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Two charges -qeach are separated by dsitance 2d. A third charge +q is kept at mid-point O. find potential energy of +q asfunction 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. |
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Answer» Solution :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. |
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