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The diagram shows three infinitely long uniform line charges placed on the $X, Y $ and $Z$ axis. The work done in moving a unit positive charge from $(1, 1, 1) $ to $(0, 1, 1) $ is equal to
$(\lambda\ ln \ 2) / 2\ \pi \varepsilon_0$
$(\lambda\ ln\ 2)\ /\pi \varepsilon_0$
$(3\ \lambda\ ln \ 2)\ / 2\ \pi \varepsilon_0$
None
Solution
$d w=\frac{\lambda q}{2 \pi \epsilon_{0} r} d z$
$w=\frac{\lambda q}{2 \pi \epsilon_{0}} \int_{r_{1}}^{r_{2}} \frac{d r}{r}$
$=\frac{\lambda q}{2 \pi \epsilon_{0}} \ln \left(\frac{r_{2}}{r_{1}}\right)$
Now, lets us it as formula $W N=W x+W y+W z$
$\left[W_{i}: \text { Work done by wire in } i \text { direction }\right]$
$=\frac{2 \lambda_{1}(1)}{2 \pi \epsilon_{0}} \ln \left(\frac{\sqrt{2}}{\sqrt{2}}\right)+\frac{3 \lambda(1)}{2 \pi \epsilon_{0}} \ln \left(\frac{1}{\sqrt{2}}\right)+\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{1}{\sqrt{2}}\right)$
$=0+\frac{\lambda}{2 \pi \epsilon_{0}}\left[\frac{-3}{2} \ln 2-\frac{1}{2} \ln 2\right]$
$=-\frac{2 \lambda \ln 2}{2 \pi \epsilon_{0}}=\frac{-\lambda \ln 2}{\pi \epsilon_{0}}$
$W_{e x t}=-W N=\frac{\lambda \ln 2}{\pi \epsilon_{0}}$
