Question: 21: Two hollow conducting spheres of radii R_{1} and R_{2}\left(R_{1} \gg R_{2}\right) have equal charges. The potential would be
(1) Equal on both the spheres
(2) Dependent on the material property of the sphere
(3) More on bigger sphere
(4) More on smaller sphere
Answer: Option (4)
Explanation:
For a conducting sphere carrying charge Q, the electric potential on its surface is given by
V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}Here, both spheres carry equal charge Q, but their radii are different,
with R_1 \gg R_2.
The potential of the bigger sphere is
V_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R_1}The potential of the smaller sphere is
V_2 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R_2}Since R_1 \gg R_2, we have \frac{1}{R_2} > \frac{1}{R_1}.
Therefore,
V_2 > V_1This shows that the potential is higher for the smaller sphere when both spheres carry the same charge.
Hence, the potential is more on the smaller sphere.