Question

A particle of mass $m$ and charge $+q$ is midway between two fixed charged particles, each having a charge $+q$ and at a distance $2L$ apart. The middle charge is displaced slightly along the line joining the fixed charges and released. The time period of oscillation is proportional to

• A. $L^{1/2}$
• B. $L$
• C. $L^{3/2}$
• D. $L^2$

Answer 1

The correct option is C $L^{3/2}$

Let the charge $q$ at the mid-point is displaced slightly to the left by $x$.


The force on the displaced charge $q$ due to charge $q$ at $\text{A}$,

$F_A=\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{(L+x{)}^2}$

The force on the displaced charge $q$ due to charge at $\text{B}$,

$F_B=\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{(L-x{)}^2}$

Net restoring force on the displaced charge $q$.

$F=F_A-F_B$

Substituting the value in the above equation,
$\Rightarrow F=\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{(L+x{)}^2}-\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{(L-x{)}^2}$

$\Rightarrow F=-\frac{1}{4\pi {\epsilon }_0}\times \frac{4q^{2}Lx}{(L^2-x^2{)}^2}$

For $x<<L,$

$\Rightarrow F=-\frac{q^{2}x}{\pi {\epsilon }_0{L}^3}=-kx$

Hence we see that $F\propto x$ and it is opposite to the direction of displacement. Therefore, the motion is $\text{SHM}$.

Now comparing the equation with the equation of $\text{SHM}$, we get

$\Rightarrow F=-\frac{q^{2}x}{\pi {\epsilon }_0{L}^3}=-kx$

Where, $k=\frac{q^2}{\pi {\epsilon }_0{L}^3}$

The time period of the oscillation of charge $q$,
$T=2\pi \sqrt{\frac{m}{k}}$

$\Rightarrow T=2\pi \sqrt{\frac{m\pi {\epsilon }_0{L}^3}{q^2}}$

Hence, $T\propto L^{3/2}$

Thus , option (c) is the correct answer.