In `C - R` circuit, answer the following two questions
Dielectric constant of the slab between plates of a capacitor is 18 and its resistivity is `(4 pi xx 10^(3)) Omega - m`. Then time constant of this capacitor when directly connected to a battery will be

Options

A. `2 mu s`
B. `3 mu s`
C. `1 mu s`
D. `9 mu s`

Answer

To find the time constant of the capacitor in the given `C - R` circuit, we will follow these steps:

### Step 1: Understand the formula for time constant
The time constant (τ) of an RC circuit is given by the formula:
\[ \tau = R \times C \]
where \( R \) is the resistance and \( C \) is the capacitance.

### Step 2: Calculate the resistance (R)
The resistance \( R \) of the material can be calculated using the formula:
\[ R = \frac{\rho \cdot l}{A} \]
where:
- \( \rho \) is the resistivity of the material,
- \( l \) is the length of the conductor,
- \( A \) is the cross-sectional area.

### Step 3: Calculate the capacitance (C)
The capacitance \( C \) of the capacitor can be calculated using the formula:
\[ C = \frac{k \cdot A \cdot \epsilon_0}{d} \]
where:
- \( k \) is the dielectric constant,
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( d \) is the distance between the plates.

### Step 4: Combine the formulas
Since the area \( A \) and length \( l \) will cancel out when substituting \( R \) and \( C \) into the time constant formula, we can simplify:
\[ \tau = R \times C = \left(\frac{\rho \cdot l}{A}\right) \times \left(\frac{k \cdot A \cdot \epsilon_0}{d}\right) \]
This simplifies to:
\[ \tau = \frac{\rho \cdot k \cdot \epsilon_0 \cdot l}{d} \]

### Step 5: Substitute the known values
Given:
- \( \rho = 4\pi \times 10^3 \, \Omega \cdot m \)
- \( k = 18 \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, F/m \)

We can substitute these values into the equation:
\[ \tau = \rho \cdot k \cdot \epsilon_0 \]
Since \( l/d \) cancels out, we can directly calculate:
\[ \tau = (4\pi \times 10^3) \cdot 18 \cdot (8.85 \times 10^{-12}) \]

### Step 6: Perform the calculation
Calculating:
1. \( 4\pi \approx 12.566 \)
2. \( 12.566 \times 10^3 \approx 12566 \)
3. \( 12566 \times 18 \approx 226188 \)
4. \( 226188 \times 8.85 \times 10^{-12} \approx 2.000 \times 10^{-6} \, s \)

Thus, the time constant \( \tau \) is approximately:
\[ \tau \approx 2 \, \mu s \]

### Final Answer
The time constant of the capacitor when directly connected to a battery is \( 2 \, \mu s \).

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