Question

A hollow prism with side ABCD as open is kept in a region of electric field E. The field is parallel to the sides. BEC and AFD. Find the flux coming out of the pyramid?

• A. $Ela$
• B. $0$
• C. $2Elacos{30}^0$
• D. None of these

Answer 1

The correct option is A

$Ela$

The pyramid has 4 sides. Not counting ABCD, as its open

Flux $=\int \overrightarrow{E}.\overrightarrow{ds}$

$=\int Eds\cos \theta$

$=E{s}_{1}cos{\theta }_1+E{s}_{2}cos{\theta }_2+E{s}_{3}cos{\theta }_3+E{s}_{4}cos{\theta }_4$

$s_1,s_2,s_3$ and $s_4$ are four areas through which the field is going through.
Here,
$s_1=$ area AFEB
$s_2=$ area DFEC
$s_3=$ area BEC
$s_4=$ area AFD

From the side view its evident that
$△$BEC is equilateral and $\angle CBE={60}^0$ so ${\varphi }_1=Ela\cos {60}^{\circ }\Rightarrow {\varphi }_1=\frac{Ela}{2}$

Similarly for the surface DFEC,

${\varphi }_2=\frac{Ela}{2}$

Now what about surface BEC. The flux line is parallel to it. This is given in the question itself. So flux will be perpendicular to the area vector.

$\theta =90$

$\cos \theta =0$

so flux through BEC = 0

Similarly flux through AFD = 0

Net flux $=\frac{Ela}{2}+\frac{Ela}{2}+0+0=Ela$