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I need Help calculating Electric Flux?

  • Thread starter bmb2009
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  • #1
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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 any one 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





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
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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 any one 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
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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
gneill
Mentor
20,793
2,773


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
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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
gneill
Mentor
20,793
2,773


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
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0


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.
 

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