Solution

$V = \frac{1}{4\pi\epsilon_0}\cdot\frac{Q}{R}$ $\frac{1}{4\pi\epsilon_0}$ = constant and Q is the same (given), so $V \propto \frac{1}{R}$ Since $R_1 \gg R_2$, the potential is more on the smaller sphere.

Concept V = kq/R

Electrostatic Potential and Capacitance in past papers

76 questions from this chapter have appeared across 27 exam years.