Electric Field Due To An Infinitely Long Charged Wire MCQ - Practice Questions with Answers

Electric field due to an infinite line charge -

Let us assume that positive electric charge $Q$ is distributed uniformly along a line, lying along the $y$-axis. We have to find the electric field at point D on the x -axis at a distance $r_0$ from the origin. Let us divide the line charge into infinitesimal segments, each of which acts as a point charge; let the length of a typical segment at height $l$ be $dl$. If the charge is distributed uniformly with the linear charge density $\lambda$, then the charge dQ in a segment of length $dl$ is $dQ=\lambda dl$. At point D , the differential electric field dE created by this element is -

$dE=\frac{dQ}{4\pi {\epsilon }_0{r}^2}=\frac{\lambda dl}{4\pi {\epsilon }_0{r}^2}=\frac{\lambda dl}{4\pi {\epsilon }_0{r}_0^2{\sec }^2\theta }$

In triangle $AOD;OA=OD\tan \theta$, i.e., $l=r_0\tan \theta$
Differentiating this equation with respect to $\theta$, we get

$=r_0{\sec }^2\theta d\theta$

Substituting the value of $dl$

$dE=\frac{\lambda d\theta }{4\pi {\epsilon }_0{r}_0}$

Field $dE$ has components $d{E}_{x}d{E}_y$ given by

$d{E}_x=\frac{\lambda \cos \theta d\theta }{4\pi {\epsilon }_0{r}_0}\text{ and }d{E}_y=\frac{\lambda \sin \theta d\theta }{4\pi {\epsilon }_0{r}_0}$

If the wire has finite length and the angles subtended by ends of wire at a point are ${\theta }_1$ and ${\theta }_2$, the limits of integration will be

$\begin{array}{rl}{E}_x& ={\int }_{-{\theta }_1}^{{\theta }_2}\frac{\lambda \cos \theta d\theta }{4\pi {\epsilon }_0{r}_0}\\ & =\frac{\lambda }{4\pi {\epsilon }_0{r}_0}(\sin {\theta }_1+\sin {\theta }_2)\\ E_y& ={\int }_{-{\theta }_1}^{+{\theta }_2}\frac{\lambda \sin \theta d\theta }{4\pi {\epsilon }_0{r}_0}\\ & =\frac{\lambda }{4\pi {\epsilon }_0{r}_0}(\cos {\theta }_1-\cos {\theta }_2)\end{array}$

Special case-
1.If the line is infinite then the ${\theta }_1={\theta }_2={90}^{\circ }$

Putting the value, we get -

Special case-
1.If the line is infinite then the ${\theta }_1={\theta }_2={90}^{\circ }$

Putting the value, we get -

$\begin{array}{rl}{E}_x& =\frac{\lambda }{2\pi {\epsilon }_o{r}_o}\\ E_y& =0\end{array}$

2.If the line is semi-infinite then the ${\theta }_1=0$, and ${\theta }_2={90}^{\circ }$

Putting the value, we get -

$\begin{array}{rl}{E}_x& =\frac{\lambda }{4\pi {\epsilon }_o{r}_o}\\ E_y& =\frac{\lambda }{4\pi {\epsilon }_o{r}_o}\end{array}$