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

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 $d l$. If the charge is distributed uniformly with the linear charge density $\lambda$, then the charge $d Q$ in a segment of length $d l$ is $d Q=\lambda d l$. At point D , the differential electric field dE created by this element is -

$
d E=\frac{d Q}{4 \pi \varepsilon_0 r^2}=\frac{\lambda d l}{4 \pi \varepsilon_0 r^2}=\frac{\lambda d l}{4 \pi \varepsilon_0 r_0^2 \sec ^2 \theta}
$

In triangle $A O D ; O A=O D \tan \theta$, i.e., $l=r_0 \tan \theta$
Differentiating this equation with respect to $\theta$, we get

$
d l=r_0 \sec ^2 \theta d \theta
$

Substituting the value of $d l$

$
d E=\frac{\lambda d \theta}{4 \pi \varepsilon_0 r 0}
$

Field $d E$ has components $d E_x d E_y$ given by

$
d E_x=\frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_0 r_0} \text { and } d E_y=\frac{\lambda \sin \theta d \theta}{4 \pi \varepsilon_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{aligned}
E_x & =\int_{-\theta_1}^{\theta_2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_0 r_0} \\
& =\frac{\lambda}{4 \pi \varepsilon_0 r_0}\left(\sin \theta_1+\sin \theta_2\right) \\
E_y & =\int_{-\theta_1}^{+\theta_2} \frac{\lambda \sin \theta d \theta}{4 \pi \varepsilon_0 r_0} \\
& =\frac{\lambda}{4 \pi \varepsilon_0 r_0}\left(\cos \theta_1-\cos \theta_2\right)
\end{aligned}
$

Special case-
1.If the line is infinite then the $\theta_1=\theta_2=90^{\circ}$

Putting the value, we get -

$
\begin{aligned}
E_x & =\frac{\lambda}{2 \pi \varepsilon_o r_o} \\
E_y & =0
\end{aligned}
$

2.If the line is semi-infinite then the $\theta_1=0$, and $\theta_2=90^{\circ}$

Putting the value, we get -

$
\begin{aligned}
E_x & =\frac{\lambda}{4 \pi \varepsilon_o r_o} \\
E_y & =\frac{\lambda}{4 \pi \varepsilon_o r_o}
\end{aligned}
$