Question

A circular surface with a radius of 0.055 m is exposed to a uniform electric field of magnitude 1.38 times 10^4 N/C. The electric flux through the surface is 78 N-m^2/C. What is the angle between the direction of the electric field and the normal to the surface?

Answer 1

Radius = r = 0.055 m

magnitude of electric field = E = 1.38 *10⁴ N/C

electric flux = $\Phi$ = 78 N m² / C

And we know $\Phi$ = E A cos $\Theta$ ,   A = area of circle = $\small \Pi$ r² = $\small \Pi$ *0.055² = 9.5033*10^-3 m²

^{$\Rightarrow$}   78 N m² /C = 1.38*10⁴ * 9.503 *10^-3 * cos $\Theta$

$\Rightarrow$ cos $\Theta$ = 0.5947

$\Rightarrow$     $\Theta$ = cos^-1(0.5947) = 53.50 degrees