Question

Two charges -q each are separated by dsitance 2d. A third charge +q is kept at mid-point O. find potential energy of +q as function of small distance x from 0 due to -q charges. Sketch PE Vs/x and convince yourself that the charge at 0 is in an unstable equilibrium.

Answer 1

let third charge +q is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by
`U=(1)/(4piepsi_(0)){(-q^(2))/((d-x))+(-q^(2))/((d+x))}`
`U=(-q^(2))/(4piepsi_(0))(2d)/((d^(2)-x^(2)))`
`(dU)/(dx)=(-q^(2)2d)/(4piepsi_(0))*(2x)/((d^(2)-x^(2))^(2))`
The system will be in equilibrium, if
`F=-(dU)/(dx)=0`
On solving, x=0 so for +q charge to be in stable/unstable equilibrium, finding second derivative of PE.
`(d^(2)U)/(dx^(2))=((-2dq^(2))/(4piepsi_(0)))[(2)/((d^(2)-x^(2))^(2))-(8x^(2))/((d^(2)-x^(2))^(3))]`
`=((2-dq^(2))/(4piepsi_(0)))(1)/((d^(2)-x^(2))^(3))[2(d^(2)-x^(2))^(2)-8x^(2)]`
at `x=0`
`(d^(2)U)/(dx^(2))=((-2dq^(2))/(4piepsi_(0)))((1)/(d^(6)))(2d^(2)),` which is lt0
This shows that system will be unstable equilibrium.