What is the Electric Flux Through a Cube and How Can it be Calculated?

In summary, the problem involves finding the electric flux through each face of a cube with given dimensions and a uniform electric field. The dot product is used to determine the parallel component of the electric field that contributes to the flux. Gauss' Law can also be used to determine the total flux through the entire cube.
  • #1
SuperCass
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Homework Statement



A cube has sides of length L = 0.500 m. It is placed with one corner at the origin as shown in Fig. 23-29. The electric field is uniform and given by E = (2.50 N/C) i - (4.40 N/C) j.
23-29.gif

Find the electric flux through each of the six cube faces S1, S2, S3, S4, S5, and S6.
Find the electric flux through the entire cube.

Homework Equations


[tex]\phi[/tex]=EAcos[tex]\theta[/tex]
[tex]\phi[/tex]=q/[tex]\epsilon[/tex]

The Attempt at a Solution



I tried solving for the area of each side and multiply it by the electric field (using the pythagoreon theorem to get it) but that's gotten be nowhere.
 
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  • #2
For the first part, remember what the dot product means. You multiply the parts of the vectors that are parallel.
With that in mind, take mind of the sign convention. The area vector always points away from the enclosed volume.

As for the second part of the question, use Gauss' Law to avoid adding many terms together. ;)
 
  • #3
RoyalCat said:
For the first part, remember what the dot product means. You multiply the parts of the vectors that are parallel.
With that in mind, take mind of the sign convention. The area vector always points away from the enclosed volume.

As for the second part of the question, use Gauss' Law to avoid adding many terms together. ;)

What do you mean about the vector part?
Since the i is positive and the j is negative, does that mean it's pointing down and out on the x axis?

(And we're just learning Gauss' Law, still trying to understand it)
 
  • #4
SuperCass said:
What do you mean about the vector part?
Since the i is positive and the j is negative, does that mean it's pointing down and out on the x axis?

(And we're just learning Gauss' Law, still trying to understand it)

That means that there are two electric fields. One that's in the [tex]+\hat i[/tex] direction, and one that's in the [tex]-\hat j[/tex] direction

Taking the dot product over anyone side, you'll see that the component of the E-Field that's parallel to the surface (Perpendicular to the Area Vector) does not contribute at all to the dot product (And therefore, does not contribute to the flux).

As for Gauss' Law:

The total flux of the E-Field through any enclosed surface is equal to the total charge enclosed by the surface divided by the dielectric constant.

In integral notation:

[tex]\oint \vec E \cdot \vec {dA} = \frac{Q_{enclosed}}{\epsilon_0}[/tex]
 

FAQ: What is the Electric Flux Through a Cube and How Can it be Calculated?

What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given area. It is represented by the symbol Φ and is measured in units of volt-meters (V∙m).

How is electric flux calculated?

The electric flux through a surface is calculated by taking the dot product of the electric field and the area vector of the surface. It can be represented by the formula: Φ = E∙A∙cosθ, where E is the electric field, A is the area, and θ is the angle between the electric field and the area vector.

What is the unit of electric flux?

The unit of electric flux is volt-meters (V∙m).

What is the difference between electric flux and electric field?

Electric flux is a measure of the amount of electric field passing through a given area, while electric field is a measure of the strength of the electric force at a point in space. Electric flux is a scalar quantity, while electric field is a vector quantity.

How does the electric flux through a cube vary with its orientation?

The electric flux through a cube depends on the orientation of the cube with respect to the electric field. If the cube is parallel to the electric field, the flux will be maximum. If the cube is perpendicular to the electric field, the flux will be zero. The flux will be minimum when the cube is at an angle of 45 degrees with the electric field.

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