Question

a fixed ring of radius R is having uniformly distributed charge Q over its circumference. A charge q of opposite nature and mass m is kept on its axis at a distance a<<R from the of ring . If charge is released from rest then minimum time after which it will pass through center of the ring will be (effect of gravity is considered to be negligible) ​

Answer 1

electric field due to ring along its axis is given by,

$E_a=\frac{kQa}{(a^2+R^2)^{3/2}}$

where a is separation between observation point to centre of ring.

it is given that, a << R

so, a/R << 1 => a²/R² <<< 1

so, a²/R² + 1 ≈ 1 .....(1)

so, $E_a=\frac{kQa}{R^3\left(\frac{a^2}{R^2}+1\right)^{3/2}}$

or, $E_a=\frac{kQa}{R^3}$[ from equation (1), ]

now, attractive force applied charge q

F = $qE_a$ = mA, here A is acceleration.

so, A = kQqa/mR³

now applying formula, S = ut + 1/2 at²

S = 0 + 1/2 At²

a = 1/2 × kQqa/mR³ t²

or, mR³/kQq = t²

or, t = √{mR³/kQq}

hence ,time taken to reach the centre of ring will be $\sqrt{\frac{mR^3}{kQq}}$

Answer 2

Answer:

Explanation:time period if oscillation .

Minimum time will be half of it.