Three point charges of 2 mC each are kept at the vertices of an equilateral triangle of side 50 cm . If the system is supplied energy at the rate 2 kW, the time taken to move one of the charges to the mid point of the line joining the other two charge is

Options

A. 18 s
B. 36 s
C. 72 s
D. 144 s

Answer

C

Explanation

According to the question,
Energy of system of three charges, ` U = ((kq_(1)q_(2))/(r))^(3)`
or `U = (k (2 xx 10^(-3) xx 2 xx 10^(-3)))/((1//2))xx3 = 24 k xx 10^(-6) J`
When one charge move to the mid point of the line joining the other two charges.
Energy of the system,
`U.(K(2 xx 2xx 10^(-6)))/((1//4))xx 2 +(k (2xx2xx10^(-6)))/(1//2)`
`U. = 40 k xx 10^(-6) T`
Net energy, `Delta U = U. - U = (40 - 24) k xx 10^(-6)`
or `Delta U = W = 16 xx 9 xx 10^(-3) J`
Given that energy supplied to the system at the rate of 2 kW.
So, ` W = 2 kW xx 1rArr 16 xx 9 xx 10^(3) = 2 xx 10^(3) xx t `
`rArr t = 72 s`
Hence, time taken to move one of the charge to the mid point of the line joining the other two charge is 72 s.