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(B) Let $n = 64$ be the number of small droplets.Let $r$ be the radius of each small droplet and $q$ be the charge on each.Let $R$ be the radius of th

Vedclass · Accepted Answer

$1 : 4$

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(B) Let $n = 64$ be the number of small droplets.Let $r$ be the radius of each small droplet and $q$ be the charge on each.Let $R$ be the radius of the large drop and $Q$ be the charge on it.Since the volume is conserved,$\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $R = n^{1/3} r$.Since the charge is conserved,$Q = nq$.The surface charge density $\sigma$ is defined as $\sigma = \frac{	ext{charge}}{	ext{area}} = \frac{q}{4\pi r^2}$.For the small droplet,$\sigma_{	ext{small}} = \frac{q}{4\pi r^2}$.For the large drop,$\sigma_{	ext{large}} = \frac{Q}{4\pi R^2} = \frac{nq}{4\pi (n^{1/3}r)^2} = \frac{nq}{4\pi n^{2/3}r^2} = n^{1/3} \frac{q}{4\pi r^2} = n^{1/3} \sigma_{	ext{small}}$.Therefore,the ratio $\frac{\sigma_{	ext{small}}}{\sigma_{	ext{large}}} = \frac{1}{n^{1/3}}$.Substituting $n = 64$,we get $\frac{\sigma_{	ext{small}}}{\sigma_{	ext{large}}} = \frac{1}{(64)^{1/3}} = \frac{1}{4}$.Thus,the ratio is $1 : 4$.

$64$ small mercury droplets,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 droplet to that of the large drop?

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