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NEET PHYSICSEasy

A parallel plate capacitor of capacitance 20 \mu F is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:

A

zero, zero

B

zero, 60 \mu A

C

60 \mu A, 60 \mu A

D

60 \mu A, zero

Step-by-Step Solution

  1. Conduction Current (IcI_c): The conduction current flowing through the connecting wires is the rate at which charge is supplied to the capacitor plates. Using the relation q=CVq = CV, we differentiate with respect to time: Ic=dqdt=d(CV)dt=CdVdtI_c = \frac{dq}{dt} = \frac{d(CV)}{dt} = C \frac{dV}{dt} Given C=20 \muF=20×10−6 FC = 20 \text{ \mu F} = 20 \times 10^{-6} \text{ F} and dVdt=3 V/s\frac{dV}{dt} = 3 \text{ V/s}: Ic=(20×10−6)×3=60×10−6 A=60 \muAI_c = (20 \times 10^{-6}) \times 3 = 60 \times 10^{-6} \text{ A} = 60 \text{ \mu A}
  2. Displacement Current (IdI_d): According to Maxwell's correction to Ampere's Circuital Law, the changing electric field between the plates constitutes a displacement current. By the property of continuity of current, the displacement current in the gap is always equal to the conduction current in the wires. Id=Ic=60 \muAI_d = I_c = 60 \text{ \mu A} Alternatively, using the formula Id=ε0dΦEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt} leads to the same result .
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