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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 new drop is (Assume, $V=0$ at infinity.)
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Verified Answer
The correct answer is:
$4^{\frac{1}{3}} V_0$
Given,
charge of spherical drop of liquid $=Q$
and potential at its surface $=V_0$
If $r$ be the radius of drop, $V_0=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$
If $R$ be the radius of big drop when two drops of radius $r$ are combined, then volume of big drop $=$ volume of both small drops
i.e.,
$$
\begin{aligned}
\frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\
R^3 & =2 r^3 \Rightarrow R=2^{1 / 3} r
\end{aligned}
$$
$\therefore$ charge on the big drop, $Q^{\prime}=Q+Q=2 Q$
$\therefore$ Potential at the surface of big (new) drop,
$$
\begin{aligned}
V^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{Q^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{2^{1 / 3} r} \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \cdot 2^{2 / 3}=V_0 \cdot 4^{1 / 3}=4^{1 / 3} \cdot V_0
\end{aligned}
$$
charge of spherical drop of liquid $=Q$
and potential at its surface $=V_0$
If $r$ be the radius of drop, $V_0=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$
If $R$ be the radius of big drop when two drops of radius $r$ are combined, then volume of big drop $=$ volume of both small drops
i.e.,
$$
\begin{aligned}
\frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\
R^3 & =2 r^3 \Rightarrow R=2^{1 / 3} r
\end{aligned}
$$
$\therefore$ charge on the big drop, $Q^{\prime}=Q+Q=2 Q$
$\therefore$ Potential at the surface of big (new) drop,
$$
\begin{aligned}
V^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{Q^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{2^{1 / 3} r} \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \cdot 2^{2 / 3}=V_0 \cdot 4^{1 / 3}=4^{1 / 3} \cdot V_0
\end{aligned}
$$
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