Question 353: Electrostatics & Capacitance

Question

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $$\mu$$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

Accepted Answer

Surface charge density of a spherical conductor is given by,
$$
\sigma=\frac{q}{4 \pi r^2}
$$
When all smaller drops combine, the radius of the bigger drop is given by
$\frac{4}{3} \pi R^3=64\left(\frac{4}{3} \pi r^3ight)$
$$
\begin{aligned}
\mathrm{R}^3 & =64 r^3 \\
\mathrm{R} & =4 r \\
\sigma^{\prime} & =\frac{64 q}{4 \pi(4 r)^2}=4 \sigma
\end{aligned}
$$
$$
	ext { Hence, } \frac{\sigma^{\prime}}{\sigma}=4: 1
$$

Answer

1 : 4

Answer

4 : 1

Answer

1 : 8

Answer

8 : 1

Electrostatics & Capacitance

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $\mu$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

Choose the correct answer: