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 solve the problem, we need to calculate the displacement current \( I_D \) for the given parallel plate capacitor. The formula for displacement current is given by:

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

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 with a dielectric is given by:

\[ C = \frac{k \cdot A \cdot \epsilon_0}{d} \]

Where:
- \( k \) is the dielectric constant (given as 10),
- \( A \) is the area of the plates (given as \( 1 \, m^2 \)),
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, F/m \)),
- \( d \) is the separation between the plates (given as \( 1 \, mm = 1 \times 10^{-3} \, m \)).

Substituting the values into the formula:

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

Calculating this gives:

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

### Step 2: Calculate the rate of change of voltage \( \frac{dV}{dt} \)

The rate of change of voltage \( \frac{dV}{dt} \) is given in the problem as:

\[ \frac{dV}{dt} = 25 \, V/s \]

### Step 3: Calculate the displacement current \( I_D \)

Now we can substitute \( C \) and \( \frac{dV}{dt} \) into the formula for displacement current:

\[ I_D = C \frac{dV}{dt} = (8.85 \times 10^{-8} \, F) \cdot (25 \, V/s) \]

Calculating this gives:

\[ I_D = 2.2125 \times 10^{-6} \, A = 2.2125 \, \mu A \]

### Final Answer

Thus, the displacement current at this instant is approximately:

\[ I_D \approx 2.21 \, \mu A \]

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