Question

When charge $+Q$ , its distance from $+q$ at A is $r_{eq}$ . Let us slightly displace $+Q$ from the Equilibrium position so that its distance from A is now $r=r_{eq}+x$ it will be found that $+Q$ oscillates about the equilibrium position in a simple harmonic manner if x is very small. Using binomial theorem$(1+P{)}^n\approx 1+np$ if $|P|<1$ , time period of Oscillations of $+Q$, if its mass is m, will be :

• A. $2\pi \sqrt{\frac{m\alpha }{R}}$
• B. $2\pi \sqrt{\frac{m{R}^3}{2\alpha }}$
• C. $2\pi \sqrt{\frac{m{R}^2}{2\alpha }}$
• D. $2\pi \sqrt{\frac{\alpha }{mR}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{m{R}^3}{2\alpha }}$

Initial distance between the charges is $r_{eq}$.

Electrostatic force acting on the charge $Q$ initially $F^{\prime }=\frac{KQq}{r_{eq}^2}$

Since charge $Q$ is displaced, so new distance between them $r=r_{eq}+x$

Electrostatic force acting on the charge $Q$, $F=\frac{KQq}{r^2}$ ....(1)

Since $r=r_{eq}+x$

On squaring both sides, we get $r^2=r_{eq}^2(1+\frac{x}{r_{eq}}{)}^2$
Or $r^{-2}=r_{eq}^{-2}(1+\frac{x}{r_{eq}}{)}^{-2}$

Using bionomial theorem $(1+P{)}^n=1+np$ for $|P|<<1$

We get $r^{-2}=r_{eq}^{-2}(1-\frac{2x}{r_{eq}})$
Here $K=\frac{1}{4\pi {ϵ}_0}$
Thus from (1), we get $F=KQq\times r_{eq}^{-2}(1-\frac{2x}{r_{eq}})$
$\therefore$ $F=\frac{KQq}{r_{eq}^2}-\frac{2KQqx}{r_{eq}^3}$
Thus change in force $\Delta F=F-F^{\prime }=\frac{-2KQqx}{r_{eq}^3}$

Comparing it with $\Delta F=-m{w}^{2}x$
We get $m{\omega }^2=\frac{2KQq}{r_{eq}^3}$
Or $\omega =\sqrt{\frac{2KQq}{m{r}_{eq}^3}}$
Time period of oscillation $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m{r}_{eq}^3}{2KQq}}$

$⟹$ $T=2\pi \sqrt{\frac{m{r}_{eq}^3}{2\alpha }}$ where $\alpha =KQq$