JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 10)
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 :
1 : 4
4 : 1
1 : 8
8 : 1
Explanation
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
$\frac{4}{3} \pi R^3=64\left(\frac{4}{3} \pi r^3\right)$
$$ \begin{aligned} \mathrm{R}^3 & =64 r^3 \\\\ \mathrm{R} & =4 r \\\\ \sigma^{\prime} & =\frac{64 q}{4 \pi(4 r)^2}=4 \sigma \end{aligned} $$
$$ \text { Hence, } \frac{\sigma^{\prime}}{\sigma}=4: 1 $$
$$ \sigma=\frac{q}{4 \pi r^2} $$
When all smaller drops combine, the radius of the bigger drop is given by
$\frac{4}{3} \pi R^3=64\left(\frac{4}{3} \pi r^3\right)$
$$ \begin{aligned} \mathrm{R}^3 & =64 r^3 \\\\ \mathrm{R} & =4 r \\\\ \sigma^{\prime} & =\frac{64 q}{4 \pi(4 r)^2}=4 \sigma \end{aligned} $$
$$ \text { Hence, } \frac{\sigma^{\prime}}{\sigma}=4: 1 $$
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