IIT JEE , AIEEE Forum - Physics , Electricity

reddevil_2009

Q1 A ring of radius R is given a charge Q and kept in Y-Z plane and an infinite wire of charge density $\lambda\\$ is kept on X- axis starting from center of ring.What will be the force of interaction (magnitude and direction both) on ring due to wire?

Q2 A thin glass rod is bent into a quarter circle of radius r and placed in XY plane + Q charge is uniformly distributed on the rod.What will be electric field vector at point O(Origin)??

karthik2007

anchitsaini

$\mbox{for Q2 I will derive the general formula for E due to an arc}\\ \\ \mbox{Consider an arc subtending an angle } \theta \mbox{ at the centre}\\ \\ \mbox{Now consider an element of the arc of length }Rd \phi \mbox{ at an angle } \phi \\ \\ \mbox{ from the vertical}<br/>$

$dE_x=\frac{1}{4\pi\epsilon_0}\frac{dq}{R^2} \\ \\Take \ \lambda = \frac{q}{R\theta} \\ \\Hence \ dE_x=\frac{1}{4\pi\epsilon_0}\frac{\lambda Rd\phi cos\phi}{R^2} \\ \\On \ integrating \ from \ -\frac{\theta}{2} \ to \ \frac{\theta}{2},\\ \\E=\frac{1}{4\pi\epsilon _0}\frac{2\lambda\sin\theta/2}{R}\\ \\=\frac{1}{4\pi\epsilon _0}\frac{q\sin\theta /2}{R^2\theta /2}<br/>\mbox{In quarter circle, }\theta = \frac{\pi}{4} \\ \\<br/>Ans=\frac{1}{4\pi^2\epsilon _0}\frac{q2\sqrt{2}}{R^2}$

$Typo-\theta=\pi / 2$

anchitsaini

nknikhilesh1

What u have written dost, typo - theta = pi/2.And ur answer and Kartic;s answer are noy matching .please explain

anchitsaini

arey
i had written theta=pi/4
in a quarter circle
which was a typo(typing error)
cos theta = pi/2

about are answers not matching, well let the qn poster check it out

reddevil_2009

hiiii anchitsaini..

Good work by u..BUT PLZ PROVIDE BOTH MAGNITUDE AND DIRECTION IN YOUR ANSWERS

karthik2007

whats the answer to the second question?

anchitsaini

Re:Electrostat

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