Calculating Electric Flux Through a Cube: A Simple Guide

[nslinker]

A charge of 12 x 10^-6 C is at the geometric center of a cube. Find the Electric Flux through one of the faces?

My class would have rather left early on friday than go over this example so now I have no clue how to find this.

 

[Doc Al]

What does Gauss's law tell you about the total flux?

 

[nslinker]

From what I've read, if the charge distribution is spread uniformly over a plane, we can determine the direction of E from it's symmetry. And if we know charge, we can find magnitude of E and vice versa.

 

[granpa]

how much flux passes through all the faces?

 

[Doc Al]

From what I've read, if the charge distribution is spread uniformly over a plane, we can determine the direction of E from it's symmetry. And if we know charge, we can find magnitude of E and vice versa.

You're not asked to find E, just the flux through each face. What's the total flux?

 

[nslinker]

For my specific problem, the total flux through a cube will be zero.

 

[Doc Al]

For my specific problem, the total flux through a cube will be zero.

What does Gauss's law tell you about the total flux from a point charge?

 

[granpa]

are you trying to visualize it? the whole point of flux lines is so you can easily visualize the field.

 

[nslinker]

That the total flux through any closed surface is proportional to the net electric charge inside the surface. And at each point on the surface, E is perpendicular to the surface, and its magnitude is the same at every point.

And yes, I've tried visualizing it but like I said before the teacher didn't get into details with it and my book doesn't give me any reassurance on what I've visualized.

 

[Doc Al]

That the total flux through any closed surface is proportional to the net electric charge inside the surface.

That's what you need to know. Now how is that flux distributed over the six faces of the cube? Which face gets the most flux?

And at each point on the surface, E is perpendicular to the surface, and its magnitude is the same at every point.

That's only true in special cases. It doesn't apply here--luckily you are not asked to find E, only the flux.

 

[granpa]

That the total flux through any closed surface is proportional to the net electric charge inside the surface.

that part you got right

And at each point on the surface, E is perpendicular to the surface, and its magnitude is the same at every point.

dont have any idea where you got this. the surface is arbitrary. why would you think then that E would be perpendicular to it?

 

[nslinker]

Well, with a cube the Area is S² and out of the 6 sides I only calculated two that did not have a Flux of 0 and I got -ES² and +ES²

 

[granpa]

would it help you to think 2 dimensionally? draw a box with a point charge in the center then draw lines of flux. then generalize the result to 3 dimensions.

hint:its much much easier than you are making it.

 

[Ithryndil]

nslinker, I think an issue you're having is that you're making the problem harder than it needs to be. Think about what flux is and what Gauss's law tells us. One thing is tells us is that the flux is the same through any closed surface with a point charge inside of it. The charge is located at the center so the flux is the same through the cube as it would be through a sphere around the point charge.

One hint is that you do not need E at all to solve this problem. There's an easier way which involves the symmetry of a cube.

 

[Doc Al]

Well, with a cube the Area is S² and out of the 6 sides I only calculated two that did not have a Flux of 0 and I got -ES² and +ES²

Nope. Forget about E or the area of the sides. You don't know E, so that will get you nowhere. (That's the hard way to solve this probem.) This problem is much simpler than all that.

For the nth time, Gauss's law will tell you how to find the total flux from that charge. What is it? (I want the actual value of the total flux.) Hint: The total flux has nothing whatsoever to do with the shape of the container--it will be the same for a cube, a sphere, or any closed surface that contains the charge.

 

[nslinker]

Okay, I have done a little more reading into the next chapter of my book and and saw that Flux = E x A, and after I put in terms I come out with Flux = q / Epsilon. But I think that's for spheres.I tend to have problems with making questions tougher than they need to be and I buckle during tests because of that.

 

[Ithryndil]

Your statement about Flux is correct. But Gauss's law tells us the flux through a surface is independent of the shape of that surface. So Gauss's law is for spheres, cubes, pyramids, etc.

 

[granpa]

you're missing the whole point of flux lines. the number of lines coming out of each change is proportional to the amount of charge. each line continues outward forever. you're simply counting the number of lines that exit the box. draw the box I told you about and count the number of lines that cross it. it should be perfectly obvious.

 

[Doc Al]

Okay, I have done a little more reading into the next chapter of my book and and saw that Flux = E x A, and after I put in terms I come out with Flux = q / Epsilon. But I think that's for spheres.

Total flux = q/epsilon is for any closed surface, not just spheres. That's what Gauss's law says. You might want to review Gauss's law: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gaulaw.html"

 

[nslinker]

Okay, so if Flux = q / Epsilon. Then it's (12x10^-6) / (8.854x10^-12), correct?

 

[Ithryndil]

That's correct, but that's for the whole cube. How much of the flux is going through each surface?

 

[Doc Al]

Okay, so if Flux = q / Epsilon. Then it's (12x10^-6) / (8.854x10^-12), correct?

Right. If that's the total, what's the flux through each face?

 

[nslinker]

2.258 x 10⁵ ?

 

[Doc Al]

2.258 x 10⁵ ?

Yes. (Use appropriate units.)

 

[nslinker]

2.258e⁵ Nm²/C

 

[Doc Al]

2.258e⁵ Nm²/C

Excellent!

 

[nslinker]

Wow...you were all right; much simpler than I ever thought it was. Thanks to all of you for putting up with my stupidity for the time being. I am sure I will need more help soon and down the road [next week probably :) ] and I would be glad to get your guys help again. Thanks again!