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: Understanding the Concept:
This problem deals with the concept of displacement current, which was introduced by James Clerk Maxwell. He postulated that a changing electric flux between the plates of a capacitor produces a magnetic field, just as a conduction current does. This effective current due to the changing electric flux is called the displacement current.
Step 2: Key Formula or Approach:
The displacement current (\(I_d\)) is defined as:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \] where \(\epsilon_0\) is the permittivity of free space and \(\frac{d\Phi_E}{dt}\) is the rate of change of electric flux.
Step 3: Detailed Explanation:
Given values:
- Rate of change of electric flux, \(\frac{d\Phi_E}{dt} = 9\pi \times 10^3 \text{ V}\cdot\text{m/s}\).
- The value of \(\epsilon_0\) is the permittivity of free space. We can use the relation from Coulomb's constant, \(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2\). From this, we can express \(\epsilon_0\) as:
\[ \epsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} \] Now, substitute the values into the formula for displacement current:
\[ I_d = \left( \frac{1}{4\pi \times 9 \times 10^9} \right) \times \left( 9\pi \times 10^3 \right) \] We can cancel the \(9\pi\) term from the numerator and the denominator.
\[ I_d = \frac{10^3}{4 \times 10^9} = \frac{1}{4} \times 10^{3-9} = 0.25 \times 10^{-6} \text{ A} \] This is equivalent to 0.25 microamperes (µA).
\[ I_d = 0.25 \text{ µA} \] Step 4: Final Answer:
The displacement current inside the capacitor is 0.25 µA. Therefore, option (B) is correct.