Eight charged water droplets, each of radius 1 mm and carrying a charge $5×10^{-10} C$, coalesce to form a single drop. The potential of the bigger drop is.

18000 V

The correct answer is Option (2) → 18000 V

Radius of each small drop: $r = 1 \, \text{mm} = 10^{-3} \, \text{m}$

Charge on each drop: $q = 5 \times 10^{-10} \, \text{C}$

Number of drops: $n = 8$

When $n$ drops coalesce:

$R^3 = n r^3 \;\;\Rightarrow\;\; R = r \, n^{1/3}$

$R = 10^{-3} \times 8^{1/3} = 10^{-3} \times 2 = 2 \times 10^{-3} \, \text{m}$

Total charge on bigger drop:

$Q = n q = 8 \times 5 \times 10^{-10} = 4 \times 10^{-9} \, \text{C}$

Potential of spherical drop:

$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}$

$V = \frac{9 \times 10^9 \cdot 4 \times 10^{-9}}{2 \times 10^{-3}}$

$V = \frac{36}{2 \times 10^{-3}} = 18 \times 10^3 \, \text{V}$

Answer: Potential of the bigger drop is $1.8 \times 10^4 \, \text{V}$