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 (I_d) is given by the formula:

Formula for displacement current:
\( I_d = \varepsilon_0 A \frac{dE}{dt} \)

where:

• \(\varepsilon_0\) = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \, \text{m}^2 \) is the permittivity of free space
• A = \( 2.5 \times 10^{-3} \, \text{m}^2 \) is the area of the plates
• \(\frac{dE}{dt}\) = \( 4 \times 10^6 \, \text{V/m/s} \) is the rate of change of the electric field

Substituting these values into the formula for I_d:

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

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