Question

A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R (R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

• A. $\frac{1}{4\pi\varepsilon_{0}} \frac{\left(R-r\right)Q}{\left(R^{2}+r^{2}\right)}$
• B. $\frac{1}{4\pi\varepsilon_{0}} \frac{\left(R+r\right)Q}{2}\left(R^{2}+r^{2}\right)$
• C. $\frac{1}{4\pi\varepsilon_{0}} \frac{\left(R+r\right)Q}{\left(R^{2}+r^{2}\right)}$
• D. $\frac{1}{4\pi\varepsilon_{0}} \frac{\left(R-r\right)Q}{2}\left(R^{2}+r^{2}\right)$

Answer 1

Answer: (C)

$\frac{1}{4\pi\varepsilon_{0}} \frac{\left(R+r\right)Q}{\left(R^{2}+r^{2}\right)}$

Answer 2

Let $q_1$ and $q_2$ be charge on two spheres of radius $'r'$ and $'R'$ respectively As, $q_1+q_2=Q$ and $\sigma_{1}+\sigma_{2}$ [Surface charge density are equal] $\therefore \frac{q_{1}}{r\pi r^{2}}=\frac{q_{2}}{4\pi R^{2}}$ So, $q_{1}=\frac{Qr^{2}}{R^{2}+r^{2}}$ and $q_{2}=\frac{QR^{2}}{R^{2}+r^{2}}$ Now, potential, $V = \frac{1}{4\pi\varepsilon_{0}}\left[\frac{q_{1}}{r}+\frac{q_{2}}{R}\right]$ $=\frac{1}{4\pi \varepsilon _{0}} \left[\frac{Qr}{R^{2}+r^{2}}+\frac{Qr}{R^{2}+r^{2}}\right]$ $=\frac{Q\left(R+r\right)}{R^{2}+r^{2}} \frac{1}{4\pi\varepsilon_{0}}$