Concentric metallic hollow spheres of radii R and 4R hold charges $$Q_1$$ and $$Q_2$$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) - V(4R) is:

We have two concentric metallic hollow spheres. The inner sphere has radius $$R$$ and carries charge $$Q_1$$. The outer sphere has radius $$4R$$ and carries charge $$Q_2$$.

Because the spheres are metallic and the charges reside only on their outer surfaces, the surface charge density of each sphere is given by the usual relation

$$\sigma = \dfrac{\text{total charge}}{\text{surface area}}.$$

For the inner sphere, the surface charge density is

$$\sigma_1=\dfrac{Q_1}{4\pi R^{2}}.$$

For the outer sphere, the surface charge density is

$$\sigma_2=\dfrac{Q_2}{4\pi (4R)^{2}}=\dfrac{Q_2}{4\pi \,16R^{2}}=\dfrac{Q_2}{64\pi R^{2}}.$$

The problem states that the two surface charge densities are equal, so

$$\sigma_1=\sigma_2\quad\Longrightarrow\quad \dfrac{Q_1}{4\pi R^{2}}=\dfrac{Q_2}{64\pi R^{2}}.$$

The factors $$4\pi R^{2}$$ cancel on both sides, leaving

$$Q_1=\dfrac{Q_2}{16}\quad\Longrightarrow\quad Q_2=16Q_1.$$

Now we evaluate the electric potential at two points: on the surface of the inner sphere (at radial distance $$r=R$$) and on the surface of the outer sphere (at $$r=4R$$). For any spherical shell, the standard electrostatic result is:

• Outside the shell, the potential behaves as if all charge were concentrated at the centre, i.e. $$V=\dfrac{kQ}{r}$$ with $$k=\dfrac{1}{4\pi\varepsilon_0}.$$
• Inside the shell, the potential is constant everywhere and equal to the value at the surface, i.e. $$V=kQ/R_{\text{shell}}.$$

Potential at $$r=R$$ (surface of the inner sphere)

1. Contribution of the inner sphere itself (point on its surface): $$V_{1\to R}=k\dfrac{Q_1}{R}.$$
2. Contribution of the outer sphere (point lies inside that shell, so potential is constant and equals the surface value of the outer shell): $$V_{2\to R}=k\dfrac{Q_2}{4R}.$$

Adding them,

$$V(R)=k\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{4R}\right).$$

Potential at $$r=4R$$ (surface of the outer sphere)

1. Contribution of the inner sphere (point is outside the inner shell): $$V_{1\to 4R}=k\dfrac{Q_1}{4R}.$$
2. Contribution of the outer sphere itself (point on its surface): $$V_{2\to 4R}=k\dfrac{Q_2}{4R}.$$

Thus,

$$V(4R)=k\left(\dfrac{Q_1}{4R}+\dfrac{Q_2}{4R}\right) =k\dfrac{Q_1+Q_2}{4R}.$$

Required potential difference

Subtracting the two results,

$$\begin{aligned} V(R)-V(4R) & = k\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{4R}\right) \;-\;k\left(\dfrac{Q_1+Q_2}{4R}\right) \\[4pt] & = k\left[\dfrac{Q_1}{R}-\dfrac{Q_1}{4R} +\dfrac{Q_2}{4R}-\dfrac{Q_2}{4R}\right] \\[6pt] & = k\left[\dfrac{4Q_1-Q_1}{4R}\right] \\[4pt] & = k\dfrac{3Q_1}{4R}. \end{aligned}$$

Finally, substituting $$k=\dfrac{1}{4\pi\varepsilon_0}$$, we obtain

$$V(R)-V(4R)=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{3Q_1}{4R} =\dfrac{3Q_1}{16\pi\varepsilon_0 R}.$$

This expression matches Option C.

Hence, the correct answer is Option C.