Question

A parallel plate capacitor with capacitance \(1\ \mu \text{F}\) is being charged at a rate such that the potential difference across it changes at \(10^{6}\ \text{V/s}\). The instantaneous displacement current between the plates is

• A. \(0\ \text{A}\)
• B. \(0.1\ \text{A}\)
• C. \(1\ \text{A}\)
• D. \(10^{-2}\ \text{A}\)

Hint:

The displacement current equals the rate of change of charge on the plates, which is C times dV by dt.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The displacement current is \(I_{d}=\varepsilon_{0}\dfrac{d\Phi_{E}}{dt}\). For a parallel-plate capacitor this equals \(\dfrac{dq}{dt}=C\dfrac{dV}{dt}\).

Step 2: Calculation
\(I_{d}=1\times10^{-6}\times10^{6}\)

Step 3: Result
\(I_{d}=1\ \text{A}\), which equals the conduction current in the connecting leads.

Final Answer:
The displacement current is 1 A. \[ \boxed{\text{(C) }1\ \text{A}} \]