How to Approach an Electro Box Problem with Nonuniform Electric Field?

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SUMMARY

The discussion revolves around solving an electrostatic problem involving a cubic box with a nonuniform electric field defined as E(x,y,z) = Kz j + Ky k, where K = 4.10 N/(C*m). The primary focus is on applying Gauss' Law to determine the electric flux through the top face of the box, given that there is no net charge inside. Participants clarify that the problem requires calculating the flux through each face of the cube using surface integrals, emphasizing the importance of understanding vector fields and surface integrals in this context.

PREREQUISITES
  • Understanding of Gauss' Law and its application in electrostatics.
  • Familiarity with vector fields and their representation.
  • Knowledge of surface integrals and their computation.
  • Basic calculus, including double integrals.
NEXT STEPS
  • Study the application of Gauss' Law in various electrostatic scenarios.
  • Learn about vector field representations and their physical significance.
  • Practice solving surface integrals, particularly in the context of electric flux.
  • Explore resources on advanced calculus topics, focusing on double integrals.
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Students studying electromagnetism, physics educators, and anyone seeking to understand electric fields and flux calculations in electrostatics.

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[SOLVED] Electro Box

1. Homework Statement :
A cubic box of side a = .290 m is placed so that its edges are parallel to the coordinate axes. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j +Ky k, where K=4.10 N/(C*m) is a constant.


2. Homework Equations :
\ointE\bulletdA = Q/E_naught



The Attempt at a Solution


I'm just inquiring as to where to begin this problem. I've only taken through a high school equivalent of first semester calculus, so I have no idea how this equation works. If anyone could point me in the right direction, I'd be very greatful!
 
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You...didn't ask a question? I'm assuming it's to find flux since that's the equation you gave

You're given a vector field in the form of E=K*zj+K*yk, j and k are the y and z unit vectors, respectively, and K. This is just given for the problem, it has no other fundamental significance. It's a simple vector field, so if you're unfamiliar, google or wiki them and try to figure out what this one looks like

The equation you have is Gauss' Law, that integral on the left that may look funky to you is a surface integral, the loop in the middle represents that it's a closed surface(which a cube is!)

http://en.wikipedia.org/wiki/Gauss's_law

That first line of math stuff is just what you've typed there, note, the derivation and understanding are useful(why flux through a closed surface is what it is)

EDIT: This is a little weird without knowing the question >_> I guess it's show that with that electric field the charge inside the box is 0? It would be weird to say "here's a non-uniform electric field" and "there's no charge inside the box" and then have one of the fundamental laws of nature point out there must be a charge inside the box. So I guess you'll do that work and find Q=0 despite E=/=0. I think. Been a while >_>

http://en.wikipedia.org/wiki/Gaussian_surface

that'll give you some help on solving it, note that you'll have six integrals for the six sides, and the integral of 1*dA over a side will be just the area of the side. I don't think it'll come up but that maybe illustrates what is meant by dA, it's the infinitismal area "vector", which, btw, points out of the surface. So the top side's dA is in the positive z direction, the bottom side's is in the negative
 
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Oh man, sorry my bad about not posting the question. Our capa problems are created...oddly. I thought I grabbed it, but I suppose not >_< The actual question it asks is:
What is the electric flux through the top face of the box? (The top face of the box is the face where z = a. Remember that we define positive flux pointing out of the box.)

Anyways, I'll look into the resources you provided. Thank you very much!
 
Do you know the answer? The way I figure, and it's been a while since I've done these, since there's no net charge in the box, there's no net flux in or out of the BOX, but the individual faces are different.

I believe you'd break down into the six integrals for each face, and each integral would represent the flux through that particular face? <--that's where I'm thinking I messed up if I did

so you'd say da=da*z(oh hey I'm going to use bolded x, y, z instead of i, j, k, I'm newschool like that)

the dot product of E and da will then result in just the z component, ky(giggle) left times da. For the top surface da is dxdy and it's now a double integral, the dx part just gives you a, and the dy part, since we have y in the integrand, leads to a net result of a^3*k/2

Lemme know if that's clearly way too much math for it to be right(it's an easy double integral but if you've never had one I probably did something wrong)or if you know the answer is wrong, so I can delete this before exposing my mistakes! :)
 
Sorry about not getting back to you sooner - our school's internet was down yesterday.

Anyways, I don't know the answer, and as of right now, and I've no idea if that math is correct or not. I think my problem lies in not understanding line integrals very well, as we've only done basic intro calc stuff. Where might be a good place to look for figuring those out?
 
I think the math is fine, I'm just not sure that it's actually the flux or anything meaningful that I found

Also, it's a surface integral, not a line integral, but similar idea. If it's just physics 2 they usually give you absurdly simple ones since they're in general not necessarily very easy, and often hard. Things like where the surface integral just ends up being the surface area of like a sphere
 
Ahhhhhhh, ok. That makes sense. It seemed really strange that they would give us math like that. Thanks for your insights!
 

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