Question

An infinite line charge of uniform linear charge density $\lambda$ is placed along the $z$-axis. The work done in moving a charge $q$ from $(a,\,0,\,0)$ to $(2a,\,0,\,0)$ is:

• A. $\dfrac{q\lambda}{2\pi\varepsilon_0}\ln 2$
• B. $\dfrac{q\lambda}{4\pi\varepsilon_0}\ln 2$
• C. $\dfrac{q\lambda}{2\pi\varepsilon_0}\ln\!\left(\tfrac{1}{2}\right)$
• D. zero

Hint:

The potential of an infinite line charge varies as $\ln r$ (not $1/r$), because its field falls off as $1/r$. So work done depends only on the ratio of distances, not their absolute values.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Work done is related to the potential difference: \( W = q(V_2 - V_1) \).
For a line charge, the electric field is radial and varies as \( 1/r \).
Key Formula or Approach:
1. Electric field: \( E = \frac{\lambda}{2\pi\epsilon_0 r} \).
2. Work: \( W = \int F \cdot dr = \int q E dr \).
Step 2: Detailed Explanation:
The work done by an external agent against the field:
\[ W = \int_a^{2a} q E dr = q \int_a^{2a} \frac{\lambda}{2\pi\epsilon_0 r} dr \] Factor out constants:
\[ W = \frac{q \lambda}{2\pi\epsilon_0} \int_a^{2a} \frac{1}{r} dr \] Integrating \( 1/r \) gives \( \ln r \):
\[ W = \frac{q \lambda}{2\pi\epsilon_0} [ \ln r ]_a^{2a} = \frac{q \lambda}{2\pi\epsilon_0} (\ln 2a - \ln a) \] \[ W = \frac{q \lambda}{2\pi\epsilon_0} \ln(2a/a) = \frac{q \lambda}{2\pi\epsilon_0} \ln 2 \] Step 3: Final Answer:
The work done is \( \frac{q\lambda}{2\pi\epsilon_0} \ln 2 \).
This matches Option (A).