Aflat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 n/c. the angle between the area vector and the electric field is 33.2°. what is the electric flux through the surface?

576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?