A spherical drop of liquid carrying charge Q | vedclass

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(B) Given,charge of each small spherical drop $= Q$ and potential at its surface $= V_0$. Let $r$ be the radius of each small drop. The potential is g

Vedclass · Accepted Answer

$4^{1/3} V_0$

Vedclass · Answer

(B) Given,charge of each small spherical drop $= Q$ and potential at its surface $= V_0$. Let $r$ be the radius of each small drop. The potential is given by $V_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.When two such drops combine to form a single large drop of radius $R$,the volume is conserved: $\frac{4}{3} \pi R^3 = 2 	imes \frac{4}{3} \pi r^3$,which gives $R^3 = 2r^3$ or $R = 2^{1/3} r$.The total charge on the new large drop is $Q' = Q + Q = 2Q$.The potential at the surface of the new drop is $V' = \frac{1}{4 \pi \varepsilon_0} \frac{Q'}{R} = \frac{1}{4 \pi \varepsilon_0} \frac{2Q}{2^{1/3} r}$.Substituting $V_0 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$,we get $V' = V_0 	imes \frac{2}{2^{1/3}} = V_0 	imes 2^{1 - 1/3} = V_0 	imes 2^{2/3} = V_0 	imes (2^2)^{1/3} = 4^{1/3} V_0$.

$A$ spherical drop of liquid carrying charge $Q$ has potential $V_0$ at its surface. If two drops of same charge and radius combine to form a single spherical drop,then the potential at the surface of the new drop is (Assume $V=0$ at infinity.)

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