Discussion Forum



1)

 Consider an electric field $E= E_{0} \widehat{x}$ where  $\overline{ E_{0}}$  is a constant. The flux through the shaded area (as shown in the figure) due to the field is 

28112021431_u1.PNG


A) $2 E_{0}a^{2}$

B) $\sqrt{2}E_{0}a^{2}$

C) $E_{0}a^{2}$

D) $\frac{E_{0}a^{2}}{\sqrt{2}}$

Answer:

Option C

Explanation:

Electric  flux , E.S , or $\phi= ES \cos \theta$

 here, $\theta$ is the angle between E and S 

 In this question $\theta= 45^{0}$ , because S is perpendicular to the surface 

 $E= E_{0}$

 $\Rightarrow$   $S= (\sqrt{2a} ) (a) = \sqrt{2} a^{2}$

 $\therefore$   $\phi = ( E_{0}) (\sqrt{2}a^{2}) \cos 45^{0}=E_{0}a^{2}$

 $\therefore$ correct option is (c)

 

(i) Question is moderatelY tough,
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is a and other side is  $\sqrt{2} a$
(iii) Electric field is uniform, whose magnitude is E0 and direction is positive x. In uniform electric field we can use , $\phi= ES \cos \theta$