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}) $