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 `2 a`.
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Explanation

(i) Method I (Derivation)
Assume all the charges are at infinity. Work done in putting charge q at corner A
`implies 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 if presence of charge at A :
`implies W_(2)=((Kq)/a-0) q=(Kq^(2))/a`
similarly work done in putting charge q at corner C in presence of charge at A and B.
`implies W_(3)=q(v_(f)-v_(i))=q [((Kq)/a+(Kq)/a)-0]=(2Kq^(2))/a`
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` Joules
(iii) Work done by electrostatic forces - Change is kinetic energy of particles.
`U_(i)-U_(f)=K_(f)-K_(i)" "implies (3Kq^(2))/a-(3Kq^(2))/(2a)=3 (1/2 mv^(2))-0 implies v= sqrt((Kq^(2))/(am))`