Question

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$and $R\left(R>r\right)$. If the surface charge densities on the two shells are equal, the electric potential at the common centre is:

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• A. $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(2R+r\right)}{R^2+r^2}Q$
• B. $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{(R+r)}{R^2+r^2}Q$
• C. $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(R+r\right)}{2\left(R^2+r^2\right)}Q$
• D. $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(R+2r\right)}{2\left(R^2+r^2\right)}Q$

Answer 1

The correct option is B

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{(R+r)}{R^2+r^2}Q$

Step 1 : Given data

Charge of the spherical shell $=Q$

Radius of the spherical shell $=R$

Charge on the surface of the spherical shell of radius $R$be $q_A$ and charge on the surface of a sphere of radius $r$ be $q_B$

the surface densities are equal

$\frac{q_A}{R^2}=\frac{q_B}{r^2}$

Step 2: To equate the charge densities

$$\begin{gathered} q_A=\frac{Q{R}^2}{R^2+r^2} \\ {q}_B=\frac{Q{r}^2}{R+r^2} \end{gathered}$$

Step 3:To find the electric potential

Potential at the common center

$$\begin{gathered} V=\frac{k{q}_A}{R}+\frac{k{q}_B}{r} \\ V=\frac{kQ\left(R+r\right)}{\left(R^2+r^2\right)} \\ V=\frac{Q\left(R+r\right)}{4{\mathrm{\pi \epsilon }}_0\left({\mathrm{R}}^2+{\mathrm{r}}^2\right)} \end{gathered}$$

Hence the potential is $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(R+r\right)}{R^2+r^2}Q$

Hence the option B is correct