A uniform non-conducting ring of mass m and radius R carries two fixed positive point charges q and q′ on its circumference. Their separation is √2R. The ring is in a vertical plane in a uniform vertically upward electric field E.Given E = 4mg/(7q), the ring is in translational equilibrium. If θ is the equilibrium angle made by the radius to charge q with the upward vertical, then θ is:
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Question

A uniform non-conducting ring of mass m and radius R carries two fixed positive point charges q and q′ on its circumference. Their separation is √2R. The ring is in a vertical plane in a uniform vertically upward electric field E.Given E = 4mg/(7q), the ring is in translational equilibrium. If θ is the equilibrium angle made by the radius to charge q with the upward vertical, then θ is:

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G1 Physics · Accepted Answer

Translational equilibrium gives (q+q′)E = mg. Using E = 4mg/(7q), q+q′ = 7q/4, so q′ = 3q/4. The chord √2R corresponds to a central angle of 90°. Torque balance about the centre gives qR sinθ = q′R cosθ. Therefore, tanθ = q′/q = 3 4 \frac{3}{4} 4 3 ​ . Hence, θ = tan⁻¹( 3 4 \frac{3}{4} 4 3 ​ ).