Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

An infinite long semicircular tube is given as shown below one half is given a surface charge density $+\sigma$ and other half is given charge density $-,\sigma $ . Electric field at a point on axis would be


$(A)\;\frac {\sigma}{\pi \in _0} \hat j \\ (B)\;\frac {\sigma}{\pi \in _0} \hat i \\ (C)\; \frac {-\sigma}{\pi \in _0} \hat i \\ (D)\;\frac {\sigma}{\pi \in _0} \hat j $

Can you answer this question?

1 Answer

0 votes
$ dE_j=0$
$dE_x=2 dE \sin \theta$
$dE=\large\frac{\sigma \times R d \theta}{2 \pi \in _0R}$
$\qquad= \large\frac{\sigma d \theta}{2 \pi \in _0}$
$E_x=\int d E_x$
$\qquad= \int \limits_0^{\pi/2} \large\frac{\sigma}{\pi \in _0}$$ \sin \theta d \theta $
$\qquad=\large\frac{\sigma}{\pi \in _0}$
Hence B is the correct answer.
answered Jan 9, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App