Question

The diagram shows three infinitely long uniform line charges placed on the X, Y and Y, axis. The work done in moving a unit positive charge from (1, 1, 1) to (0, 1, 1) is equal to

• A. $\frac{\left(\lambda ln2\right)}{2\pi {\epsilon }_0}$
• B. $\frac{\left(\lambda ln2\right)}{\pi {\epsilon }_0}$
• C. $\frac{\left(3\lambda ln2\right)}{2\pi {\epsilon }_0}$
• D. None

Answer 1

The correct option is B $\frac{\left(\lambda ln2\right)}{\pi {\epsilon }_0}$

$dw=\frac{\lambda q}{2\pi {ϵ}_{0}r}dz$

$w=\frac{\lambda q}{2\pi {ϵ}_0}{\int }_{r_1}^{r_2}\frac{dr}{r}$

$=\frac{\lambda q}{2\pi {ϵ}_0}ln\left(\frac{r_2}{r_1}\right)$

Now, lets us it as formula $WN=Wx+Wy+Wz$

[$W_i$ : Work done by wire in $i$ direction]

$=\frac{{2\lambda }_1\left(1\right)}{2\pi {ϵ}_0}ln\left(\frac{\sqrt{2}}{\sqrt{2}}\right)+\frac{3\lambda \left(1\right)}{2\pi {ϵ}_0}ln\left(\frac{1}{\sqrt{2}}\right)+\frac{\lambda }{2\pi {ϵ}_0}ln\left(\frac{1}{\sqrt{2}}\right)$

$=0+\frac{\lambda }{2\pi {ϵ}_0}[\frac{-3}{2}ln2-\frac{1}{2}ln2]$

$=-\frac{2\lambda ln2}{2\pi {ϵ}_0}=\frac{-\lambda ln2}{\pi {ϵ}_0}$

$W_{ext}=-WN=\frac{\lambda ln2}{\pi {ϵ}_0}$