Question 787: Electrostats & Capacitance

Question

' n ' small spherical drops of same size which are charged to ' $V$ ' volt each coalesce to form a single big drop. The potential of the big drop is

Accepted Answer

Let the radius of each small drop be $r$ and the potential of each drop is $V$.
Step 1: Potential of a small drop
The potential of a conducting sphere with charge $q$ and radius $r$ is
$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$
Step 2: Volume of $n$ drops and single big drop
The total volume before and after coalescing must be the same.
Volume of $n$ small drops:
$$n 	imes \frac{4}{3}\pi r^3$$
Let $R$ be the radius of the big drop. Its volume is:
$$\frac{4}{3}\pi R^3$$
Equating volumes:
$$n 	imes \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$
$$\Rightarrow n r^3 = R^3$$
$$\Rightarrow R = n^{1/3} r$$
Step 3: Charge on each small drop
Let charge on each small drop be $q$.
Using the formula for potential:
$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$
$$\Rightarrow q = 4\pi\epsilon_0 V r$$
Total charge after coalescence:
$$Q = n q = n \cdot 4\pi\epsilon_0 V r$$
Step 4: Potential of big drop
Potential of big drop:
$$V' = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$$
Substitute $Q$ and $R$:
$$V' = \frac{1}{4\pi\epsilon_0} \cdot \frac{n \cdot 4\pi\epsilon_0 V r}{n^{1/3} r}$$
Simplifying,
$$V' = \frac{n \cdot 4\pi\epsilon_0 V r}{4\pi\epsilon_0 n^{1/3} r}$$
$$V' = V \cdot n^{2/3}$$
Step 5: Final answer
The potential of the big drop is
$$\boxed{n^{2/3} V}$$
Correct option: D

Answer

$\frac{V}{n}$

Answer

$n \cdot V$

Answer

$n^{1 / 3} \cdot V$

Answer

$\mathrm{n}^{2 / 3} \cdot \mathrm{~V}$

Electrostats & Capacitance

' n ' small spherical drops of same size which are charged to ' V ' volt each coalesce to form a single big drop. The potential of the big drop is

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