Question

Two identical positive point charges $Q$ each fixed at a distance $d=2a$ apart in air. A point charge $q$ lies at mid-point on the joining the charges. Now $q$ executes simple harmonic motion, if:
(i) $q$ is positive and it is given a small displacement along the line joining the charges.
(ii) $q$ is negative and it is given a small displacement. If frequencies of oscillations are $n_a$ and $n_b$, in case (i) and case (ii) then the ratio $\frac{\sqrt{2}{n}_a}{n_b}$ is:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

Two identical positive point charge$=Q$

Fixed distance $d=2a$

Apart in air

A point charge$=q$

lie at mid point on the joining the charge

Now, $q$ executes simple harmonic motion

If

$\left(i\right)q$ be the $(+ve)$ and it is given a small displacement along the line

$\left(ii\right)q$ be the $(-ve)$ and it is given a small displacement. If the frequency of oscillator are $n_a$ and $n_b$ in case $\left(i\right)$ and case $\left(ii\right)$ then the ratio $\frac{\sqrt{2}{n}_a}{n_b}$ is

The particle is going to undergo SHM. I can easily say that at points $+q$ and $-q$ the force on the particle is the maximum and is

$=\frac{Qqd}{4\pi {ϵ}_{0}ta{n}^2\theta }=\frac{Qq\left(2a\right)}{4\pi {ϵ}_{0}ta{n}^2\theta }=\frac{Qqa}{2\pi {ϵ}_{0}ta{n}^2\theta }{n}_a=f_a=\frac{1}{2\pi }\sqrt{\frac{q}{d/2}}=\frac{1}{2\pi }\sqrt{\frac{Q}{d/2}}{n}_b=f_b=\frac{1}{2\pi }\sqrt{\frac{Q}{d/2}}=\frac{1}{2\pi }\sqrt{\frac{Q}{d}}\frac{\sqrt{2}{n}_a}{n_b}=\frac{Q}{q}=3\times 2=6$