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(A) Let the radius of each small drop be $r$ and the charge be $q$. The potential of each small drop is $V = \frac{q}{C}$,where $C = 4\pi \varepsilon_

Vedclass · Accepted Answer

$2^{2/3}\ V$

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(A) Let the radius of each small drop be $r$ and the charge be $q$. The potential of each small drop is $V = \frac{q}{C}$,where $C = 4\pi \varepsilon_0 r$.When two identical drops coalesce to form a larger drop of radius $R$ and charge $Q$,the total volume is conserved:$\frac{4}{3}\pi R^3 = 2 	imes \frac{4}{3}\pi r^3 \implies R^3 = 2r^3 \implies R = 2^{1/3}r$.The total charge on the larger drop is $Q = 2q$.The capacitance of the larger drop is $C' = 4\pi \varepsilon_0 R = 4\pi \varepsilon_0 (2^{1/3}r) = 2^{1/3}C$.The potential of the larger drop $V'$ is given by:$V' = \frac{Q}{C'} = \frac{2q}{2^{1/3}C} = \frac{2}{2^{1/3}} 	imes \frac{q}{C} = 2^{1 - 1/3} V = 2^{2/3} V$.

Two identical charged spherical drops,each at a potential $V$,coalesce to form a single larger drop. If the capacitance of each small drop is $C$,find the potential of the larger drop.

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