Question

If the rate of change of electric field across the plates of a parallel plate capacitor is E and the displacement current is I, then the area of one plate of the capacitor is (\(\epsilon_0\) is permittivity of free space)

• A. \( \frac{I}{2\epsilon_0 E} \)
• B. \( \frac{2I}{\epsilon_0 E} \)
• C. \( I\epsilon_0 E \)
• D. \( \frac{I}{\epsilon_0 E} \)

Hint:

The notation in this problem is a bit confusing. "Rate of change of electric field is E" should be interpreted as \(dE/dt\). The displacement current \(I_d = \epsilon_0 d\Phi_E/dt\) is Maxwell's crucial addition to Ampere's law, showing that a changing electric field creates a magnetic field, just like a real current does.

Answer 1

The Correct Option is D

This question involves Maxwell's equations, specifically the concept of displacement current.
The displacement current (\(I_d\)) is defined as \( I_d = \epsilon_0 \frac{d\Phi_E}{dt} \), where \(\Phi_E\) is the electric flux.
For a parallel plate capacitor with plate area A, the electric field E between the plates is uniform (ignoring fringing effects).
The electric flux through a surface of area A parallel to the plates is \( \Phi_E = E \cdot A \).
Now, let's find the rate of change of this electric flux.
\( \frac{d\Phi_E}{dt} = \frac{d(EA)}{dt} \). Since the area A is constant, this becomes \( A \frac{dE}{dt} \).
The question states that the rate of change of the electric field is E. This is a slightly confusing notation. It means \( \frac{dE}{dt} \). Let's use this in our equation.
So, \( \frac{d\Phi_E}{dt} = A \cdot \left(\frac{dE}{dt}\right) \).
Substitute this back into the formula for displacement current.
\( I_d = \epsilon_0 \left( A \frac{dE}{dt} \right) \).
The problem states the displacement current is I and the rate of change of the electric field is E. So, \( I_d = I \) and \( \frac{dE}{dt} = E \).
The equation becomes \( I = \epsilon_0 A E \).
We need to find the area A. Rearranging the formula:
\( A = \frac{I}{\epsilon_0 E} \).