There is a parallel plate capacitor of capacitance \( 3\,\mu\mathrm{F} \). The voltage between the plates of parallel plate capacitor is changing at the rate of 8 V/s. The displacement current produced is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 24\,\mu\mathrm{A} \)
The displacement current is the current produced by a changing electric flux, \( I_{d} = \varepsilon_{0}\,d\Phi_{E}/dt \). Between the plates of a capacitor it equals the conduction current in the wires, \( dq/dt \).
\( \displaystyle I_{d} = \frac{dq}{dt} = \frac{d(CV)}{dt} = C\,\frac{dV}{dt} \)
\( \displaystyle I_{d} = 3\times10^{-6}\,\mathrm{F} \times 8\,\mathrm{V\,s^{-1}} = 24\times10^{-6}\,\mathrm{A} = 24\,\mu\mathrm{A} \)
The capacitance is fixed, so the rate of change of voltage alone sets the current.