In the figure shown S is a large nonconducting sheet of uniform charge density σ. A rod R of length λ and mass ‘m’ is parallel to the sheet and hinged at its mid point. The linear charge densities on the upper and lower half of the rod are shown in the figure. If the angular acceleration of the rod just after it is released is then write value of n
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A large nonconducting sheet with uniform charge density σ creates a uniform electric field E = σ/(2ε₀) perpendicular to the sheet. The hinged rod has two halves with linear charge densities +λ and -λ. The electric force on each half is F = (λL/2) * E, where L is the rod length. These equal and opposite forces create a net torque τ about the hinge.
The torque is τ = F * (L/4) + F * (L/4) = (F * L)/2. Substituting F gives τ = (λL²σ)/(8ε₀).
Using τ = Iα, where I = mL²/12 is the moment of inertia of a rod about its center, we get α = τ/I = (λL²σ)/(8ε₀) * (12/(mL²)) = (3σλ)/(2mε₀).
Comparing with the given angular acceleration nσλ/(2mε₀), we find n = 3.
Final Answer: 3