Question

An electric field (E=4.5N/C) passes through a flat surface area (0.30m) with an angle of 45°. What will be the magnitude of the electric flux?

Answer 1

The electric flux

$$\Phi=EA\cos\theta\\ =\rm 4.5\:N/C*0.3\:m^2*\cos45^\circ=0.95\:N\cdot m^2/C$$