Intensity of electric field at a point at a perpendicular distance .r. from an infinite line charge, having linear charge density `.lambda.` is given by :

Options

A. A) `E=((1)/(4pi epsi_(0)))(lambda)/(r)`
B. B) `E=((1)/(4pi epsi_(0)))(2lambda)/(r)`
C. C) `E=((1)/(4pi epsi_(0)))(lambda)/(r^(2))`
D. D) `E=((1)/(4pi epsi_(0))) (2lambda)/(r^(2))`

Answer

To find the intensity of the electric field at a point at a perpendicular distance \( r \) from an infinite line charge with linear charge density \( \lambda \), we can use Gauss's Law. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have an infinite line charge with a linear charge density \( \lambda \). We want to find the electric field intensity \( E \) at a distance \( r \) from this line charge.

### Step 2: Choose a Gaussian Surface
To apply Gauss's Law, we choose a cylindrical Gaussian surface of radius \( r \) and length \( L \) that is coaxial with the line charge. The symmetry of the problem suggests that the electric field will be uniform and directed radially outward from the line charge.

### Step 3: Apply Gauss's Law
Gauss's Law states:
\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux through the Gaussian surface, \( q_{\text{enclosed}} \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Calculate the Electric Flux
The electric field \( E \) is constant over the curved surface of the cylinder, and there is no contribution from the ends of the cylinder (the electric field is perpendicular to the area vector at those points). Thus, the electric flux through the curved surface is:
\[ \Phi_E = E \cdot (2\pi r L) \]
where \( 2\pi r L \) is the surface area of the curved side of the cylinder.

### Step 5: Calculate the Charge Enclosed
The charge enclosed by the Gaussian surface is given by:
\[ q_{\text{enclosed}} = \lambda \cdot L \]
where \( \lambda \) is the linear charge density.

### Step 6: Substitute into Gauss's Law
Substituting the expressions for electric flux and charge enclosed into Gauss's Law gives:
\[ E \cdot (2\pi r L) = \frac{\lambda L}{\epsilon_0} \]

### Step 7: Solve for Electric Field \( E \)
Now, we can solve for \( E \):
\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]

### Conclusion
Thus, the intensity of the electric field at a point at a perpendicular distance \( r \) from an infinite line charge with linear charge density \( \lambda \) is given by:
\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]