Three point charges of 0.1 C each are placed at the corners of an equilateral triangle with side 1m. If this system is supplied energy at the rate of 1 kW, how much time will be required to move one of the charges on the mid-point of the line joining the two others ?

Answer

50 h.

Explanation

Initial electrostatic energy of system of three charges is

`U_i = 3 (1)/(4 pi epsilon_0) (q^2)/(r) = 3 xx 9 xx 10^9 xx (0.1)^2/(1) = 2.7 xx 10^8 J`
When charge A is moved to new position A', which is midpoint of side BC, then final electrotatic energy is
`U_f = (1)/(4 pi epsilon_0) ((q_1 q_2)/(r_(12))+(q_2 q_3)/(r_(23))+ (q_2 q_1)/(r_(13)))`
=`(1)/(4 pi epsilon_0) [(q^2)/((A' B)) + (q^2)/((B C)) + (q^2)/((A'C))]`
But `A' B = A' C = r // 2 = 0.5 m`. Therefore,
`U_f = 9 xx 10^9 [((0.1)^2)/(0.5) + (0.1)^2/(1) + (0.1)^2/(0.5)] = 4.5 xx 10^8 J`
Work done `W = U_f - U _i`
=`4.5 xx 10^8 - 2.7 xx 10^8 = 1.8 xx 10^8 J`
As `W = P t`
or `t = (W)/P =(1.8 xx 10^8)/(10^3)`
=`1.8 xx 10^5 s = (1.8 xx 10^5)/(3600) h = 50 h`.