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Question

Physics Question on Electromagnetic waves

A parallel plate capacitor of capacitance 20μF20\mu\,F is being charged by a voltage source whose potential is changing at the rate of 3V/s3 \,V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :

A

60μA60\mu\,A, zero

B

zero, zero

C

zero, 60μA60\mu\,A

D

60μA60\mu\,A 60μA60\mu\,A

Answer

60μA60\mu\,A 60μA60\mu\,A

Explanation

Solution

Capacitance of capacitor C=20μFC =20\mu F
=20×106F= 20 \times10^{-6} F
Rate of change of potential (dVdt)=3v/s \left(\frac{dV}{dt}\right) =3v/s
q=CVq=CV
dqdt=CdVdt\frac{dq}{dt} =C \frac{dV}{dt}
ic=20×106×3i_{c} = 20\times10^{-6} \times3
=60×106A=60μA= 60\times10^{-6} A = 60\mu A
As we know that id=ic=60μAi_{d} =i_{c} = 60 \mu A