A spherical drop of mercury of radius 400 mm has a potential

A spherical drop of mercury of radius 4.00 mm has a potential 320 V at its surface. Another spherical drop of mercury of radius 3.00 mm has a potential of 400 V at its surface. The two drops merge to form a single larger spherical drop. What is the potential at the surface of that drop?

Solution

here,

Volume of spherical drop1 = V1 = 4pir^3/3

V1 = 4/3 *3.14 * (4*10^-3)^3

V1 = 2.679 *10^-7 m^3

Potential V = Kq/r

charge q1 = Vr/k

q1 = (320 * 4 *10^-3)/(9 *10^9)

q1 = 0.142 nC

when they merged,

V2 = 4/3 * 3.14 * (3*10^-3)^3

V2 = 1.134 *10^-7 m^3

new charge q2 = Vr/k

q2 = (400 * 3*10^-3)/(9*10^9)

q2 = 0.133 nC

when they merge,

total volume V = V1+V2 = (2.679 + 1.134) *10^-7

V =   2.813 *10^-7 m^3

so V = 4/3 * 3.14 * R^3 = 2.813 *10^-7

R = 4 mm

so new charge q = q1+q2 = 0.142 nC + 0.133 nC = 0.275 nC

so Vnet = (9*10^9 * 0.275 *10^-9)/(4*10^-3)

Vnet =   618.75 Volts (answer)