512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ______V.

Explanation:

The potential on a charged particle with charge Q and radius R is given as,

V = \(\frac{Q}{4π \epsilon_o R}\)     -----(1)

Given,

No of mercury drops = 512,  Potential of each drop = 2 V

Let us assume the charge on the smaller drop is q and the radius is r.

The potential on one charge particle = \(\frac{q}{4π \epsilon_o r}\)

⇒\(\frac{q}{4π \epsilon_o r}\) = 2 ⇒\(\frac{1}{4π \epsilon_o }\)= \(\frac{2r}{q }\)  ----(2)

After joining the drops volume will be the same as the sum of the volume of individual drops. i.e.

\(\frac{4}{3 }\)πR³ = \(\frac{4}{3 }\)πr3 × 512

⇒R = 8r  ----(3)

From equations (1), (2) & (3) we have-

∴ Potential on all 512 charge particles i.e single bigger drop = \(\frac{512q}{4π \epsilon_o R}\)= \(\frac{512q}{q \times 8r}\times 2r\) = 128 V.

Hence the correct value of potential is 128 V.