Question

A positive charge Q is distributed uniformly over a circular ring of radius R. A particle of mass m, and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x << R, find the time period of oscillation of the particle if it is released from there.

Answer 1

Force of the particle is given by, F = $q \times E=q \frac{K Q x}{R^{3}}$.

Time period of oscillation of the particle is given by, $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m R^{3}}{Q q}}$.

Explanation:

Step 1:

Given in the question :

A positive charge Q is evenly distributed across a circular radius R ring.

At a distance of  "x" from its  centre, a particle of negative charge q and mass m are put on its axis  

Step 2:

We know that ,

Electric field E, at P due to the charged ring  

              E   $=\frac{K Q x}{\left(R^{2}+x^{2}\right)^{\frac{3}{2}}}=\frac{K Q_{x}}{R^{3}}$

Force experienced F = $q \times E=q \frac{K Q x}{R^{3}}$

Now, amplitude = x  

So ,

                        $T=2 \pi \sqrt{\frac{x}{q \frac{K Q x}{m R^{3}}}}$

                       $T=2 \pi \sqrt{\frac{m R^{3} x}{q K Q x}}$

                       $T=2 \pi \sqrt{\frac{4 \pi \varepsilon_{0} m R^{3}}{Q q}}$

                       $T=\sqrt{\frac{4 \pi^{2} \times 4 \pi \varepsilon_{0} m R^{3}}{Q q}}$

                       $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m R^{3}}{Q q}}$

Time period of particle oscillation $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m R^{3}}{Q q}}$

Thus, the force is F = $q \times E=q \frac{K Q x}{R^{3}}$ and time period is $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m R^{3}}{Q q}}$.

Answer 2

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Please refer the above attachment