I need Help calculating Electric Flux?

In summary, by symmetry, the electric flux through a cube is equal to the electric flux through a sphere that encloses the charge.
  • #1
bmb2009
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**I need Help calculating Electric Flux?

Homework Statement



"A 2.51 μC point charge is placed at the center of a cube with a volume of 7.27 m3. Calculate the electric flux through one side of the cube."

Homework Equations



Eflux = E*Surface Area

The Attempt at a Solution



I tried using the E field of a point charge and since it's in the center... it is .96861 from anyone side of the cube. From this i said electric flux was = E*Surface area but I don't get the right answer? Can anyone walk me through the steps? Help is much appreciated!

 
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  • #2


bmb2009 said:

Homework Statement



"A 2.51 μC point charge is placed at the center of a cube with a volume of 7.27 m3. Calculate the electric flux through one side of the cube."

Homework Equations



Eflux = E*Surface Area

The Attempt at a Solution



I tried using the E field of a point charge and since it's in the center... it is .96861 from anyone side of the cube. From this i said electric flux was = E*Surface area but I don't get the right answer? Can anyone walk me through the steps? Help is much appreciated!

Homework Statement


Integrating the field over the surface area will be a pain, since the field is not constant (varies with distance from the center) and its angle with respect to the surface changes, too.

As a hint: take a good look at Gauss' Law and what it says about the flux leaving an enclosed surface versus the charge within that surface. Symmetry is your friend!
 
  • #3


I still have no idea what to do... To my understanding Gauss' law allows us to evaluate the E flux through a Gaussian surface, which isn't actually a real physical structure, with ease because the E field is perpendicular to the Gaussian surface at all points. But I still don't know how to take that and apply it to the cube? Because the surface to even make Gausses law effective would be the sphere wouldn't it? because a sphere would be symmetric to the point charge?
 
  • #4


bmb2009 said:
I still have no idea what to do... To my understanding Gauss' law allows us to evaluate the E flux through a Gaussian surface, which isn't actually a real physical structure, with ease because the E field is perpendicular to the Gaussian surface at all points. But I still don't know how to take that and apply it to the cube? Because the surface to even make Gausses law effective would be the sphere wouldn't it? because a sphere would be symmetric to the point charge?

If you look closely at Gauss' law, it states nothing about the field being perpendicular to the enclosing surface. It just so happens that we often choose it to be so in order to make calculations easier. In words, Gauss' Law states:

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

Does that help?
 
  • #5


A little...I ended up finding the answer... but I'm still not sure why.

Is the Gaussian surface of the sphere enclosing both the cube and the point charge so that when you calculate the flux through the Gaussian sphere you just divide by 6 to get the flux through 1 side of the cube?
 
  • #6


bmb2009 said:
A little...I ended up finding the answer... but I'm still not sure why.

Is the Gaussian surface of the sphere enclosing both the cube and the point charge so that when you calculate the flux through the Gaussian sphere you just divide by 6 to get the flux through 1 side of the cube?

The idea is that it doesn't matter what shape the Gaussian surface takes... it can be a sphere or a cylinder, or a dodecahedron, some amorphous blob, or a cube. As long as it is a closed surface, the entire electric flux passes through it, and Gauss' Law holds.

A cube describes a perfectly good Gaussian surface. So, by symmetry, since each face of a cube is exactly the same as far as a centrally located charge is concerned, each face must pass the same amount of flux. A cube has six faces, so...
 
  • #7


Further point, consider the relation between the flux and the enclosed charge.

We know the equation
[itex]\Phi \epsilon_{0} = q_{enc}[/itex]
holds for any Gaussian surface. So we know that [itex] \Phi [/itex] must be the same for any Gaussian surface enclosing the charge in question. In this case a spherical Gaussian surface is the most convient surface to use to calculate a the flux produced by the point charge due to symmetry.

Once we calculate the flux through the sphere we observe from the above equation that the total flux through the cube will equal the total flux through the sphere. Since the flux is radically symmetric because the point charge is placed at the exact center of the cube then the flux through all sides of the cube are identical. Thus, we divide the total flux by six giving the answer.
 

1. What is electric flux?

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

2. How do I calculate electric flux?

The formula for calculating electric flux is ΦE = E * A * cos(θ), where E is the strength of the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface. You can also calculate it by integrating the electric field over the surface.

3. What is the unit of electric flux?

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

4. Can electric flux be negative?

Yes, electric flux can be negative. This means that the electric field is entering the surface at a certain angle, rather than leaving it. This is why it is important to pay attention to the direction of the electric field and the angle between it and the surface when calculating electric flux.

5. How is electric flux related to Gauss's law?

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). This relationship is represented by the formula ΦE = Q/ε0, where Q is the charge enclosed and ε0 is a constant value.

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