In order to establish an instantaneous displacement current of `1 mA` in the space between the plates of `2 mu F` parallel plate capacitor, the rate of change of potential difference needed to be applied is

Options

A. `100 V s^(-1)`
B. `200V s^(-1)`
C. `300Vs^(-1)`
D. `500 V s^(-1)`

Explanation

To solve the problem of finding the rate of change of potential difference required to establish an instantaneous displacement current of 1 mA in a parallel plate capacitor of capacitance 2 µF, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Displacement Current**:
The displacement current \( I_d \) is related to the rate of change of electric flux. According to Maxwell's equations, the displacement current can be expressed as:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux.

2. **Express Electric Flux**:
The electric flux \( \Phi_E \) through the capacitor can be expressed as:
\[ \Phi_E = E \cdot A \]
where \( E \) is the electric field between the plates and \( A \) is the area of the plates.

3. **Relate Electric Field to Potential Difference**:
The electric field \( E \) between the plates of a capacitor is related to the potential difference \( V \) and the distance \( d \) between the plates:
\[ V = E \cdot d \quad \Rightarrow \quad E = \frac{V}{d} \]

4. **Substitute Electric Field into the Flux Equation**:
Substitute \( E \) into the electric flux equation:
\[ \Phi_E = \frac{V}{d} \cdot A \]

5. **Differentiate Electric Flux**:
Taking the derivative of electric flux with respect to time gives:
\[ \frac{d\Phi_E}{dt} = \frac{A}{d} \frac{dV}{dt} \]

6. **Substitute into the Displacement Current Equation**:
Now substitute \( \frac{d\Phi_E}{dt} \) back into the displacement current equation:
\[ I_d = \epsilon_0 \frac{A}{d} \frac{dV}{dt} \]

7. **Express Capacitance**:
Recall that the capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
Therefore, we can rewrite the equation for displacement current as:
\[ I_d = C \frac{dV}{dt} \]

8. **Rearranging for \( \frac{dV}{dt} \)**:
Rearranging gives:
\[ \frac{dV}{dt} = \frac{I_d}{C} \]

9. **Substituting Known Values**:
Now substitute the given values:
- \( I_d = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)
- \( C = 2 \, \mu\text{F} = 2 \times 10^{-6} \, \text{F} \)

Thus,
\[ \frac{dV}{dt} = \frac{1 \times 10^{-3}}{2 \times 10^{-6}} = \frac{1}{2} \times 10^{3} = 500 \, \text{V/s} \]

### Final Answer:
The rate of change of potential difference needed to establish an instantaneous displacement current of 1 mA is:
\[ \frac{dV}{dt} = 500 \, \text{V/s} \]