Question

Find the flux of the electric field through surface $\text{ABCD}$ of the inclined plane as shown in figure.

• A. $Eab\cos \theta$
• B. $Eab\sin \theta$
• C. $Eab\tan \theta$
• D. $Eab\cot \theta$

Answer 1

The correct option is B $Eab\sin \theta$

Given:
$\text{AB}=a;\text{BC}=b$
So, area of surface $\text{ABCD},A=ab$

From above figure,
angle $\left(\alpha \right)$ between electric field lines and area vector is $({90}^{\circ }-\theta )$

Electric flux is given by

$\varphi =\overrightarrow{E}.\overrightarrow{A}=EA\cos \alpha$

$\Rightarrow \varphi =Eab\cos ({90}^{\circ }-\theta )$

$\Rightarrow \varphi =Eab\sin \theta$