A thin semicircular ring of radius 35 cm carries a uniformly distributed total charge Q. The magnitude of the electric field at the centre of the semicircle is 100 V·m⁻¹. Taking ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻² and π = 3.14, what is Q in nanocoulombs?
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Question

A thin semicircular ring of radius 35 cm carries a uniformly distributed total charge Q. The magnitude of the electric field at the centre of the semicircle is 100 V·m⁻¹. Taking ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻² and π = 3.14, what is Q in nanocoulombs?

Options

G1 Physics · Accepted Answer

For a uniformly charged semicircular ring, the electric field at its centre is E = 2kQ ÷ (πR²), where k = 1 ÷ (4πε₀) ≈ 9×10⁹ N·m²·C⁻². Given, E = 100 V·m⁻¹, R = 35 cm = 0.35 m. Therefore, Q = EπR² ÷ (2k). Substitute: Q = 100×3.14×(0.35)² ÷ (2×9×10⁹) ≈ 2.14×10⁻⁹ C. Hence, Q ≈ 2.14 nC. Correct option: A.