Question

A thin fixed ring of radius $1m$ has a positive charge $1\times {10}^{-5}C$ uniformly distributed over it. A particle of mass $0.9gm$ having a negative charge of $1\times {10}^{-6}C$ is placed on the axis at a distance of $1cm$ from the centre of the ring. Assuming that the oscillations has small amplitude, the time period of oscillations is:

• A. $0.23s$
• B. $0.39s$
• C. $0.49s$
• D. $0.63s$

Answer 1

The correct option is D $0.63s$

We know $E$ on the axis of the ring is:

$E=\frac{kqx}{(r^2+x^2{)}^{3/2}}$

For small amplitudes $r>>x$:

$\Rightarrow E=\frac{kqx}{r^3}$

We can see that force is always applied in the opposite direction of displacement i.e:

$F=\frac{-kqx}{r^3}(1\times {10}^{-6})C$

$\Rightarrow a=\frac{-kqx(1\times {10}^{-6})}{m{r}^3}$

We know for a particle in S. H. M: $a=-{\omega }^{2}x$

$\therefore {\omega }^2=\frac{kq(1\times {10}^{-6})}{m{r}^3}=\frac{9\times {10}^9\times 1\times {10}^{-5}\times 1\times {10}^{-6}}{0\cdot 9\times {10}^{-3}\times 1^3}=100$

$\Rightarrow \omega =10$

Now, $T=\frac{2\pi }{w}=\frac{2\times 3\cdot 14}{10}=0.63s$