Question

A flat disk 1.0 m in diameter is oriented so that the plane of the disk makes an angle of π/6 radians with a uniform electric field. If the field strength is 487.0 N/C, find the electric flux through the surface. A) 1 10m N·m2/C B) 420 N·m2/C C) 61 π N·m2/C D) 1700/T N·m2/C

Answer 1

Answer:

Option c is correct

Explanation:

Given:

The diameter of the disk, (d) = 1 m

The radius of the disk, (r) = d/ 2 = 0.5 m

Thus,

the area of the disk (A) = πr² = π  × (0.5)² = 0.25π m²

The electric field, (E) = 487.0 N/C

The angle between plane of the disk and electric field, Θ =$\frac{\pi}{6} \ \textup{rad}$ = 180°/6 = 30°

Thus,

the angle between normal to the plane of the disk and electric field (∅) will be

Θ' = 90° - 30° = 60°

Now,

The electric flux through the surface is given as:

= EA cosΘ'

on substituting the values in the above equation we get,

= 487 × (0.25π) × cos 60 °

or

Electric flux = 60.875π N.m²/ C  ≈ 61π N.m²/c

Hence, Option c is correct