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: {Using Maxwell's Displacement Current Formula}
The displacement current is given by: \[ I_d = \varepsilon_0 \frac{d\Phi}{dt} \] Since: \[ \frac{d\Phi}{dt} = A \frac{dE}{dt} \] Step 2: {Finding the Area of Plates}
\[ A = \pi R^2 = 3.14 \times (0.1)^2 = 3.14 \times 10^{-2} { m}^2 \] Step 3: {Calculating Displacement Current}
\[ I_d = \varepsilon_0 A \frac{dE}{dt} \] \[ = (8.85 \times 10^{-12}) \times (3.14 \times 10^{-2}) \times (6 \times 10^{13}) \] \[ I_d = 16.67 { A} \] Thus, the correct answer is \( 16.67 \) A.

Answer 2

Step 1: Understand displacement current
Displacement current \( I_d \) is given by Maxwell’s extension to Ampere’s Law:
\[ I_d = \varepsilon_0 \cdot \frac{d\Phi_E}{dt} \]
Where \( \Phi_E \) is the electric flux:
\[ \Phi_E = E \cdot A \]
So,
\[ \frac{d\Phi_E}{dt} = A \cdot \frac{dE}{dt} \]
Thus,
\[ I_d = \varepsilon_0 \cdot A \cdot \frac{dE}{dt} \]

Step 2: Use the given values
- Radius \( R = 0.1 \) m
- Area \( A = \pi R^2 = \pi (0.1)^2 = 0.01\pi \) m²
- \( \frac{dE}{dt} = 6 \times 10^{13} \) V/m·s
- \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m

Step 3: Plug into the formula
\[ I_d = 8.85 \times 10^{-12} \cdot 0.01\pi \cdot 6 \times 10^{13} \]
\[ I_d = 8.85 \cdot 0.01 \cdot \pi \cdot 6 \times 10^{1} \]
\[ I_d \approx 8.85 \cdot 0.01 \cdot 3.1416 \cdot 60 \]
\[ I_d \approx (8.85 \cdot 0.01 \cdot 188.496) \approx 16.67 \text{ A} \]

Step 4: Final Answer
The displacement current is:
16.67 A