A displacement current of 4.0 A is established between the plates of a 6 μF parallel-plate capacitor. If the rate of change of potential difference across the capacitor is α×10⁶ V·s⁻¹, what is the value of α?
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Question

A displacement current of 4.0 A is established between the plates of a 6 μF parallel-plate capacitor. If the rate of change of potential difference across the capacitor is α×10⁶ V·s⁻¹, what is the value of α?

Options

G1 Physics · Accepted Answer

Displacement current for a capacitor is I = C(dV ÷ dt). Therefore, dV ÷ dt = I ÷ C. Substitute: I = 4.0 A, C = 6×10⁻⁶ F. Thus, dV ÷ dt = 4 ÷ (6×10⁻⁶) = (2 ÷ 3)×10⁶ V·s⁻¹ ≈ 0.67×10⁶ V·s⁻¹. Therefore α = 0.67. Correct option: B.