A very long uniformly charged cylinder (radius R) has a surface charge density `sigma` . A very long uniformly charged line charge( linear charge density `lambda`) is placed along the cylinder axis. If electric field intensity vector outside the cylinder is zero then :

Options

A. `lambda=Rsigma`
B. `lambda=-Rsigma`
C. `lambda=2piRsigma`
D. `lambda=-2piRsigma`

Answer

To solve the problem, we need to analyze the electric fields produced by both the charged cylinder and the charged line along the cylinder's axis. We will use Gauss's law and the principle of superposition.

### Step-by-Step Solution:

1. **Identify the Electric Field due to the Charged Cylinder:**
The electric field \( E_{\text{cylinder}} \) outside a uniformly charged cylinder with surface charge density \( \sigma \) can be calculated using Gauss's law. For a cylindrical Gaussian surface of radius \( r \) (where \( r > R \)), the electric field is given by:
\[ E_{\text{cylinder}} = \frac{\sigma R}{\epsilon_0 r} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Electric Field due to the Charged Line:**
The electric field \( E_{\text{line}} \) due to a long line charge with linear charge density \( \lambda \) at a distance \( r \) from the line charge is given by:
\[ E_{\text{line}} = \frac{\lambda}{2\pi \epsilon_0 r} \]

3. **Set Up the Condition for Zero Electric Field Outside the Cylinder:**
According to the problem, the total electric field outside the cylinder is zero. Therefore, we can write:
\[ E_{\text{cylinder}} - E_{\text{line}} = 0 \]
This implies:
\[ E_{\text{cylinder}} = E_{\text{line}} \]

4. **Substitute the Expressions for the Electric Fields:**
Substituting the expressions from steps 1 and 2 into the equation gives:
\[ \frac{\sigma R}{\epsilon_0 r} = \frac{\lambda}{2\pi \epsilon_0 r} \]

5. **Simplify the Equation:**
We can cancel \( \epsilon_0 r \) from both sides (assuming \( r \neq 0 \)):
\[ \sigma R = \frac{\lambda}{2\pi} \]

6. **Solve for \( \lambda \):**
Rearranging the equation gives:
\[ \lambda = 2\pi \sigma R \]

### Final Answer:
The value of the linear charge density \( \lambda \) is:
\[ \lambda = 2\pi \sigma R \]