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(D) Let $r$ be the radius and $q$ be the charge of each small drop. Let $R$ be the radius and $Q$ be the charge of the large drop.Since $64$ drops coa

Vedclass · Accepted Answer

$1:4$

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(D) Let $r$ be the radius and $q$ be the charge of each small drop. Let $R$ be the radius and $Q$ be the charge of the large drop.Since $64$ drops coalesce,the total charge $Q = 64q$.The volume of the large drop is equal to the sum of the volumes of $64$ small drops: $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$,which gives $R^3 = 64r^3$,so $R = 4r$.The surface charge density $\sigma$ is defined as $\sigma = \frac{	ext{charge}}{	ext{area}} = \frac{q}{4\pi r^2}$.For a small drop,$\sigma_{small} = \frac{q}{4\pi r^2}$.For the large drop,$\sigma_{large} = \frac{Q}{4\pi R^2} = \frac{64q}{4\pi (4r)^2} = \frac{64q}{4\pi (16r^2)} = \frac{4q}{4\pi r^2}$.The ratio $\frac{\sigma_{small}}{\sigma_{large}} = \frac{q / 4\pi r^2}{4q / 4\pi r^2} = \frac{1}{4}$.

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small drop to that of the large drop?

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