Question

If the rate of change in electric flux between the plates of a capacitor is \(9\pi \times 10^3 \text{ Vms}^{-1}\), then the displacement current inside the capacitor is

• A. 0.36 µA
• B. 0.25 µA
• C. 3.14 µA
• D. 4 µA

Hint:

Instead of memorizing the numerical value of \(\epsilon_0\), it's often easier and less prone to error to remember the relation \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \). This is particularly useful when the problem contains factors of \(\pi\) or 9, as they often cancel out.

Answer 1

The Correct Option is B

Step 1: Formula for Displacement Current. Maxwell's definition of displacement current (\(I_d\)) is: \[ I_d = \varepsilon_0 \frac{d\phi_E}{dt} \] Where \(\varepsilon_0\) is the permittivity of free space (\( \approx 8.854 \times 10^{-12} \, \text{F/m} \)) and \(\frac{d\phi_E}{dt}\) is the rate of change of electric flux.
Step 2: Use the relationship with Coulomb's constant. We know that \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2}\). So, \(\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} = \frac{1}{36\pi \times 10^9}\).
Step 3: Calculation. Given \(\frac{d\phi_E}{dt} = 9\pi \times 10^3\). \[ I_d = \left( \frac{1}{36\pi \times 10^9} \right) \times (9\pi \times 10^3) \] \[ I_d = \frac{9\pi}{36\pi} \times \frac{10^3}{10^9} \] \[ I_d = \frac{1}{4} \times 10^{-6} \, \text{A} \] \[ I_d = 0.25 \times 10^{-6} \, \text{A} = 0.25 \, \mu\text{A} \]