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 0.9 g and having a negative charge of $1\times {10}^{-}6$ C is placed on the axis at a distance of 1 cm from the centre of the ring. Calculate the time period of oscillations.

• A. $0.5secs$
• B. $9.28secs$
• C. $0.628secs$
• D. $0.1secs$

Answer 1

Answer: (C)

$0.628secs$

The negatively charged particle kept on the axis will get attracted towards the positively charged ring. As it approaches the center, the inertia of the particle will make the charge to move to the other side of the ring. Simultaneously, the attractive force due to the ring will prevent it from going to infinity. Thus, it goes to one more extreme and returns back. Thus the charge starts oscillation about the center of the ring.

The electric field at a distance x from the centre on the axis of a ring, distanct x <<R is given by $E=\frac{kQx}{R^3}$. Net force on negatively charged particle would be qE and towards the centre of the ring. Hence, we can write $F=-\frac{kQqx}{R^3}$

Acceleration of the charge is $a=F/m=-\frac{kQqx}{m{R}^3}=-w^{2}x$.

Time period of oscillations is $T=2/w=2\left(m{R}^3/kQq\right)$. Substituting the values , we get

T=0.628 s