Question

TW Equal point charges are fixed at x=-a and x=+a on the x-axis. Another point charge Q is plac at the origin. The change in the electricla potential energdy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to A)x B) x² D]1/x (2002) C) x3

NEET/Medical Exams

Physics

Solution

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(2) Initially according to figure (i) potential energy of ( Q ) is ( U_{i}=frac{2 k q Q}{a} ldots ldots . ) (i) According to figure (ii) when charge ( Q ) is displaced by small distance ( x ) then it's potential energy now ( U_{f}=k q Qleft[frac{1}{(a+x)}+frac{1}{(a-x)}right]=frac{2 k q Q a}{left(a^{2}-x^{2}right)} ldots ldots . ) (ii) Hence change in potential energy ( Delta U=U_{f}-U_{i}=2 k q Qleft[frac{a}{a^{2}-x^{2}}-frac{1}{a}right]=frac{2 k q Q x^{2}}{left(a^{2}-x^{2}right)} ) since ( x<