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: Concept
The electric field due to an infinite line charge at perpendicular distance $r$ is $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$, directed radially outward.

Step 2: Meaning
Work done in moving charge $q$ from $r_i = a$ to $r_f = 2a$: \[W = q\int_{a}^{2a} E\,dr = \frac{q\lambda}{2\pi\varepsilon_0}\int_{a}^{2a}\frac{dr}{r}.\]

Step 3: Analysis
\[W = \frac{q\lambda}{2\pi\varepsilon_0}\Big[\ln r\Big]_{a}^{2a} = \frac{q\lambda}{2\pi\varepsilon_0}(\ln 2a - \ln a) = \frac{q\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{2a}{a}\right).\]

Step 4: Conclusion
$W = \dfrac{q\lambda}{2\pi\varepsilon_0}\ln 2$. Final Answer: (A)