Question

Two spherical conductors \[{A_1}\] and \[{A_2}\] of radii \[{r_1}\] and \[{r_2}\]\[\left( {{r_2} > {r_1}} \right)\]are placed concentrically in air. \[{A_1}\]is given a charge \[ + Q\] while \[{A_2}\] is earthed. Then the equivalent capacitance of the system is:

A. \[\dfrac{{4\pi {\varepsilon _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}\]
B. \[4\pi {\varepsilon _0}\left( {{r_1} + {r_2}} \right)\]
C. \[4\pi {\varepsilon _0}{r_2}\]
D. \[4\pi {\varepsilon _0}{r_1}\]

Answer 1

Hint: Formula for capacitance is given by, \[C = \dfrac{Q}{V}\]. Here, \[Q\] is the charge. Apply Gauss’s law to obtain the value of the electric field \[E\] in terms of \[\dfrac{Q}{V}\].

Complete step by step solution:
Given, the radii of the conductors \[{A_1}\]and \[{A_2}\] are \[{r_1}\]and \[{r_2}\] respectively.
Here, \[{r_2} > {r_1}\]
By Gauss's law the electric field at a point \[r\] in between the shells is given by,
\[E4\pi {r^2} = \dfrac{Q}{{{\varepsilon _0}}}\]
Rearrange the above equation in terms of electric field \[E\].
\[E = \dfrac{Q}{{4\pi {r^2}{\varepsilon _0}}}\]

Now, the potential difference between the shells is given by,
\[\int_0^V {dV} = - \int_{{r_2}}^{{r_1}} {Edr} \]

Place the value of \[E = \dfrac{Q}{{4\pi {r^2}{\varepsilon _0}}}\] in the above equation and simplify.
\[
  \int_0^V {dV} = - \int_{{r_2}}^{{r_1}} {\dfrac{Q}{{4\pi {r^2}{\varepsilon _0}}}dr} \\
  V = - \dfrac{Q}{{4\pi {\varepsilon _0}}}\int_{{r_2}}^{{r_1}} {\dfrac{{dr}}{{{r^2}}}} \\
  V = - \dfrac{Q}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_2}}} - \dfrac{1}{{{r_1}}}} \right) \\
  V = \dfrac{Q}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}}} \right) \\
  V = \dfrac{Q}{{4\pi {\varepsilon _0}}}\dfrac{{{r_2} - {r_1}}}{{{r_1}{r_2}}} \\
 \] …… (i)

Capacitance is the ratio of a system's change in electrical charge and the resulting change in its electrical potential.
The formula for capacitance is given by,
\[C = \dfrac{Q}{V}\] …… (ii)
Rearrange the equation (i) in terms of \[\dfrac{Q}{V}\] and we get,
\[\dfrac{Q}{V} = \dfrac{{4\pi {\varepsilon _0} \times {r_1}{r_2}}}{{{r_2} - {r_1}}}\]

Place the above value in equation (ii)
\[C = \dfrac{{4\pi {\varepsilon _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}\]

Hence, the correct answer is option A.

Additional information:
Gauss’s law: The electrical flux sum of a closed surface is equal to the enclosed charge separated by permittivity. The electric flux through a region is known as the electric field multiplied in a plane perpendicular to the field by the area of the surface predicted. Simply put, Gauss' law states that the overall electric charge within a container can be determined by measuring the net electric field flux flowing out of a sealed surface covering the volume.

Note: In this question we are asked to determine the value of capacitance which is given by, \[C = \dfrac{Q}{V}\]. The electrical flux sum of a closed surface is equal to the enclosed charge separated by permittivity. i.e. \[E4\pi {r^2} = \dfrac{Q}{{{\varepsilon _0}}}\]. The potential difference between the shells is given by, \[\int_0^V {dV} = - \int_{{r_2}}^{{r_1}} {Edr} \]. Solve the equation in terms of \[\dfrac{Q}{V}\].