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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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The electric field intensity due to a semi-infinite wire  at a point A which is at a distance of 10 cm as shown in the figure: 

(Given: charge density, \lambda = 3.0 \times 10^{-5} \frac{C}{m^2}    )

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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}{dl=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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