A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N. m2/C D) 280 N. m2/C

electric flux is 280 Nm²/C

so correct option is D 280 Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength * area * cos∅ ... 1

here area = πr² = π (0.50) ²

put here all value in equation 1

electric flux = field strength * area * cos∅

electric flux = 713 * π (0.50) ² * cos60

we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk

so we use cos60

electric flux = 280 Nm²/C

so correct option is D 280 Nm²/C