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
Answer:
15.24 m/s in the downward direction
Explanation:
Given that the initial upward velocity of the ball is 24 m/s.
Assuming that the upward direction is positive.
As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.
Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.
Given that u=24 m/s, t=4 seconds, .
So, the final velocity is
m/s
Here, the negative sign means the final velocity is in the downward direction.
Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.