Two charges - q each are separated by distanc · 2d. A third charge + q is kept at mid point C Find potential energy of + q as a function small distance x from O due to - q charges Sketch P.E. v/s x and convince yourself that the, charge at O is in an unstable equilibrium.

Explanation

Suppose + q charge is displaced towards - , and hence total potential energy or system
`U= k[(-q^(2))/((d-x))+(-d^(2))/((d-x))]`
` = -kq^(2)[(d+x+d-x)/(d^(2)-x^(2))]`
`=-kq^(2)[(2d)/(d^(2)-x^(2))]`
`:.` By differentiating U w.r.t.x
`(dU)/(dx)=-kq^(2)xx2d((-2x)/((d^(2)-x^(2))^(2)))`
If `(dU)/(dx)=0` then F =0
`:. 0 =(4kq^(2)dx)/((d^(2)-x^(2))^(2)):. x=0`
Hence +q is in stable and unstable equilibrium .
By differentiating again w.r.t.x
`(d^(2)U)/(dx^(2))=[(-2dq^(2))/(4pi in_(0))][(2)/((d^(2)-x^(2))^(2))-(8x^(2))/((d^(2)-x^(2))^(3))]`
`=((-2dq^2)/(4piin_(0)))-(1)/((d^(2)-x^(2))^(3))-[2(d^(2)-x^(2))-8x^(2)]`
At x= 0 ` x lt lt d implies ` By neglecting x
`(d^(2)U)/(dx^(2))=((-2dq^(2))/(4piin_(0)))((1)/(d^(6)))[2d^(2)]`
`(d^(2)U)/(dx^(2))lt 0`
Hence system is in unstable equilibrium .