Question

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

Answer 1

Step 1: Calculate the volume

Let v and V be the volume of small and large drop respectively.

r and R be the radius of small and large drop respectively.

q and Q be the charge on small and large drop respectively.

u and U be the potential of small and large drop respectively.

Volume of small drop, $v=\frac{4}{3}{\mathrm{\pi r}}^3$

Volume of large drop,$V=\frac{4}{3}{\mathrm{\pi R}}^3$

Volume of 512 small drops$=512\times \frac{4}{3}{\mathrm{\pi r}}^3$

Step 2: Calculate the radius of large drop

Volume of 512 small drops $=$ Volume of large drop

$$\begin{gathered} 512\times \frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\pi R^3 \\ \Rightarrow R=8r \end{gathered}$$

Step 3: Calculate the charge on large drop

Charge on large drop$=$$512\times$charge on small drop

$Q=512q$

Step 4: Calculate potential

Potential of small drop, $u=\frac{kq}{r}=2V$

Potential of large drop, $U=\frac{kQ}{R}=\frac{k\left(512q\right)}{8r}=64\frac{kq}{r}$

$$\begin{gathered} \Rightarrow U=64\times 2 \\ \Rightarrow U=128V \end{gathered}$$

Thus, the potential of the large drop is $\mathbf{128}V$.