A large parallel plate capacitor, whose plates have an area of `1m^2` and are separated from each other by 1 mm, is being charged at a rate of `25Vs^(-1)` . If the dielectric constant 10, then the displacement current at this instant is

Options

A. `25muA`
B. `11muA`
C. `2.2muA`
D. `1.1muA`

Answer

To find the displacement current in a parallel plate capacitor, we can use the formula for displacement current \( I_d \):

\[ I_d = C \frac{dV}{dt} \]

where:
- \( C \) is the capacitance of the capacitor,
- \( \frac{dV}{dt} \) is the rate of change of voltage across the capacitor.

### Step 1: Calculate the Capacitance \( C \)

The capacitance \( C \) of a parallel plate capacitor can be calculated using the formula:

\[ C = \frac{\varepsilon_0 \cdot A \cdot k}{d} \]

where:
- \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space),
- \( A = 1 \, \text{m}^2 \) (area of the plates),
- \( k = 10 \) (dielectric constant),
- \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) (separation between the plates).

Substituting the values into the capacitance formula:

\[ C = \frac{(8.85 \times 10^{-12}) \cdot (1) \cdot (10)}{1 \times 10^{-3}} \]

Calculating this gives:

\[ C = \frac{8.85 \times 10^{-11}}{1 \times 10^{-3}} = 8.85 \times 10^{-8} \, \text{F} \]

### Step 2: Calculate the Displacement Current \( I_d \)

Now we can substitute \( C \) and \( \frac{dV}{dt} \) into the displacement current formula. Given that \( \frac{dV}{dt} = 25 \, \text{V/s} \):

\[ I_d = C \cdot \frac{dV}{dt} = (8.85 \times 10^{-8}) \cdot (25) \]

Calculating this gives:

\[ I_d = 2.2125 \times 10^{-6} \, \text{A} \]

### Step 3: Convert to Microamperes

To express the displacement current in microamperes, we convert:

\[ I_d = 2.2125 \times 10^{-6} \, \text{A} = 2.2125 \, \mu\text{A} \approx 2.2 \, \mu\text{A} \]

### Final Answer

Thus, the displacement current at this instant is approximately:

\[ \boxed{2.2 \, \mu\text{A}} \]