Question

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

Answer 1

electric field due to infinitely long wire at a point which lies perpendicular distance from its length is given by, $E=\frac{\lambda}{2\pi\epsilon_0 r}$

so, workdone by charge q at that point , $dW=\frac{\lambda q}{2\pi\epsilon_0r}dr$

or, $\int{dW}=\frac{\lambda}{2\pi\epsilon_0}\int\limits^{r_2}_{r_1}\frac{dr}{r}$

or, $W=\frac{\lambda q}{2\pi\epsilon_0}ln\frac{r_2}{r_1}$

because workdone is scalar quantity.

so, $W_N=W_x+W_y+W_z$

along x-axis, $r_2=\sqrt{y_2^2+z_2^2}=\sqrt{2}$

$r_1=\sqrt{y_1^2+z_1^2}=\sqrt{2}$

Similarly, along Y-axis , $r_2=1,r_1=\sqrt{2}$

along z-aixs , $r_2=1,r_1=\sqrt{2}$

now, $W_N=\frac{(2\lambda)(1)}{2\pi\epsilon_0}ln\frac{\sqrt{2}}{\sqrt{2}}+\frac{(3\lambda)(1)}{2\pi\epsilon_0}ln\frac{1}{\sqrt{2}}+\frac{(\lambda)(1)}{2\pi\epsilon_0}ln\frac{1}{\sqrt{2}}$

= 0 + $\frac{\lambda}{2\pi\epsilon_0}\left[-\frac{3}{2}ln2-\frac{1}{2}ln2\right]$

= $-\frac{\lambda ln2}{\pi\epsilon_0}$

so, workdone by external agent to charge q = -$W_N$

= $\frac{\lambda ln2}{\pi\epsilon_0}$

Answer 2

Solution in attachment

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