Two conducting spheres A and B of radii 'a' and 'b', respectively are charged to have the same electric potential. The ratio of surface charge densities of A and B is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \frac{b}{a} \)
The potential of a charged conducting sphere is \( V = kQ/r \), and its surface charge density is the charge spread over the surface, \( \sigma = Q/4\pi r^{2} \). At the same potential, the charge is proportional to the radius.
\( \displaystyle Q = \frac{V r}{k} \;\Rightarrow\; Q \propto r \)
\( \displaystyle \sigma = \frac{Q}{4\pi r^{2}} \propto \frac{r}{r^{2}} = \frac{1}{r} \)
\( \displaystyle \frac{\sigma_A}{\sigma_B} = \frac{b}{a} \)
So the smaller sphere carries the larger charge density, which is why charge crowds at the sharp points of a conductor.