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 :
Solution
Surface charge density of a spherical conductor is given by,
<br/><br/>$\sigma=\frac{q}{4 \pi r^2}$
<br/><br/>When all smaller drops combine, the radius of the bigger drop is given by
<br/><br/>$\frac{4}{3} \pi R^3=64\left(\frac{4}{3} \pi r^3\right)$
<br/><br/>$$
\begin{aligned}
\mathrm{R}^3 & =64 r^3 \\\\
\mathrm{R} & =4 r \\\\
\sigma^{\prime} & =\frac{64 q}{4 \pi(4 r)^2}=4 \sigma
\end{aligned}
$$
<br/><br/>$\text { Hence, } \frac{\sigma^{\prime}}{\sigma}=4: 1$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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