The diagram shows three infinitely long unifo | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$d w=\frac{\lambda q}{2 \pi \epsilon_{0} r} d z$

$w=\frac{\lambda q}{2 \pi \epsilon_{0}} \int_{r_{1}}^{r_{2}} \frac{d r}{r}$

$=\frac{\lambda q}{

Vedclass · Accepted Answer

$(\lambda\  ln\  2)\  /\pi \varepsilon_0$

Vedclass · Answer

$d w=\frac{\lambda q}{2 \pi \epsilon_{0} r} d z$

$w=\frac{\lambda q}{2 \pi \epsilon_{0}} \int_{r_{1}}^{r_{2}} \frac{d r}{r}$

$=\frac{\lambda q}{2 \pi \epsilon_{0}} \ln \left(\frac{r_{2}}{r_{1}}ight)$

Now, lets us it as formula $W N=W x+W y+W z$

$\left[W_{i}: 	ext { Work done by wire in } i 	ext { direction }ight]$

$=\frac{2 \lambda_{1}(1)}{2 \pi \epsilon_{0}} \ln \left(\frac{\sqrt{2}}{\sqrt{2}}ight)+\frac{3 \lambda(1)}{2 \pi \epsilon_{0}} \ln \left(\frac{1}{\sqrt{2}}ight)+\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{1}{\sqrt{2}}ight)$

$=0+\frac{\lambda}{2 \pi \epsilon_{0}}\left[\frac{-3}{2} \ln 2-\frac{1}{2} \ln 2ight]$

$=-\frac{2 \lambda \ln 2}{2 \pi \epsilon_{0}}=\frac{-\lambda \ln 2}{\pi \epsilon_{0}}$

$W_{e x t}=-W N=\frac{\lambda \ln 2}{\pi \epsilon_{0}}$

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

Similar Questions

On moving a charge of $20$ coulombs by $2 \;cm , 2 \;J$ of work is done, then the potential difference between the points is (in $volt$)

Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$ both the particles are found to be separated by distance $r_2$. The speed of each particle is

In a region, electric field varies as $E = 2x^2 -4$ where $x$ is the distance in metre from origin along $x-$ axis. A positive charge of $1\,\mu C$ is released with minimum velocity from infinity for crossing the origin, then