Concentric metallic hollow spheres of radii $R$ and 4R hold charges $Q_{1}$ and $\mathrm{Q}_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V(R)-V(4 R)$ is:

(C) $\frac{3 Q_{1}}{16 \pi \in_{0} R}$

(A) $\frac{3 Q_{2}}{4 \pi \epsilon_{0} R}$

(B) $\frac{Q_{2}}{4 \pi \epsilon_{0} R}$

(D) $\frac{3 Q_{1}}{4 \pi \in_{0} R}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Electrostatics

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 (4 R)

is:

Answer: (C)

Equal surface charge densities give

\frac{Q _{1}}{4 π R ^{2}} = \frac{Q _{1} + Q _{2}}{4 π ( 4 R ) ^{2}}

, which simplifies to

16 Q_{1} = Q_{1} + Q_{2}

, so

Q_{2} = 15 Q_{1}

. The potential of each sphere is the sum of the contributions from both shells: at the inner sphere

V (R) = \frac{k Q _{1}}{R} + \frac{k Q _{2}}{4 R}

, and at the outer sphere

V (4 R) = \frac{k Q _{1}}{4 R} + \frac{k Q _{2}}{4 R}

. Subtracting, the

Q_{2}

terms cancel and

V (R) - V (4 R) = \frac{k Q _{1}}{R} - \frac{k Q _{1}}{4 R} = \frac{3 k Q _{1}}{4 R} = \frac{3 Q _{1}}{16 π ϵ _{0} R}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option