A positively charged rod lies along X-axis in such a manner that one end of the rod is at the origin and the other end at `x =-oo`. The linear charge density for the rod is `lambda`. Electric field intensity at x =a is

Options

A. `lambda/(4piepsilon_0a)hati`
B. `-lambda/(4piepsilon_0a)hati`
C. `lambda/(4piepsilon_0a^2)hati`
D. `-lambda/(4piepsilon_0a^2) hati`

Answer

To find the electric field intensity at point \( x = a \) due to a positively charged rod lying along the negative x-axis, we can follow these steps:

### Step 1: Understand the Configuration
The rod is positioned along the x-axis from the origin (0,0) to \( x = -\infty \). The linear charge density of the rod is given as \( \lambda \).

### Step 2: Define a Small Charge Element
Consider a small element of the rod at position \( x \) with a length \( dx \). The charge \( dq \) on this small element can be expressed as:
\[ dq = \lambda \, dx \]

### Step 3: Determine the Distance to the Point of Interest
The point where we want to calculate the electric field is at \( x = a \). The distance from the charge element at position \( x \) to the point \( x = a \) is:
\[ r = a - x \]

### Step 4: Write the Expression for the Electric Field due to \( dq \)
The electric field \( dE \) produced by the small charge \( dq \) at the point \( x = a \) is given by Coulomb's law:
\[ dE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{(a - x)^2} \]
Substituting \( dq \):
\[ dE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\lambda \, dx}{(a - x)^2} \]

### Step 5: Integrate to Find the Total Electric Field
To find the total electric field \( E \) at \( x = a \), we need to integrate \( dE \) from \( x = -\infty \) to \( x = 0 \):
\[ E = \int_{-\infty}^{0} dE = \int_{-\infty}^{0} \frac{\lambda}{4 \pi \epsilon_0} \cdot \frac{1}{(a - x)^2} \, dx \]
This can be simplified to:
\[ E = \frac{\lambda}{4 \pi \epsilon_0} \int_{-\infty}^{0} \frac{1}{(a - x)^2} \, dx \]

### Step 6: Evaluate the Integral
To evaluate the integral, we can use the substitution \( u = a - x \), which gives \( du = -dx \). When \( x = -\infty \), \( u = \infty \) and when \( x = 0 \), \( u = a \):
\[ E = \frac{\lambda}{4 \pi \epsilon_0} \int_{\infty}^{a} \frac{-1}{u^2} \, du \]
This integral evaluates to:
\[ E = \frac{\lambda}{4 \pi \epsilon_0} \left[ \frac{1}{u} \right]_{\infty}^{a} = \frac{\lambda}{4 \pi \epsilon_0} \left( 0 - \frac{1}{a} \right) = -\frac{\lambda}{4 \pi \epsilon_0 a} \]

### Step 7: Determine the Direction of the Electric Field
Since the rod is positively charged, the electric field at point \( x = a \) will point towards the rod (i.e., in the negative x-direction). Thus, we can express the electric field vectorially as:
\[ E = -\frac{\lambda}{4 \pi \epsilon_0 a} \hat{i} \]

### Final Answer
The electric field intensity at \( x = a \) is:
\[ E = -\frac{\lambda}{4 \pi \epsilon_0 a} \hat{i} \]