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(A) Let the two charges $q$ be at $x = -a$ and $x = +a$. The third charge $q$ is at the origin $(x = 0)$.When the charge is displaced by a small dista

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Perform simple harmonic motion about the equilibrium position

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(A) Let the two charges $q$ be at $x = -a$ and $x = +a$. The third charge $q$ is at the origin $(x = 0)$.When the charge is displaced by a small distance $x$ along the $x$-axis, the net force on it is $F = F_1 - F_2 = \frac{kq^2}{(a-x)^2} - \frac{kq^2}{(a+x)^2}$.$F = kq^2 \left[ \frac{(a+x)^2 - (a-x)^2}{(a^2-x^2)^2} \right] = kq^2 \left[ \frac{4ax}{(a^2-x^2)^2} \right]$.Since $x Thus, $F \approx \frac{4kq^2x}{a^3}$.Since the force $F$ is proportional to the displacement $x$ and directed towards the equilibrium position (restoring force), the particle will perform Simple Harmonic Motion $(S.H.M)$.

Two equal charges $q$ are placed at a distance of $2a$ on the $x$-axis. A third charge $q$ of mass $m$ is placed at the midpoint between them. If this charge is displaced by a small distance $x$ from the equilibrium position, the particle will:

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