Question

A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

Answer 1

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V  

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.