Given data

Two concentric metallic hollow spheres have radii: $R \quad \text{and} \quad 4R$ with charges: $Q_1 \quad \text{and} \quad Q_2$

Their surface charge densities are equal.

Use equality of surface charge densities

Surface charge density of a sphere is: $\sigma = \frac{Q}{4\pi a^2}$ where $a$ is the radius.

So for the inner sphere: $\sigma_1 = \frac{Q_1}{4\pi R^2}$

For the outer sphere: $\sigma_2 = \frac{Q_2}{4\pi (4R)^2} = \frac{Q_2}{64\pi R^2}$

Given $\sigma_1 = \sigma_2$, $\frac{Q_1}{4\pi R^2} = \frac{Q_2}{64\pi R^2}$

Hence, $Q_2 = 16Q_1$

Potential at $r=R$

Potential at the surface of inner sphere is due to both spheres.

• Due to inner sphere at its surface: $V_{11} = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R}$

• Due to outer sphere, inside a spherical shell potential is constant and equal to potential at its surface: $V_{12} = \frac{1}{4\pi\varepsilon_0}\frac{Q_2}{4R}$

Therefore, $V(R) = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{R} + \frac{Q_2}{4R}\right)$

Potential at $r=4R$

At the surface of outer sphere:

• Due to inner sphere: $V_{21} = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{4R}$

• Due to outer sphere at its own surface: $V_{22} = \frac{1}{4\pi\varepsilon_0}\frac{Q_2}{4R}$

Therefore, $V(4R) = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{4R} + \frac{Q_2}{4R}\right)$

Find the potential difference

$V(R)-V(4R) = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{R} + \frac{Q_2}{4R} - \frac{Q_1}{4R} - \frac{Q_2}{4R}\right)$

The $Q_2$ terms cancel: $V(R)-V(4R) = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{R} - \frac{Q_1}{4R}\right)$

$V(R)-V(4R) = \frac{1}{4\pi\varepsilon_0}\cdot \frac{3Q_1}{4R}$

$V(R)-V(4R) = \frac{3Q_1}{16\pi\varepsilon_0 R}$

Match with options

This corresponds to: $\boxed{\text{C}}$

Compare with stored correct answer

Stored correct answer is C, which matches the derived answer.