Question

Two positive equal charges are placed symmetrically about the origin on the x-axis and kept fixed as shown. Another positive charge Q is placed at the origin. The charge Q can be displaced by a very small distance $\delta$ to $P_1$ along the x-axis or to $P_2$ along the y-axis. Then,

• A. to the lowest order, the force on Q at $P_2$ is proportional to $\delta$.
• B. to the lowest order, the force on Q at $P_1$ is proportional to ${\delta }^2$.
• C. the charge Q is in stable equilibrium with respect to small displacements along y-axis.
• D. the charge Q is in stable equilibrium with respect to small displacements along the x-axis.

Answer 1

Answer: (A), (D)

The correct options are
A to the lowest order, the force on Q at $P_2$ is proportional to $\delta$.
D the charge Q is in stable equilibrium with respect to small displacements along the x-axis.

For $P_1$:-

$F=\frac{qQ}{4\pi {ϵ}_o(a+\delta {)}^2}-\frac{qQ}{4\pi {ϵ}_o(a-\delta {)}^2}$

$⟹F=\frac{qQ}{4\pi {ϵ}_o}\{\frac{1}{(a+\delta {)}^2}-\frac{1}{(a-\delta {)}^2}\}$

$⟹F=\frac{qQ}{4\pi {ϵ}_o}\left\{\frac{-4a\delta }{(a^2-{\delta }^2{)}^2}\right\}$

$⟹F=-\frac{qQ}{\pi {ϵ}_o{a}^3}\delta$

Since, $F\propto (-\delta )$

Hence, Q will undergo a simple harmonic motion and is thus stable equilibrium for small displacement along X-axis.

For $P_2$:-

$F=\frac{2Qq}{4\pi {ϵ}_0{r}^2}\cos \theta$ where r is distance between $q$ and $Q$ and $\theta$ is angle between $Y-axis$ and $r$.

$⟹F=\frac{2Qq}{4\pi {ϵ}_o(a^2+{\delta }^2)}.\frac{\delta }{\sqrt{a^2+{\delta }^2}}$

$⟹F=\frac{Qq}{2\pi {ϵ}_o{a}^3}\delta$

$⟹F\propto \delta$

The force is along $+ve$ Y direction and hence is away from original position and hence is unstable equilibrium.

Hence, option -(A) and (D) are correct.