A spherical drop of liquid carrying charge, Q has potential `V_0` at its surface. If two drops of same charge and radius combine to form of single spherical drop, then the potential at the surface of new drop is (Assume, V = 0 at infinity.)

Options

A. ` 2^(1/3) V_0`
B. `4^(1/3) V_0`
C. `6^(1/3)V_0`
D. `2^(-1/3) V_0`

Answer

B

Explanation

Given,
charge of spherical drop of liquid = Q
and potential at its surface ` = V_0`
If r be the radius of drop , `V_0 = (1)/(4 pi epsi_0) - Q/r " " ..(i)`
If R be the radius of big drop when two drops of radius r are combined, then volume of big drop = volume of both small drops
i.e., `4/3 piR^3 = 4/3 pir^3 + 4/3 pir^3`
` R^3 = 2r^3 rArr R = 2^(1//3) r`
` therefore ` charge on the big drop Q'=Q+Q=2Q
` therefore` Potential at the surface of big (new) drop.
`V' = (1)/(4 pi epsi_0) (Q')/(R) = (1)/(4 pi epsi_0) (2Q)/(2^(1//3) r)`
` = (1)/(4 pi epsi_0) Q/r. 2^(2//3) = V_0 . 4^(1//3) = 4^(1//3) . V_0`