Question

A parallel plate capacitor with plate area A and plate separation \(d\) is charged by constant current \(I\). A plane surface of area \(\frac{\mathrm{A}}{2}\), parallel to the plates is drawn simultaneously between the plates. The displacement current through this area is

• A. I
• B. \(\frac{\mathrm{I}}{2}\)
• C. \(\frac{\mathrm{I}}{4}\)
• D. \(\frac{\mathrm{I}}{8}\)

Hint:

Displacement current density is uniform between the plates, so it scales with area.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Between the plates of a charging capacitor the displacement current equals the conduction current in the wires, \( I_d=I \), and it is spread uniformly over the plate area \( A \).

Step 2: Current Density
\( J_d=\dfrac{I}{A} \)

Step 3: Smaller Surface
For a parallel surface of area \( \dfrac{A}{2} \):
\( I'=J_d\times\dfrac{A}{2}=\dfrac{I}{2} \)

Step 4: Compare the options
This matches option (B).

Final Answer:
The displacement current through the smaller surface is half of I. \[ \boxed{\text{(B) }\dfrac{\mathrm{I}}{2}} \]