A leaky parallel plate capacitor is filled completely with a material having dielectric constant `K = 5` and electrical conductivity `sigma= 7.4 xx 10^-12 Omega^-1 m^-1`. If the charge on the capacitor at the instant `t = 0` is `q_0 = 8.55 muC`, then calculate the leakage current at the instant `t = 12 s`.

Explanation

To solve the problem of finding the leakage current at the instant \( t = 12 \) seconds for a leaky parallel plate capacitor filled with a dielectric material, we will follow these steps:

### Step 1: Identify Given Values
- Dielectric constant, \( K = 5 \)
- Electrical conductivity, \( \sigma = 7.4 \times 10^{-12} \, \Omega^{-1} \, \text{m}^{-1} \)
- Initial charge on the capacitor, \( q_0 = 8.55 \, \mu C = 8.55 \times 10^{-6} \, C \)

### Step 2: Calculate the Resistivity
The resistivity \( \rho \) is given by the reciprocal of the conductivity:
\[ \rho = \frac{1}{\sigma} = \frac{1}{7.4 \times 10^{-12}} \approx 1.3514 \times 10^{11} \, \Omega \cdot m \]

### Step 3: Calculate the Time Constant \( \tau \)
The time constant \( \tau \) for a leaky capacitor can be calculated using the formula:
\[ \tau = R \cdot C \]
Where:
- \( R = \frac{\rho d}{A} \)
- \( C = K \cdot \epsilon_0 \cdot \frac{A}{d} \)

Since \( d \) and \( A \) will cancel out when calculating \( R \cdot C \), we can simplify:
\[ \tau = \rho \cdot K \cdot \epsilon_0 \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, F/m \).

Substituting the values:
\[ \tau = \left(1.3514 \times 10^{11}\right) \cdot 5 \cdot \left(8.85 \times 10^{-12}\right) \]
Calculating this gives:
\[ \tau \approx 5 \cdot 1.3514 \cdot 8.85 \approx 5 \cdot 1.196 \approx 5.98 \approx 6 \, seconds \]

### Step 4: Calculate the Initial Leakage Current \( I_0 \)
The initial leakage current \( I_0 \) can be calculated using:
\[ I_0 = \frac{q_0}{\tau} \]
Substituting the values:
\[ I_0 = \frac{8.55 \times 10^{-6}}{6} \approx 1.425 \times 10^{-6} \, A = 1.425 \, \mu A \]

### Step 5: Calculate the Leakage Current at \( t = 12 \) seconds
The leakage current \( I(t) \) at time \( t \) can be calculated using the formula:
\[ I(t) = I_0 \cdot e^{-\frac{t}{\tau}} \]
Substituting \( t = 12 \, seconds \):
\[ I(12) = 1.425 \cdot e^{-\frac{12}{6}} = 1.425 \cdot e^{-2} \]
Calculating \( e^{-2} \approx 0.1353 \):
\[ I(12) \approx 1.425 \cdot 0.1353 \approx 0.193 \, \mu A \]

### Final Answer
The leakage current at the instant \( t = 12 \) seconds is approximately \( 0.193 \, \mu A \).

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