The drawing shows an edge-on view of two planar surfaces thatintersect and are mutually perpendicular. Surface 1 has an area of 1.7 m^2, while surface 2 san area of 3.2 m^2. The electric field in the drawing is uniform andhas a magnitude of 250 N/C. Find the electrical flux through surface 1 and surface 2
I have tried to do the problem multiple amounts of time but I never seem to get the correct answer. For surface 1 its suppose to be 350 Nm^2/c and surface 2 is suppose to be 460 Nm^2/c. This is what I have gotten so far:
1: EACos(pheta) 250(1.7)(cos 55)= 9.4028 Nm^2/c
2: EACose(pheta) 250(3.2)(cos 35)= -722.95
Please help with anything you can!
close, you need to change your theta.
A) 250(1.7)cos35
B) 250(3.2)cos55
this will work, just did it for my homework and they are correct. My E and A are different but the theta was the same.
changed my login from the above entry.
A) 250(1.7)cos35 = 348.13 Nm2/C
B) 250(3.2)cos55 = 458.86 Nm2/C
Hopefully this works for you
To find the electric flux through a surface, you need to use the formula:
Electric flux = Electric field strength x Area x cos(angle)
For surface 1:
Given:
Electric field strength (E) = 250 N/C
Area (A) = 1.7 m^2
To find the angle (θ) between the electric field and the normal vector to the surface, you mentioned in the question that the surfaces are mutually perpendicular. Therefore, the angle between them would be 90 degrees.
Plugging in the values into the formula:
Electric flux (through surface 1) = 250 N/C * 1.7 m^2 * cos(90°)
= 250 N/C * 1.7 m^2 * 0
= 0 Nm^2/C
For surface 2:
Given:
Electric field strength (E) = 250 N/C
Area (A) = 3.2 m^2
Similarly, for surface 2, the angle (θ) between the electric field and the normal vector to the surface would be 90 degrees.
Plugging in the values into the formula:
Electric flux (through surface 2) = 250 N/C * 3.2 m^2 * cos(90°)
= 250 N/C * 3.2 m^2 * 0
= 0 Nm^2/C
Therefore, the correct answer would be that the electric flux through both surface 1 and surface 2 is 0 Nm^2/C, assuming the angle between the electric field and the normal vector to the surfaces is 90 degrees.
To find the electrical flux through each surface, you need to use the formula:
Φ = E * A * cos(θ)
Where:
Φ is the electrical flux through the surface,
E is the magnitude of the electric field,
A is the area of the surface,
and θ is the angle between the electric field and the surface normal.
Let's calculate the electrical flux through surface 1 first:
Given:
E = 250 N/C (magnitude of the electric field)
A1 = 1.7 m^2 (area of surface 1)
θ1 = 90° (since the surfaces are mutually perpendicular)
Substituting the values into the formula:
Φ1 = (250 N/C) * (1.7 m^2) * cos(90°)
cos(90°) = 0
Φ1 = (250 N/C) * (1.7 m^2) * 0
Φ1 = 0
It seems that there is an error in your calculations for surface 1. The electrical flux through surface 1 should be zero since the angle between the electric field and the surface normal is 90°.
Now let's calculate the electrical flux through surface 2:
Given:
E = 250 N/C (magnitude of the electric field)
A2 = 3.2 m^2 (area of surface 2)
θ2 = 0° (since the surfaces are mutually perpendicular)
Substituting the values into the formula:
Φ2 = (250 N/C) * (3.2 m^2) * cos(0°)
cos(0°) = 1
Φ2 = (250 N/C) * (3.2 m^2) * 1
Φ2 = 800 Nm^2/C
So, the electrical flux through surface 2 is 800 Nm^2/C, not 460 Nm^2/C as you mentioned.
In summary:
The electrical flux through surface 1 is 0 Nm^2/C.
The electrical flux through surface 2 is 800 Nm^2/C.
It's important to double-check your calculations, paying attention to the angles and trigonometric functions used.