A non-conducting ring of mass m and radius r is lying at rest in the vertical XY plane on a smooth non conducting horizontal XZ plane. Charge +q and –q are distributed uniformly on the ring. On the two sides of the vertical diameter of the ring. A constant and uniform electric field E  is set up along the x-direction. The ring is given a small rotation about the vertical diameter of the ring and released. Find the period of oscillation of the ring.?

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Problem Statement

A non-conducting ring of mass m and radius r is lying at rest in the vertical XY plane on a smooth non conducting horizontal XZ plane. Charge q and –q are distributed uniformly on the ring. On the two sides of the vertical diameter of the ring. A constant and uniform electric field E  is set up along the x-direction. The ring is given a small rotation about the vertical diameter of the ring and released. Find the period of oscillation of the ring.

Solution

The electric field E will exert a torque on the ring causing it to oscillate about the vertical diameter. The period of oscillation can be found using the equation:

T = 2π * √(I/τ)

Step 1: Find the moment of inertia

The moment of inertia of the ring about the vertical diameter can be found using the formula:

I = (1/2) * m * r^2

Step 2: Find the torque exerted by the electric field

The torque exerted by the electric field can be found using the formula:

τ = qEr * sin(θ)

where q is the charge on the ring, E is the electric field, r is the radius of the ring, θ is the angle between the electric field and the vertical diameter, and r is the radius of the ring.

Since the ring is released from rest, the initial angle θ is 0. The torque exerted by the electric field will be maximum when the ring is at its maximum displacement from the equilibrium position, which occurs when θ = π/2.

Substituting the values in the formula:

τ = qEr

Step 3: Find the period of oscillation

Substituting the values of I and τ in the formula for the period of oscillation:

T = 2π * √(I/τ) = 2π * √((m * r^2)/(qEr))

Therefore, the period of oscillation of the ring is given by:

T = 2π * √((m * r^2)/(qEr))

Answer

HEY Aayan

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