To establish an instantaneous current of 2 A through a `1 mu F `capacitor, the potential difference across the capacitor plates should be changed at the rate of

Options

A. `2 xx 10^4 V//s`
B. `4 xx 10^(6) V//s`
C. `2 xx 10^(6) V//s`
D. `4 xx 10^4 V//s`

Answer

To solve the problem, we need to find the rate of change of potential difference (dV/dt) across a capacitor when an instantaneous current of 2 A flows through it. We are given the capacitance of the capacitor as 1 µF (microfarad).

### Step-by-Step Solution:

1. **Understand the relationship between current, capacitance, and voltage:**
The current (I) flowing through a capacitor is related to the capacitance (C) and the rate of change of voltage (dV/dt) across the capacitor by the formula:
\[ I = C \frac{dV}{dt} \]

2. **Rearranging the formula:**
To find the rate of change of voltage (dV/dt), we can rearrange the formula:
\[ \frac{dV}{dt} = \frac{I}{C} \]

3. **Substituting the known values:**
We know:
- \( I = 2 \, \text{A} \)
- \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)

Now substituting these values into the rearranged formula:
\[ \frac{dV}{dt} = \frac{2 \, \text{A}}{1 \times 10^{-6} \, F} \]

4. **Calculating dV/dt:**
Performing the calculation:
\[ \frac{dV}{dt} = 2 \times 10^{6} \, \text{V/s} \]

5. **Conclusion:**
Therefore, the potential difference across the capacitor plates should be changed at the rate of \( 2 \times 10^{6} \, \text{V/s} \).

### Final Answer:
The rate of change of potential difference (dV/dt) is \( 2 \times 10^{6} \, \text{V/s} \).