Three equal charges q are placed at the corners of an equilateral triangle of side a.
(i) Find out potential energy of charge system.

(ii) Calculate work required to decrease the side of triangle to a/2.
(iii) If the charges are released from the shown position and each of them has same mass m then find the speed of each particle when they lie on triangle of side 2a.

Explanation

(i) Method I (Derivation)
Assume all the charges are at infinity initially.

work done in putting charge q at corner A
`W_(1) = q(v_(f) - v_(i)) = q(0 -0)`
Since potential at A is zero in absence of charges, work done in putting q at corner B in presence of charge at A :
`W_(2) = ((Kq)/(a) - 0) = (Kq^(2))/(a)`
Similarly work done in putting charge q at corner C in presence of charge at A and B.
`W_(3) = q(v_(f) - v_(i)) = q[((Kq)/(a) + (Kq)/(A))-0]`
So net potential energy `PE = W_(1) + W_(2) + W_(3)`
`= 0 + (Kq^(2))/(a) + (2Kq^(2))/(a) = (3Kq^(2))/(a)`
Method II (using direct formula)
`U = U_(12) + U_(13) + U_(23) = (Kq^(2))/(a) + (Kq^(2))/(a) + (Kq^(2))/(a) = (3Kq^(2))/(a)`
(ii) Work required to decrease the sides
`W= U_(f) - U_(i) = (3Kq^(2))/(a//2) - (3Kq^(2))/(a) = (3Kq^(2))/(a)`
(iii) Work done by electrostatic force = change is kinetic energy of particles.
`U_(i) - U_(f) = K_(f) - K_(i)`
`rArr (3Kq^(2))/(a) - (3Kq^(2))/(2a) = 3((1)/(2) mv^(2)) - 0`
`rArr v = sqrt((Kq^(2))/(a m))`