The diameter of a spherical liquid drop is 2 mm and its charge is `5 xx 10^(-6)` esu.
If two such liquid drops coalesce to form a bigger drop, what will be the potential on its surface ?

Explanation

Let the radius of the bigger drop be R.
According to the question,
`(4)/(3)pi R^3 = 2xx(4)/(3)pi(0.1)^3 " or ", R^3 = 2 xx(0.1)^3`
or, `R =0.1 xx2^(1/3)= 0.1 xx 1.26= 0.126` cm
Total charge, `Q = 2 xx 5 xx 10^(-6)= 10^(-5)` esu of charge.
Potential on the surface of the bigger liquid drop,
`V = (Q)/(C )= (10^(-5))/(0.126) = 7.94 xx 10^(-5)` statV
`= 7.94 xx 10^(-5) xx 300 V = 0.0238 V`.