What is the flux through side B of the box?

In summary, the problem involves finding the flux through side B of an empty cardboard box placed in an electric field with a strength of 1200 N/C. The box is tilted up at an angle of 13° from the horizontal, with sides A, B, and C facing out of the page and sides D and E facing into the page. The dimensions of the sides are given, and the equation for electric flux is provided. The key to solving the problem is determining the cross sectional area of side B that is presented to the electric field. This can be done by visualizing the field as a stream of particles and finding the area that the side presents in the direction of the stream. The given angle theta can be used to calculate
  • #1
eMac
17
0

Homework Statement


http://i885.photobucket.com/albums/ac52/HummusNFalafel/showmepl.jpg
A empty cardboard box is placed in an electric field pointing upwards with a strength of 1200 N/C as shown above. The bottom of the box is tilted up by an angle Θ = 13° from the horizontal. Side A of the box faces out of the page and is 11 cm high and 16 cm long. Sides B and C are 11 cm high and 11 cm deep. Sides D and E are 16 cm long and 11 cm deep.


b) What is the flux through side B of the box?

Homework Equations



Electric Flux = EAcos(theta)


The Attempt at a Solution



I do not know what angle to use for any of the sides. I tried 90-13 to get 77 but that doesn't work as well. I am not sure if I'm using the correct area as well. I am multiplying base x height but I am not sure if that is the correct equation to use. Any help will be much appreciated, thanks in advance.
 
Physics news on Phys.org
  • #2
It boils down to determining the cross sectional area of the side in question that is presented to the field. So in this case the field is running directly upwards (in the positive z-axis direction. If you were to look down from above the xy plane (from some height up the z axis), what cross sectional area would side B occupy?
 
  • #3
gneill said:
It boils down to determining the cross sectional area of the side in question that is presented to the field. So in this case the field is running directly upwards (in the positive z-axis direction. If you were to look down from above the xy plane (from some height up the z axis), what cross sectional area would side B occupy?

I'm not sure if I understand, are you referring to the field in 3-D.
 
  • #4
find the area vector of side B and use the flux formula
flux = E.S (dot product)
 
  • #5
eMac said:
I'm not sure if I understand, are you referring to the field in 3-D.

Certainly. It's a 3D box and you are looking for the flux through one particular side. So what cross sectional area does the side present to the field? Think of it as determining the area of a shadow projection of the side onto the xy plane (since the field is the "light" moving strictly vertically in this case).
 
  • #6
gneill said:
Certainly. It's a 3D box and you are looking for the flux through one particular side. So what cross sectional area does the side present to the field? Think of it as determining the area of a shadow projection of the side onto the xy plane (since the field is the "light" moving strictly vertically in this case).

I'm really not following your explanation. Would the area just be base x height?
 
  • #7
eMac said:
I'm really not following your explanation. Would the area just be base x height?

That's the area of the side that you would see if you looked at it straight-on. The field is not "looking" at the side straight-on; It's viewing it from an angle. You want to find the cross sectional area that the side presents to the field.

Picture the electric field as a stream of particles. In the midst of this stream is the side of the box that you're interested in, a rectangle of some area A. The amount of the stream that is intercepted by the side depends upon the angle that the side makes with the direction of flow of the stream. If the side is perpendicular to the stream, presenting its whole face to it, then the area of the stream that's intercepted is A. If the side is at some angle to the flow, then it presents a smaller cross section and proportionately less of the particles of the stream are intercepted.

You want to find the cross sectional area of that the side presents in the direction of the "stream" that is the electric field, and multiply that area by the field strength. This will be the total flux passing through the side.
 
  • #8
gneill said:
That's the area of the side that you would see if you looked at it straight-on. The field is not "looking" at the side straight-on; It's viewing it from an angle. You want to find the cross sectional area that the side presents to the field.

Picture the electric field as a stream of particles. In the midst of this stream is the side of the box that you're interested in, a rectangle of some area A. The amount of the stream that is intercepted by the side depends upon the angle that the side makes with the direction of flow of the stream. If the side is perpendicular to the stream, presenting its whole face to it, then the area of the stream that's intercepted is A. If the side is at some angle to the flow, then it presents a smaller cross section and proportionately less of the particles of the stream are intercepted.

You want to find the cross sectional area of that the side presents in the direction of the "stream" that is the electric field, and multiply that area by the field strength. This will be the total flux passing through the side.

That's my main problem, I'm not sure what angle the side makes with the stream.
 
  • #9
eMac said:
That's my main problem, I'm not sure what angle the side makes with the stream.

So you've been given the angle theta which is 13 degrees.

jmUKi.jpg


You'll need to find the cross sectional area which will be given by the distance x multiplied by the height of the box which I believe is given by the question.

The angle that you need to work out the length of x is also in the problem. Look carefully at the diagram and try to picture the box rotating and the angles being made in those two triangles.
 

Related to What is the flux through side B of the box?

What is electric flux through a box?

Electric flux through a box is a measure of the total number of electric field lines passing through the box. It is calculated by multiplying the electric field strength by the area of the box perpendicular to the electric field.

How is electric flux through a box related to electric charge?

The electric flux through a box is directly proportional to the amount of electric charge enclosed within the box. This means that if the charge inside the box increases, the electric flux through the box will also increase.

What is the unit of electric flux through a box?

The unit of electric flux through a box is volts times meters (V*m). This is because electric flux is calculated by multiplying the electric field strength (in volts per meter) by the area (in square meters).

What factors affect the electric flux through a box?

The electric flux through a box is affected by the strength of the electric field, the size and shape of the box, and the amount of electric charge enclosed within the box. It is also affected by the angle at which the electric field lines intersect the box's surface.

How is the direction of electric flux through a box determined?

The direction of electric flux through a box is determined by the direction of the electric field lines passing through the box. The flux is considered positive if the field lines are entering the box, and negative if the field lines are leaving the box.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
11K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
7K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
6K
Back
Top