electrodynamics

Question

Consider two concentric spherical shells, of radii a and b. Suppose the inner one carries a charge q , and the outer one a charge -q (both of them uniformly distributed over the surface). Calculate the energy of this configuration, (a) using Eq. 2.45, and (b) using Eq. 2.47 and the results of Ex. 2.9.

Step-by-Step Solution

Answer

(a) The energy of the given configuration is $\boxed{W=\frac{q^2}{8{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{a}-\frac{1}{b}\end{array}\right)}.$

(b)The energy of the given configuration is $\boxed{W=\frac{q^2}{8{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{a}-\frac{1}{b}\end{array}\right)}.$

Step 1: Determine Gauss Law

Consider the diagram for the condition.

                                                                   

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

 

Consider the electric field at point r between the two concentric spherical shells of radii a and b is expressed as follows using Gauss' law

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, q is the charge enclosed in Gaussian surface

Step 2: Determine the expression for the energy.

(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:

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

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}$for E and$4{\mathrm{\pi r}}^2\mathrm{dr}$for$d\tau$in above equation.

$W=\frac{{\epsilon }_0}{2}\int \left(\begin{array}{c}\frac{q}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{r^2}\end{array}\right) 4{\mathrm{\pi r}}^2\mathrm{dr}$

Integrate the above equation with limits a and b.

$$\begin{gathered} W=\frac{{\epsilon }_0}{2}{\left(\begin{array}{c}\frac{q}{4{\mathrm{\pi \epsilon }}_0}\end{array}\right)}^2{\int }_a^b{\left(\begin{array}{c}\frac{1}{r^2}\end{array}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}{\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{r}}^{-2}\mathrm{dr} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}{\left(\begin{array}{c}-\frac{1}{\mathrm{r}}\end{array}\right)}_{\mathrm{a}}^{\mathrm{b}} \\ =\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\end{array}\right) \end{gathered}$$

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

Step 3: Determine the total energy for the given configuration.

(b)

 

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

$$\begin{gathered} W=\frac{{\epsilon }_0}{2}\int E^{2}d\tau \\ =\frac{{\epsilon }_0}{2}\int {(E_1+E_2)}^{2}d\tau \\ =\frac{{\epsilon }_0}{2}\int (E_1+E_2+2E_1{E}_2)d\tau \\ =W_1+W_2+{\epsilon }_0\int E_1{E}_{2}d\tau \end{gathered}$$

Here,$W_1$is the 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 a and $E_2$Electric field due to spherical shell of radius b.

 

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

$E_1= \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}$

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

$W_1=\frac{{\epsilon }_0}{2}{\int }_a^{\alpha }{E_1}^{2}d\tau$

Substitute$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}$for $E_1$and $4{\mathrm{\pi r}}^2\mathrm{dr}$ for$d\tau$in above equation and integrate.$$\begin{gathered} W_1=\frac{{\epsilon }_0}{2}{\int }_a^{\infty }{\left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}\end{array}\right)}^2\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ =\frac{{\mathrm{\epsilon }}_0}{2}{\left(\begin{array}{c}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\end{array}\right)}^2{\int }_{\mathrm{a}}^{\infty }{\left(\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr} \\ =\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0}{\left(-\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}_{\mathrm{a}}^{\infty } \\ =\frac{1}{8{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}} \end{gathered}$$

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

$E_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{-q}{r^2}$

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

$W_2=\frac{{\epsilon }_0}{2}{\int }_b^{\infty }{E_2}^{2}d\tau$

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{-q}{r^2}$for $E_2$and$4{\mathrm{\pi r}}^2\mathrm{dr}$for$d\tau$in above equation and integrate the equation.

$$\begin{gathered} W_2=\frac{{\epsilon }_0}{2}{\int }_b^{\infty }{\left(\begin{array}{c}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\end{array}\frac{q}{r^2}\right)}^2\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ =\frac{{\mathrm{\epsilon }}_0}{2}{\left(\begin{array}{c}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\end{array}\right)}^2{\int }_{\mathrm{b}}^{\infty }{\left(\begin{array}{c}\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^{2}4{\mathrm{\pi r}}^2\mathrm{dr} \\ = \frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}{\left(\begin{array}{c}-\frac{1}{{\mathrm{r}}^2}\end{array}\right)}^2 \\ =\frac{1}{8{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{b}} \end{gathered}$$

 The product of two fields is now calculated as follows:

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

Integrate the above equation,

$$\begin{gathered} {\int }_b^{\infty }{E}_1-E_{2}d\tau ={\int }_b^{\infty }-{\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)}^2\frac{q^2}{r^4}d\tau \\ =-{\left(\begin{array}{c}\frac{q}{4{\mathrm{\pi \epsilon }}_0}\end{array}\right)}^2\left(4\mathrm{\pi }\right){\int }_b^{\infty }\frac{1}{r4}\left(r^2\right)dr \\ =-{\left(\begin{array}{c}\frac{q}{4{{\mathrm{\pi \epsilon }}_0}^2}\end{array}\right)}^2{\int }_b^{\infty }\left(r^{-2}\right)dr \\ =-{\left(\begin{array}{c}\frac{q}{4{{\mathrm{\pi \epsilon }}_0}^2}\end{array}\right)}^2{\left(-\frac{1}{r}\right)}_b^{\infty } \\ =-\frac{q^2}{4{{\mathrm{\pi \epsilon }}_0}^{2}b} \end{gathered}$$

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

Substitute the$\frac{1}{8{\mathrm{\pi \epsilon }}_0}\frac{q^2}{a}$for$W_1$and$\frac{1}{8{\mathrm{\pi \epsilon }}_0}\frac{q^2}{b}$for$W_2$and$-\frac{q^2}{4{{\mathrm{\pi \epsilon }}_0}^2\mathrm{b}}$for${\int }_b^{\infty }{E}_1.E_{2}d\tau$in equation.

$$\begin{gathered} W=W_1+W_2+{\epsilon }_0{\int }_b^{\infty }{E}_1.E_{2}d\tau . \\ W=\frac{q^2}{8{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{a}+\frac{1}{b}-\frac{2}{b}\right] \\ =\frac{q^2}{8{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{a}-\frac{1}{b}\right] \end{gathered}$$

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