Question

The electric field in space between the plates of a parallel plate capacitor (each of area \( 2.5 \times 10^{-3} \, \text{m}^2 \)) is changing at the rate of \( 4 \times 10^6 \, \text{Vm}^{-1}\text{s}^{-1} \). The displacement current between the plates of the capacitor is :

• A. \( 1.8 \times 10^{-5} \, \text{A} \)
• B. \( 3.47 \times 10^{-6} \, \text{A} \)
• C. \( 8.85 \times 10^{-8} \, \text{A} \)
• D. \( 6.32 \times 10^{-4} \, \text{A} \)

Answer 1

The Correct Option is C

The displacement current, denoted as \( I_d \), is defined by the following equation:

Displacement Current Formula:
\( I_d = \varepsilon_0 A \frac{dE}{dt} \)

The components are defined as:

• \(\varepsilon_0\): Permittivity of free space, with a value of \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \, \text{m}^2 \).
• A: The area of the plates, measuring \( 2.5 \times 10^{-3} \, \text{m}^2 \).
• \(\frac{dE}{dt}\): The rate of change of the electric field, calculated as \( 4 \times 10^6 \, \text{V/m/s} \).

Upon substituting these values into the displacement current formula:

\( I_d = (8.85 \times 10^{-12})(2.5 \times 10^{-3})(4 \times 10^6) \)

Calculated Displacement Current:
\( I_d = 8.85 \times 10^{-8} \, \text{A} \)