Question

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.)

• A. $2^{\frac{1}{3}} V_{0}$
• B. $4^{\frac{1}{3}} V_{0}$
• C. $6^{\frac{1}{3}} V_{0}$
• D. $2^{\frac{-1}{3}} V_{0}$

Answer 1

Answer: (B)

Given, charge of spherical drop of liquid $=Q$
and potential at its surface $=V_{0}$
If $r$ be the radius of drop, $V_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$ $\ldots( i )$
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., $\frac{4}{3} \pi R^{3} =\frac{4}{3} \pi r^{3}+\frac{4}{3} \pi r^{3}$
$R^{3} =2 r^{3} \Rightarrow R=2^{1 / 3} r$
$\therefore $ charge on the big drop, $Q^{\prime}=Q+Q=2 Q$
$\therefore $ Potential at the surface of big (new) drop,
$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} $
$=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r} \cdot 2^{2 / 3}=V_{0} \cdot 4^{1 / 3}=4^{1 / 3} \cdot V_{0}$