Ask Questions, Get Answers


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 $

1 Answer

$ 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