Question

Two equal negative charges $-q$ are placed at $(0,a)$ and $(0,-a)$. A positive charge $q$ is released from rest at the point $x(<<a)$ on the x-axis, then the frequency of the motion is

• A. $\sqrt{\frac{q^2}{\pi {ϵ}_{o}m{a}^3}}$
• B. $\sqrt{\frac{q^2}{2\pi {ϵ}_{o}m{a}^3}}$
• C. $\sqrt{\frac{2q^2}{\pi {ϵ}_{o}m{a}^3}}$
• D. $\sqrt{\frac{q^2}{4\pi {ϵ}_{o}m{a}^3}}$

Answer 1

The correct option is B $\sqrt{\frac{q^2}{2\pi {ϵ}_{o}m{a}^3}}$

$F=\frac{q^2}{4\pi {ϵ}_o{r}^2}$

$⟹F=\frac{q^2}{4\pi {ϵ}_o(x^2+a^2)}$

As we can see from figure, the net force will be the vector sum of two F force and net force will be along negative X-axis.

$F_{net}=2F\cos \theta$

$⟹F_{net}=2\frac{q^2}{4\pi {ϵ}_o(a^2+x^2)}\frac{x}{\sqrt{x^2+a^2}}$

$⟹F_{net}=\frac{2q^{2}x}{4\pi {ϵ}_o(x^2+a^2{)}^{3/2}}$

Since, $x<<a$

Hence, $x^2+a^2\approx a^2$

Hence, $F_{net}=\frac{2q^{2}x}{4\pi {ϵ}_o{a}^3}$

$F_{net}\propto x$

Hence, the charge q will undergo simple harmonic motion.

$F_{net}=m{\omega }^{2}x=\frac{2q^{2}x}{4\pi {ϵ}_o{a}^3}$

$⟹\omega =\sqrt{\frac{q^2}{2\pi {ϵ}_{o}m{a}^3}}$

Answer-(B)