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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

10 Questions around this concept.

Solve by difficulty

Two parallel line charges $+\lambda$ and $-\lambda$ are placed with a separation R in free space. The net electric field exactly mid way between the two line charge is -

A

zero

B

$\frac{2 \lambda}{\pi \epsilon_0 K}$

C

$\frac{\lambda}{\pi \epsilon_0 R}$

D

$\frac{\lambda}{4 \pi \epsilon \nu R}$

Concepts Covered - 1

Electric field due to an infinite line charge

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 $d Q=\lambda d l$ . At point D, the differential electric field dE created by this element is -

$d E=\frac{d Q}{4 \pi \varepsilon_{0} r^{2}}=\frac{\lambda d l}{4 \pi \varepsilon_{0} r^{2}}=\frac{\lambda d l}{4 \pi \varepsilon_{0} r_{0}^{2} \sec ^{2} \theta}$

$\begin{array}{l}{\text { In triangle } A O D ; O A=O D \tan \theta, \text { i.e., } l=r_{0} \tan \theta} \\ \\ {\text { Differentiating this equation with respect to } \theta, \text { we get }} \\ \\ {\text { } \begin{array}{l}{=r_{0} \sec ^{2} \theta d \theta} \\ \\ {\text { Substituting the value of } d l} \\ \\ {\qquad d E=\frac{\lambda d \theta}{4 \pi \varepsilon_{0} r_{0}}}\end{array}}\end{array}$

$\begin{array}{l}{\text { Field } d E \text { has components } d E_{x} d E_{y} \text { given by }} \\ \\ {\qquad d E_{x}=\frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_{0} r_{0}} \text { and } d E_{y}=\frac{\lambda \sin \theta d \theta}{4 \pi \varepsilon_{0} r_{0}}}\end{array}$

$\begin{array}{l}{\text { If the wire has finite length and the angles subtended by }} \\ {\text { ends of wire at a point are } \theta_{1} \text { and } \theta_{2} \text { , the limits of integration }} \\ {\text { will be}} \\ {\qquad \begin{aligned} E_{x} &=\int_{-\theta_1}^{\theta_2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_{0} r_{0}} \\ \\ &=\frac{\lambda}{4 \pi \varepsilon_{0} r_{0}}\left(\sin \theta_{1}+\sin \theta_{2}\right) \\ \\ E_{y} &=\int_{- \theta_{1}}^{+\theta_{2}} \frac{\lambda \sin \theta d \theta}{4 \pi \varepsilon_{0} r_{0}} \\ &=\frac{\lambda}{4 \pi \varepsilon_{0} r_{0}}\left(\cos \theta_{1}-\cos \theta_{2}\right) \end{aligned}}\end{array}$

Special case-

1.If the line is infinite then the $\\ \theta_1 = \theta_2 = 90^o$

Putting the value, we get -

$E_x = \frac{\lambda}{2 \pi \varepsilon_or_o}$

$E_y = 0$

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

Putting the value, we get -

$E_x = \frac{\lambda}{4 \pi \varepsilon_or_o}$

$E_y = \frac{\lambda}{4 \pi \varepsilon_or_o}$

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Electric field due to an infinite line charge

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