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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 functionsmall 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. |
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Answer» Solution :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 . |
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