Question

A parallel plate capacitor consists of two circular plates of radius \( R = 0.1 \) m. They are separated by a short distance. If the electric field between the capacitor plates changes as: \[ \frac{dE}{dt} = 6 \times 10^{13} \frac{V}{m \cdot s} \] then the value of the displacement current is:

• A. 15.25 A
• B. 6.25 A
• C. 16.67 A
• D. 4.69 A

Hint:

Displacement current plays a crucial role in Maxwell's equations, bridging the gap between capacitors in AC circuits.

Answer 1

The Correct Option is C

Step 1: {Apply Maxwell's Displacement Current Formula}
The formula for displacement current is:\[I_d = \varepsilon_0 \frac{d\Phi}{dt}\]Given that:\[\frac{d\Phi}{dt} = A \frac{dE}{dt}\]Step 2: {Determine the Area of the Plates}
\[A = \pi R^2 = 3.14 \times (0.1)^2 = 3.14 \times 10^{-2} { m}^2\]Step 3: {Calculate the Displacement Current}
Substitute the values into the formula:\[I_d = \varepsilon_0 A \frac{dE}{dt}\]\[= (8.85 \times 10^{-12}) \times (3.14 \times 10^{-2}) \times (6 \times 10^{13})\]\[I_d = 16.67 { A}\]Therefore, the displacement current is \( 16.67 \) A.