Question

There is a square wire frame ABCD of side equal to b and whose centre is at a distance OP = $\frac{b}{2}$ from an infinite line charge (parallel to BC and AD) with linear charge density $\lambda$ then [plane of the frame is perpendicular to line OP.]

• A. flux through ABCD is $\frac{\lambda b}{{\epsilon }_0}$
• B. flux through ABCD is $\frac{\lambda b}{4{\epsilon }_0}$
• C. electric field across AB is uniform
• D. electric field across BC is uniform

Answer 1

Answer: (B), (D)

The correct options are
B

flux through ABCD is $\frac{\lambda b}{4{\epsilon }_0}$

D

electric field across BC is uniform

dA = bdx; $x=\frac{b}{2}tan\theta ;E=\frac{2k\lambda }{r}$

$d\varphi =EdAcos\theta =\frac{2k\lambda }{\left(\frac{b}{2cos\theta }\right)}\times \left[\frac{b}{2}se{c}^2\theta \right]cos\theta d\theta ;\varphi ={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}2k\lambda bd\theta$

$k\lambda b\pi =\frac{\lambda b}{4{\epsilon }_0}$