A flat surface of area 3.20 m2 is rotated in a uniform electric field of magnitude e = 6.20 x 105 n/c. determine the electric fl
ux through this area (a) when the electric field is perpendicular to the surface and (b) when the electric field is parallel to the surface.
1 answer:
The electric flux on a surface of area A in a uniform electric field E is given by
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.
When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0
Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.
When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0
Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
You might be interested in