Question

Electric field in the given region is vertically downwards. An open hemisphere of radius $a$ is placed as shown. Find the net flux of electric field through the curved surface of hemisphere.

• A. 0
• B. $E\pi a^2$
• C. $2E\pi a^2$
• D. $4E\pi a^2$

Answer 1

The correct option is B

$E\pi a^2$

Flux is $\int \overrightarrow{E}.\overrightarrow{dA}$

If you take a small $dA$ area element that makes some angle $\theta$ with $\overrightarrow{E}$.

The flux through that area element we know will be $d\varphi =EdAcos\theta$

Another way to look at it is

$E\left(dAcos\theta \right)$

If you keep doing that you will find that component of entire hemispheres area will be circle.

So the area becomes $E\pi a^2$.

Try visualizing this with how Aanand showed you the perpendicular component of the net reduced to a smaller rectangle when he tilted the net. Here, the same component of a hemisphere projected perpendicular to the $\overrightarrow{E}$ vector becomes the circle.

Alternate method:

No. of field lines $\propto$ Electric Field

The stronger the field, the more the number of lines.

Since flux is effectively the number of lines passing through an area,

Here since the entire flux through the open hemisphere should pass through the circle first, if I find the flux through that circular opening (imagine we’re placing a circular area there to calculate the flux through it).

$\varphi =\overrightarrow{E}.\overrightarrow{A}$

$\varphi =E.\pi a^{2}cos{0}^0$

$\varphi =E\pi a^2$