A cylindrical capacitor has has two co-axial cylinders of length 20 cm and radii 2r and r, inner cylinder is gliven a charge 10 `mu`C and and outer cylinder a charge of - 10 `mu`C. the potential difference between the two cylinders will be

Options

A. `(0.1 " In" 2 )/(4 pi epsilon_(0)) `m V
B. `( " In" 2 )/(4 pi epsilon_(0)) `m V
C. `(10 " In" 2 )/(4 pi epsilon_(0)) `m V
D. `(0.01 " In" 2 )/(4 pi epsilon_(0)) `m V

Answer

To find the potential difference between the two cylinders of a cylindrical capacitor, we can follow these steps:

### Step 1: Identify the given values
- Length of the cylindrical capacitor, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Charge on the inner cylinder, \( Q = 10 \, \mu\text{C} = 10 \times 10^{-6} \, \text{C} \)
- Radius of the inner cylinder, \( r \)
- Radius of the outer cylinder, \( R_2 = 2r \)

### Step 2: Calculate the capacitance of the cylindrical capacitor
The formula for the capacitance \( C \) of a cylindrical capacitor is given by:

\[ C = \frac{2 \pi \epsilon_0 L}{\ln\left(\frac{R_2}{R_1}\right)} \]

Here, \( R_1 = r \) and \( R_2 = 2r \). Thus, we can substitute these values into the formula:

\[ C = \frac{2 \pi \epsilon_0 (0.2)}{\ln\left(\frac{2r}{r}\right)} = \frac{2 \pi \epsilon_0 (0.2)}{\ln(2)} \]

### Step 3: Calculate the potential difference \( V \)
The potential difference \( V \) between the two cylinders can be calculated using the formula:

\[ V = \frac{Q}{C} \]

Substituting the expression for \( C \):

\[ V = \frac{10 \times 10^{-6}}{\frac{2 \pi \epsilon_0 (0.2)}{\ln(2)}} \]

This simplifies to:

\[ V = \frac{10 \times 10^{-6} \ln(2)}{2 \pi \epsilon_0 (0.2)} \]

### Step 4: Substitute the value of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) is approximately \( 8.854 \times 10^{-12} \, \text{F/m} \). Substituting this value into the equation:

\[ V = \frac{10 \times 10^{-6} \ln(2)}{2 \pi (8.854 \times 10^{-12}) (0.2)} \]

### Step 5: Calculate the final value
Now we can calculate the numerical value of \( V \):

1. Calculate \( 2 \pi (8.854 \times 10^{-12}) (0.2) \):
\[ 2 \pi (8.854 \times 10^{-12}) (0.2) \approx 1.113 \times 10^{-12} \]

2. Then calculate \( V \):
\[ V \approx \frac{10 \times 10^{-6} \ln(2)}{1.113 \times 10^{-12}} \]

3. Using \( \ln(2) \approx 0.693 \):
\[ V \approx \frac{10 \times 10^{-6} \times 0.693}{1.113 \times 10^{-12}} \approx 6.22 \times 10^{6} \, \text{V} \]

### Final Result
The potential difference between the two cylinders is approximately \( 6.22 \times 10^{6} \, \text{V} \) or \( 0.1 \ln(2) / (4 \pi \epsilon_0) \, \text{mV} \).