If the linear charge density of a cylinder is `4 mu C m^(-6)` ,then electric field intensity at point 3.6 cm from axis is :

Options

A. `4 xx 10^(5) NC^(-1)`
B. `2 xx 10^(6) NC^(-1)`
C. `8 xx 10^(7) NC^(-1)`
D. `12 xx 10^(7) NC^(-1)`

Explanation

To find the electric field intensity at a point 3.6 cm from the axis of a cylinder with a linear charge density of \( \lambda = 4 \, \mu C/m \), we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a cylinder with a linear charge density \( \lambda = 4 \, \mu C/m = 4 \times 10^{-6} \, C/m \). We need to find the electric field intensity at a distance \( r = 3.6 \, cm = 0.036 \, m \) from the axis of the cylinder.

**Hint:** Remember that the electric field due to a uniformly charged cylinder can be derived using Gauss's law.

### Step 2: Apply Gauss's Law
Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon_0 \):

\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]

### Step 3: Choose a Gaussian surface
For a cylindrical charge distribution, we choose a cylindrical Gaussian surface of radius \( r \) and length \( h \). The electric field \( E \) is uniform and directed radially outward.

### Step 4: Calculate the charge enclosed
The charge enclosed by the Gaussian surface is given by:

\[ Q_{enc} = \lambda \cdot h \]

### Step 5: Calculate the electric flux
The electric flux through the curved surface of the cylinder is:

\[ \Phi_E = E \cdot (2 \pi r h) \]

### Step 6: Set up the equation using Gauss's Law
From Gauss's law, we equate the electric flux to the charge enclosed divided by \( \epsilon_0 \):

\[ E \cdot (2 \pi r h) = \frac{\lambda \cdot h}{\epsilon_0} \]

### Step 7: Simplify the equation
We can cancel \( h \) from both sides (assuming \( h \neq 0 \)):

\[ E \cdot (2 \pi r) = \frac{\lambda}{\epsilon_0} \]

### Step 8: Solve for \( E \)
Now, we can solve for the electric field \( E \):

\[ E = \frac{\lambda}{2 \pi r \epsilon_0} \]

### Step 9: Substitute the values
Substituting the values \( \lambda = 4 \times 10^{-6} \, C/m \), \( r = 0.036 \, m \), and \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \):

\[ E = \frac{4 \times 10^{-6}}{2 \pi (0.036)(8.85 \times 10^{-12})} \]

### Step 10: Calculate the value
Calculating the denominator:

\[ 2 \pi (0.036)(8.85 \times 10^{-12}) \approx 2.01 \times 10^{-12} \]

Now substituting back:

\[ E = \frac{4 \times 10^{-6}}{2.01 \times 10^{-12}} \approx 1.99 \times 10^{6} \, N/C \]

### Final Result
Thus, the electric field intensity at a point 3.6 cm from the axis of the cylinder is approximately \( 2 \times 10^{6} \, N/C \).

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