Let the volume of one drop be V and radius R and after combination of two drops, new volume be V′. Hence, V′=V+V=2V ⇒
4
3
πR′3=2×
4
3
πR3 ⇒R′3=2R3⇒R′=21∕3R Capacitance of spherical capacitor, C=4πε0R C∝R . . . (i) Now, the new capacitance C′∝R′, C′∝21∕3R . . . (ii) On dividing Eq. (ii) by Eq. (i), we get