A capacitor is having a capacity of 2pF. Electric field across the capacitor is changing with a value of `10^(12)` V/s. The displacement current is

Options

A. 2A
B. 3A
C. 6A
D. 9A

Answer

To find the displacement current in the given capacitor, we can follow these steps:

### Step 1: Understand the formula for displacement current
The displacement current \( I_d \) can be calculated using the formula derived from Maxwell's equations:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux.

### Step 2: Define 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 and \( A \) is the area of the capacitor plates.

### Step 3: Differentiate the electric flux
Taking the time derivative of the electric flux, we have:
\[ \frac{d\Phi_E}{dt} = A \frac{dE}{dt} \]
Since the area \( A \) is constant, we can factor it out.

### Step 4: Substitute into the displacement current formula
Substituting the expression for \( \frac{d\Phi_E}{dt} \) into the displacement current formula gives:
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

### Step 5: Relate electric field to voltage
The relationship between electric field \( E \), voltage \( V \), and distance \( D \) is given by:
\[ V = E \cdot D \]
Thus, the rate of change of voltage can be expressed as:
\[ \frac{dV}{dt} = D \frac{dE}{dt} \]
From this, we can express \( \frac{dE}{dt} \) as:
\[ \frac{dE}{dt} = \frac{1}{D} \frac{dV}{dt} \]

### Step 6: Substitute back into the displacement current formula
Substituting \( \frac{dE}{dt} \) into the displacement current formula gives:
\[ I_d = \epsilon_0 A \left(\frac{1}{D} \frac{dV}{dt}\right) \]

### Step 7: Use the capacitance relation
The capacitance \( C \) of the capacitor is defined as:
\[ C = \frac{\epsilon_0 A}{D} \]
Thus, we can rewrite the displacement current as:
\[ I_d = C \frac{dV}{dt} \]

### Step 8: Substitute known values
Given:
- \( C = 2 \, \text{pF} = 2 \times 10^{-12} \, \text{F} \)
- \( \frac{dV}{dt} = 10^{12} \, \text{V/s} \)

Now substituting these values:
\[ I_d = (2 \times 10^{-12} \, \text{F}) \times (10^{12} \, \text{V/s}) = 2 \, \text{A} \]

### Final Answer
The displacement current \( I_d \) is:
\[ \boxed{2 \, \text{A}} \]

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