An infinite line of charge `lamda` per unit length is placed along the y-axis. The work done in moving a charge q from `A(a,0)` to `B (2a,0)` is

Options

A. `(q lamda)/(2piepsilon_0) `"ln" 2`
B. `(qlamda)/(2piepsilon_0)"ln" (1/2)`
C. `(qlamda)/4piepsilon_0) "ln" sqrt2`
D. `(qlamda)/(4piepsilon_0) "ln"2`

Explanation

To solve the problem of finding the work done in moving a charge \( q \) from point \( A(a, 0) \) to point \( B(2a, 0) \) in the electric field created by an infinite line charge with linear charge density \( \lambda \), we can follow these steps:

### Step 1: Understand the Electric Field due to an Infinite Line Charge
The electric field \( E \) at a distance \( x \) from an infinite line charge with linear charge density \( \lambda \) is given by the formula:
\[ E = \frac{2k\lambda}{x} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

### Step 2: Set Up the Work Done Formula
The work done \( W \) in moving a charge \( q \) from point \( A \) to point \( B \) can be expressed as:
\[ W = q(V_B - V_A) \]
where \( V_B \) and \( V_A \) are the electric potentials at points \( B \) and \( A \) respectively.

### Step 3: Relate Work Done to Electric Field
The work done can also be expressed in terms of the electric field:
\[ W = -q \int_{A}^{B} E \cdot dx \]
Since the movement is along the x-axis, we can simplify this to:
\[ W = -q \int_{a}^{2a} E \, dx \]

### Step 4: Substitute the Electric Field into the Integral
Substituting the expression for \( E \):
\[ W = -q \int_{a}^{2a} \frac{2k\lambda}{x} \, dx \]

### Step 5: Solve the Integral
We can factor out the constants:
\[ W = -q \cdot 2k\lambda \int_{a}^{2a} \frac{1}{x} \, dx \]
The integral of \( \frac{1}{x} \) is \( \ln x \):
\[ W = -q \cdot 2k\lambda \left[ \ln x \right]_{a}^{2a} \]
Calculating the definite integral gives:
\[ W = -q \cdot 2k\lambda \left( \ln(2a) - \ln(a) \right) \]
This simplifies to:
\[ W = -q \cdot 2k\lambda \ln\left(\frac{2a}{a}\right) = -q \cdot 2k\lambda \ln(2) \]

### Step 6: Substitute \( k \) Back
Substituting \( k = \frac{1}{4\pi\epsilon_0} \):
\[ W = -q \cdot 2 \cdot \frac{1}{4\pi\epsilon_0} \cdot \lambda \ln(2) \]
This simplifies to:
\[ W = -\frac{q\lambda}{2\pi\epsilon_0} \ln(2) \]

### Final Result
Thus, the work done in moving the charge \( q \) from point \( A \) to point \( B \) is:
\[ W = -\frac{q\lambda}{2\pi\epsilon_0} \ln(2) \]