A cylindrical capacitor has two co-axial cylinders of length 20 cm and radii 1.5 cm and 1.6 cm. The outer cylinder is earthed and inner cylinder is given a charge `4 muC.` The capacitance of the system is (neglect end effect)

Options

A. `2.8xx10^(-8)F`
B. `4.2xx10^(-14)F`
C. `1.7xx10^(-10)F`
D. `3.4xx10^(-12)F`

Explanation

To find the capacitance of a cylindrical capacitor with two coaxial cylinders, we will follow these steps:

### Step 1: Identify the given values
- Length of the cylinders (L) = 20 cm = 0.2 m
- Radius of the inner cylinder (R2) = 1.5 cm = 0.015 m
- Radius of the outer cylinder (R1) = 1.6 cm = 0.016 m
- Charge on the inner cylinder (Q) = 4 µC = 4 × 10^(-6) C

### Step 2: Write the formula for capacitance of a cylindrical capacitor
The capacitance (C) of a cylindrical capacitor is given by the formula:
\[ C = \frac{2 \pi \epsilon_0 L}{\log_{10}(B/A)} \]
where:
- \( \epsilon_0 \) = permittivity of free space = \( 8.854 \times 10^{-12} \, \text{F/m} \)
- A = radius of the inner cylinder (R2)
- B = radius of the outer cylinder (R1)

### Step 3: Substitute the values into the formula
Substituting the values into the capacitance formula:
\[ C = \frac{2 \pi (8.854 \times 10^{-12}) (0.2)}{\log_{10}(0.016/0.015)} \]

### Step 4: Calculate the logarithmic term
Calculate the logarithmic term:
\[ \log_{10}(0.016/0.015) = \log_{10}(1.0667) \approx 0.028 \]

### Step 5: Substitute the logarithmic value back into the formula
Now substitute this value back into the capacitance formula:
\[ C = \frac{2 \pi (8.854 \times 10^{-12}) (0.2)}{0.028} \]

### Step 6: Perform the calculations
Calculating the numerator:
\[ 2 \pi (8.854 \times 10^{-12}) (0.2) \approx 1.113 \times 10^{-12} \]

Now divide by the logarithmic term:
\[ C \approx \frac{1.113 \times 10^{-12}}{0.028} \approx 3.975 \times 10^{-11} \, \text{F} \]

### Step 7: Final result
Thus, the capacitance of the cylindrical capacitor system is approximately:
\[ C \approx 3.975 \times 10^{-11} \, \text{F} \]