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, $V_0=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$ $\dots (i)$
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., $\frac{4}{3}\pi R^3=\frac{4}{3}\pi r^3+\frac{4}{3}\pi r^3$
$R^3=2r^3\Rightarrow R=2^{1/3}r$
$\therefore$ charge on the big drop, $Q^{\prime }=Q+Q=2Q$
$\therefore$ Potential at the surface of big (new) drop,
$V^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{Q^{\prime }}{R}=\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{2^{1/3}r}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}\cdot 2^{2/3}=V_0\cdot 4^{1/3}=4^{1/3}\cdot V_0$