Aflat 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

resonant frequency