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A charged particle having a charge of - 31C is placed close to a sheet of charge having acharge density 5 * 10 Cm The force between the particle and sheet of charge is1) 84 7 N (attraction)2) 84.7 N (repulsion)3) 0.847 N (attractive)4) 0.847 N (repulsion) |
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Answer» Answer: The electric field due to charge plate with charge density (σ) is GIVEN by, E = σ/2ϵ₀Now the force between charge plate and charge Q is, F = qEsubstituting the value of E in the equation, F = qσ/2ϵ₀Here,
∴ F = (-31) × {(5×10¹⁰)/(2 × 8.85 × 10⁻¹²)} ⇒ F = -875.706 N |
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