A small mercury drop is charged such that its surface charge density is 2 μC/m2. Exactly 125 such drops are combined to form a big drop. What is the surface charge density of the big drop?

For one drop, if the radius is r and the charge is q, thenσ=q4πr2=2μC/cm2When 125 drops are combined, the charge on it becomes 125 q and if its radius is R, then125×43πr3=43πR3⇒R=5rCharge density of the big drop isσ1=125q4πR2=125q4π×25r2=5q4πr2=(5×2)μC/cm2