Two equal negative charges -q are placed at p | vedclass

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Question

(B) The force on the positive charge $q$ at position $(x, 0)$ due to the two negative charges at $(0, a)$ and $(0, -a)$ is directed towards the origin

Vedclass · Accepted Answer

It will oscillate but not execute $S.H.M.$

Vedclass · Answer

(B) The force on the positive charge $q$ at position $(x, 0)$ due to the two negative charges at $(0, a)$ and $(0, -a)$ is directed towards the origin.The magnitude of the force is $F = 2 \cdot \frac{1}{4\pi\epsilon_0} \cdot \frac{q^2}{(x^2 + a^2)} \cdot \cos	heta$, where $\cos	heta = \frac{x}{\sqrt{x^2 + a^2}}$.Thus, $F = \frac{2kq^2x}{(x^2 + a^2)^{3/2}}$.For $S.H.M.$, the force must be proportional to the displacement $(F \propto -x)$.Since the force is not linear with respect to $x$ for all values, the motion is oscillatory but not simple harmonic motion $(S.H.M.)$.

Two equal negative charges $-q$ are placed at points $(0, a)$ and $(0, -a)$ on the $Y$-axis. A positive charge $q$ is released from rest at point $(2a, 0)$. What will be the motion of this charge?

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