Question

A uniform electric field of magnitude 3.50x104 N/C makes an angle of 47 ° with a plane surface of area 1.63x10-2 m2 What is the electric flux through this surface? N m2/C Submit Request Ans

Answer 1

Solution:

Given data:

Electric field of magnitude( E ) = 2.90×10^4 N/C

Angle ( $\Theta$ ) =  $47^{0}$

Plane surface of area ( A ) =  1.43×10^−2 $m^{2}$

Now, we have to find out the:

The electric flux through this surface :

We already know the formula for the  electric flux is,

Electric flux  $\Phi$ = EA COS ($\Theta$)

  $\Phi$ = ( 2.90×10^4 N/C ) ( 1.43×10^−2 $m^{2}$) COS( $47^{0}$)

= 414.7 N.$m^{2}$/ C (0.68189 )

= 282.8247  N.$m^{2}$/ C

= $2.82\times 10^{2}$   N.$m^{2}$/ C

$\therefore$ Electric flux  $\Phi$  through this surface = $2.82\times 10^{2}$   N.$m^{2}$/ C