To establish an instantaneous displacement current of `2A` in the space between two parallel plates of `1muF` capacitor, the potential difference across the capacitor plates will have to be changed at the rate of

Options

A. `4xx10^(4)V//s`
B. `4xx10^(6)V//s`
C. `2xx10^(4)V//s`
D. `2xx10^(6)V//s`

Explanation

To find the rate at which the potential difference across the capacitor plates must change to establish a displacement current of 2 A in a capacitor of capacitance \(1 \mu F\), we can follow these steps:

### Step 1: Understand the relationship between displacement current and changing electric field
The displacement current \(I_d\) is given by the formula:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \(\Phi_E\) is the electric flux.

### Step 2: Relate electric flux to potential difference
The electric flux \(\Phi_E\) can be expressed in terms of the electric field \(E\) and the area \(A\):
\[ \Phi_E = E \cdot A \]
For a capacitor, the electric field \(E\) can also be expressed in terms of the potential difference \(V\) and the distance \(d\) between the plates:
\[ E = \frac{V}{d} \]
Thus, we can rewrite the electric flux as:
\[ \Phi_E = \frac{V}{d} \cdot A \]

### Step 3: Substitute into the displacement current formula
Substituting \(\Phi_E\) into the displacement current formula gives:
\[ I_d = \epsilon_0 \cdot A \cdot \frac{d}{dt}\left(\frac{V}{d}\right) \]
Since \(d\) is constant, we can simplify this to:
\[ I_d = \frac{\epsilon_0 \cdot A}{d} \cdot \frac{dV}{dt} \]

### Step 4: Recognize that \(\frac{\epsilon_0 \cdot A}{d}\) is the capacitance \(C\)
The capacitance \(C\) of a capacitor is given by:
\[ C = \frac{\epsilon_0 \cdot A}{d} \]
Thus, we can rewrite the equation for displacement current as:
\[ I_d = C \cdot \frac{dV}{dt} \]

### Step 5: Solve for \(\frac{dV}{dt}\)
Rearranging the equation to find \(\frac{dV}{dt}\):
\[ \frac{dV}{dt} = \frac{I_d}{C} \]

### Step 6: Substitute known values
Given \(I_d = 2 \, A\) and \(C = 1 \, \mu F = 1 \times 10^{-6} \, F\):
\[ \frac{dV}{dt} = \frac{2 \, A}{1 \times 10^{-6} \, F} \]

### Step 7: Calculate the result
Calculating the above expression:
\[ \frac{dV}{dt} = 2 \times 10^{6} \, V/s \]

### Final Answer
The rate at which the potential difference across the capacitor plates must change is:
\[ \frac{dV}{dt} = 2 \times 10^{6} \, V/s \]

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