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 :
[tex]V=\dfrac{kq}{r}[/tex]
r is the radius of the drop
[tex]r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m[/tex]
(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,
[tex]\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[/tex]
Now the charge on the new drop is 2q. New potential is given by :
[tex]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[/tex]
Hence, the radius of the drop is [tex]7.17\times 10^{-4}\ m[/tex] and the potential at the surface of the new drop is 856.79 V.
