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Circular Kinematics

Two spherical...

Question

Two spherical conductors $A_{1}$ and $A_{2}$ of radii $r_{1}$ and $r_{2} (r_{2} > r_{1})$ 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

\frac{4 π ϵ_{0} r_{1} r_{2}}{r_{2} - r_{1}}

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B

4 π ϵ_{0} (r_{1} + r_{2})

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C

4 π ϵ_{0} r_{2}

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D

4 π ϵ_{0} r_{1}

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Solution

The correct option is A $\frac{4 π ϵ_{0} r_{1} r_{2}}{r_{2} - r_{1}}$
By Gauss's law the electric field at a distance r in between the shells is
$E .4 π r^{2} = \frac{Q}{ϵ_{0}} \Rightarrow E = \frac{Q}{4 π ϵ_{0} r^{2}}$
The potential difference between shells is $\int_{0}^{V} d V = - \int_{r_{2}}^{r_{1}} E d r$
or $V = - \frac{Q}{4 π ϵ_{0}} \int_{r_{2}}^{r_{1}} \frac{d r}{r^{2}} = \frac{Q}{4 π ϵ_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$
or $V = \frac{Q}{4 π ϵ_{0}} . \frac{r_{2} - r_{1}}{r_{1} r_{2}}$
$∴$ Capacitance, $C = \frac{Q}{V} = \frac{4 π ϵ_{0} r_{1} r_{2}}{r_{2} - r_{1}}$

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