vikas askiitian expert
Last Activity: 14 Years ago
let the vertices of triangle are a,b,c ...
ans for part 1)
total electric potential energy=kq2 /d +kq2 /d+kq2/d (sum of potential energy of each pair)
(PE)total =3kq2 /d
ans for part 2)
when two charges are idential & let other is Q..
now total potential energy=KQq/d +KQq/d+Kq2 /d (sum of potential energy of each pair of charges)
work required to place these charges at corner of equilateral triangle
is equal to potential energy of system
so
total potential energy=total work=0
2kQq/d=-kq2/d
Q=-q/2 ans