Question

One hundred twenty five small liquid drops, each carrying a charge of 0.5 µC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.

Answer 1

Data : n = 125, q = 0.5 × 10^-6 C, d = 0.1 m
The radius of each small drop, r = d/2 = 0.05 m
The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is
R = \(\sqrt[3]{n} r\) = \(\sqrt[3]{125}\) (0.05) = 5 × 0.05 = 0.25 m
The charge on the larger drop,
Q = nq = 125 × (0.5 × 10^-6) C
∴ The electric potential of the surface of the larger drop,
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}\) = (9 × 10⁹) × \(\frac{125 \times\left(0.5 \times 10^{-6}\right)}{0.25}\)
= 9 × 125 × 2 × 10³ = 2.25 × 10⁶ V