Question

9 -2 points SerPSET9 24 P001 Aflat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E = 5.90 x 10s NC (a) Determine the electric flux through this area when the electric field is perpendicular to the surface (b) Determine the electric flux through this area when the electric field is parallel to the surface.

Answer 1

Flat surface area of body A= $2.6 m^2$

Uniform electric field E=$5.9 *10^5 N/c$

Let the $\Phi$ be the flux through the area.

a) when the electric field is perpendicular to the surface i.e $\Theta$=90

Now the electric flux ( $\Phi$)=EA cos ( $\Theta$)

$\Phi$= $5.9*10^5 * 2.6 * cos (90) = 0 web$ or  $0 N-m^2/C$

Hence 0 web is answer when the surface is perpendicular to the electric field.

b)when the electric field is parallel to the surface then $\Theta$=0°

Therefore

$\Phi$=$5.9*10^5*2.6*cos(0)= 15.34 web$ . Or . $15.34 N-m^2/C$

Hence $\Phi$ is 15.34 web when the surface is parallel to the electric field .