A spherical drop of water carries a charge of `10xx10^(-12) C` and has a potential of 100 V at its surface. If eight such charged drops combine to form a single drop, what will be the potential at the surface of the new drop?

Explanation

If R be the radius of the large drop, then
`(4)/(3) pi R^3 = 8xx(4)/(3)pi r^3`
or, `R= 2r = 2xx0.09 = 0.18 cm`
Total charge, `Q= 8xx3xx10^(-2)= 24xx10^(-2)` statC
Potential at the surface of the new drop,
`V= (Q)/(C )= (Q)/(R )= (24xx10^(-2))/(0.18)= (4)/(3) = 1.33` statV
`= 1.33xx300 X = 400V`.