JEE Main 2023 — Electrostatics Question with Solution

The problem involves 64 identical drops, each charged up to a potential of $$10\mathrm{mV}$$, that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.

The potential of each drop is given by:

$$V=\frac{Kq}{r}$$

where $$K$$ is the Coulomb constant, $$q$$ is the charge on each drop, and $$r$$ is the radius of each drop.

The radius of the bigger drop is:

$$R=4r$$

since the 64 identical drops combine to form a bigger drop.

The total charge on the 64 identical drops is:

$$Q=64q$$

The potential of the bigger drop is:

$$V_{bigger}=\frac{KQ}{R}=\frac{K(64q)}{4r}=16\frac{Kq}{r}=16V=16\times 10\mathrm{mV}=160\mathrm{mV}$$

Therefore, the potential of the bigger drop is $$160\mathrm{mV}$$.