Question

(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<

# 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

Solution