A rod lies along the X-axis with one end at the origin and the other at `x rarr oo`. It carries a uniform charge `lambda Cm^(-1)`. The electric field at the point `x=-a` on the axis will be

Options

A. `E=lambda/(4pi epsi_(0)a) (-hat(i))`
B. `E=lambda/(4pi epsi_(0)a) (hat(i))`
C. `E=lambda/(2pi epsi_(0)a) (-hat(i))`
D. `E=lambda/(2pi epsi_(0)a) (hat(i))`

Answer

To find the electric field at the point \( x = -a \) due to a uniformly charged rod extending from the origin to infinity along the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a rod lying along the x-axis from \( x = 0 \) to \( x = \infty \) with a uniform charge density \( \lambda \) (in \( \text{C/m} \)).
- We want to find the electric field at the point \( P \) located at \( x = -a \).

2. **Consider an Element of Charge**:
- Take a small element of the rod \( dx \) at a position \( x \) (where \( x \) is from \( 0 \) to \( \infty \)).
- The charge \( dq \) on this small element is given by:
\[ dq = \lambda \, dx \]

3. **Calculate the Electric Field Contribution**:
- The electric field \( dE \) at point \( P \) due to the charge \( dq \) is given by Coulomb's law:
\[ dE = \frac{k \, dq}{r^2} \]
- Here, \( r \) is the distance from the charge element to point \( P \). Since point \( P \) is at \( x = -a \), the distance \( r \) can be expressed as:
\[ r = x + a \]
- Therefore, the electric field contribution becomes:
\[ dE = \frac{k \, \lambda \, dx}{(x + a)^2} \]

4. **Determine the Direction of the Electric Field**:
- Since the rod has a positive charge, the electric field at point \( P \) will point towards the left (negative x-direction).

5. **Integrate to Find the Total Electric Field**:
- The total electric field \( E \) at point \( P \) is obtained by integrating \( dE \) from \( x = 0 \) to \( x = \infty \):
\[ E = \int_{0}^{\infty} dE = \int_{0}^{\infty} \frac{k \, \lambda \, dx}{(x + a)^2} \]

6. **Perform the Integration**:
- The integral can be solved as follows:
\[ E = k \lambda \int_{0}^{\infty} \frac{dx}{(x + a)^2} \]
- The integral \( \int \frac{dx}{(x + a)^2} \) evaluates to \( -\frac{1}{x + a} \), so we compute:
\[ E = k \lambda \left[-\frac{1}{x + a}\right]_{0}^{\infty} = k \lambda \left(0 - \left(-\frac{1}{a}\right)\right) = \frac{k \lambda}{a} \]

7. **Final Result**:
- The total electric field at point \( P \) is:
\[ E = \frac{k \lambda}{a} \]
- Since the direction is towards the left, we can express it as:
\[ E = -\frac{k \lambda}{a} \hat{i} \]