The rate of change of voltage of a parallel plate capacitor if the instantaneous displacement current of 1A is established between the two plates of a 1µF parallel plate capacitor

Options

A. `10^(6)` v/s
B. 10 v/s
C. `10^(8)` v/s
D. `10^(–6)` v/s

Answer

To solve the problem of finding the rate of change of voltage across a parallel plate capacitor with a given displacement current and capacitance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Displacement current (\(I_d\)) = 1 A
- Capacitance (\(C\)) = 1 µF = \(1 \times 10^{-6}\) F

2. **Use the Relationship of Displacement Current**:
The displacement current can be expressed in terms of the electric field and the area of the capacitor plates using Maxwell's equations:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \(\Phi_E\) is the electric flux.

3. **Express Electric Flux**:
The electric flux (\(\Phi_E\)) is given by:
\[ \Phi_E = E \cdot A \]
where \(E\) is the electric field and \(A\) is the area of the plates.

4. **Substitute Electric Flux into the Displacement Current Equation**:
From the previous equations, we can substitute:
\[ I_d = \epsilon_0 \frac{d(E \cdot A)}{dt} \]
Since the area \(A\) is constant, we can simplify this to:
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{dE}{dt} = \frac{I_d}{\epsilon_0 A} \]

6. **Relate Electric Field to Voltage**:
The relationship between electric field \(E\) and voltage \(V\) across the capacitor is:
\[ V = E \cdot d \]
where \(d\) is the separation between the plates. Thus:
\[ \frac{dV}{dt} = \frac{dE}{dt} \cdot d \]

7. **Substituting for \(\frac{dE}{dt}\)**:
Substituting the expression for \(\frac{dE}{dt}\) into the equation for \(\frac{dV}{dt}\):
\[ \frac{dV}{dt} = \left(\frac{I_d}{\epsilon_0 A}\right) \cdot d \]

8. **Using the Capacitance Formula**:
The capacitance \(C\) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
Rearranging gives:
\[ \epsilon_0 = \frac{C \cdot d}{A} \]

9. **Substituting \(\epsilon_0\) into the Voltage Change Rate**:
Substitute \(\epsilon_0\) back into the equation for \(\frac{dV}{dt}\):
\[ \frac{dV}{dt} = \frac{I_d \cdot d}{\frac{C \cdot d}{A}} = \frac{I_d \cdot A}{C} \]

10. **Final Calculation**:
Substitute the known values:
\[ \frac{dV}{dt} = \frac{1 \, \text{A} \cdot A}{1 \times 10^{-6} \, \text{F}} \]
Since \(A\) cancels out, we find:
\[ \frac{dV}{dt} = \frac{1}{1 \times 10^{-6}} = 10^6 \, \text{V/s} \]

### Final Answer:
The rate of change of voltage across the parallel plate capacitor is \(10^6 \, \text{V/s}\).