The figures below depict two situations in wh | vedclass

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(C) Let the distance of the equilibrium position from each line charge be $r$. The electric field due to a line charge at distance $d$ is $E = \frac{\

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Charge $+q$ executes simple harmonic motion while charge $-q$ continues moving in the direction of its displacement.

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(C) Let the distance of the equilibrium position from each line charge be $r$. The electric field due to a line charge at distance $d$ is $E = \frac{\lambda}{2\pi\epsilon_0 d}$.For charge $+q$ (Case $I$): If it is displaced by $x$ towards the right, the net force is $F = qE_{left} - qE_{right} = q \frac{\lambda}{2\pi\epsilon_0(r-x)} - q \frac{\lambda}{2\pi\epsilon_0(r+x)} = \frac{q\lambda}{2\pi\epsilon_0} \left( \frac{1}{r-x} - \frac{1}{r+x} \right) = \frac{q\lambda}{2\pi\epsilon_0} \left( \frac{2x}{r^2-x^2} \right) \approx \frac{q\lambda x}{\pi\epsilon_0 r^2}$. Since the force is proportional to $-x$ (restoring force), the charge $+q$ executes simple harmonic motion $(SHM)$.For charge $-q$ (Case $II$): If it is displaced by $x$ towards the right, the force from the left line charge is attractive (towards left) and the force from the right line charge is attractive (towards right). The net force is $F = qE_{right} - qE_{left} = \frac{q\lambda}{2\pi\epsilon_0(r+x)} - \frac{q\lambda}{2\pi\epsilon_0(r-x)} = -\frac{q\lambda x}{\pi\epsilon_0 r^2}$. Since the force is in the direction of displacement, it will continue moving away from the equilibrium position.

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