64 small drops of mercury, each of radius r | vedclass

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Question

(d) $\frac{{{\sigma _{small}}}}{{{\sigma _{Big}}}} = \frac{q}{Q} 	imes \frac{{{R^2}}}{{{r^2}}} = \frac{q}{{(nq)}} 	imes \frac{{{{({n^{1/3}}r)}^2}}}{

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(d) $\frac{{{\sigma _{small}}}}{{{\sigma _{Big}}}} = \frac{q}{Q} 	imes \frac{{{R^2}}}{{{r^2}}} = \frac{q}{{(nq)}} 	imes \frac{{{{({n^{1/3}}r)}^2}}}{{{r^2}}}$$ = {n^{ - 1/3}} = {(64)^{ - 1/3}} = \frac{1}{4}$

$64$ small drops of mercury, each of radius $ r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is

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