# An infinite line of charge λ per unit length is placed along the y-axis. The work done in moving a charge…..

Q. An infinite line of charge λ 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

(a) $\displaystyle \frac{q\lambda}{2\pi\epsilon_0} ln2$

(b) $\displaystyle \frac{q\lambda}{2\pi\epsilon_0} ln \frac{1}{2}$

(c) $\displaystyle \frac{q\lambda}{4\pi\epsilon_0} ln\sqrt2$

(d) $\displaystyle \frac{q\lambda}{4\pi\epsilon_0} ln2$

Ans: (b)

Sol: $\displaystyle W= -\int_{A}^{B}\vec{F}.\vec{dr}$

$\displaystyle = -q \int_{a}^{2a}Edr$

$\displaystyle = -q\int_{a}^{2a}\frac{\lambda}{2\pi \epsilon_0 r}dr$

$\displaystyle = -q \frac{\lambda}{2\pi \epsilon_0 }\int_{a}^{2a}\frac{dr}{r}$

$\displaystyle = – \frac{q \lambda}{2\pi \epsilon_0 }[ln r]_{a}^{2a}$

$\displaystyle = – \frac{q \lambda}{2\pi \epsilon_0 }[ln 2a – ln a]$

$\displaystyle = – \frac{q \lambda}{2\pi \epsilon_0 }ln2$

$\displaystyle = \frac{q \lambda}{2\pi \epsilon_0 }(ln\frac{1}{2})$