Calculating Electric Fields on the Surface of a Charged Cube

AI Thread Summary
The discussion focuses on calculating the electric field on the surface of a charged outer cube surrounding a smaller charged inner cube. Participants suggest using Gauss's law to simplify the integral of the electric field over the surface. It is noted that while the total electric flux can be calculated, the lack of symmetry in the outer cube's surface means that the electric field cannot be determined at all points without further integration. The conversation emphasizes the importance of symmetry in applying Gauss's law effectively. Ultimately, the consensus is that additional calculations are necessary to find the electric field at specific points on the outer cube's surface.
Violagirl
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Homework Statement



A cube of side a has a cube of side a/2 centered within it. The inner cube has a total charge Q that is uniformly distributed over its surface. A) For the surface of the outer cube, find:

s E * dA

B) Is this sufficient information to find the electric fields at points on the surface of the outer cube? Explain.

Homework Equations



E = F/Q

(?)

Not sure with a cube otherwise.

The Attempt at a Solution



Not sure how to start it. Drew out diagram of situation. See attached document.
 

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Thanks for your response! Ok so looking at it again:

∫E * dA

Since E is a constant, it can be pulled out:

E ∫dA = (a) (a/2) = a2/2

If I'm doing this right, then I just need to find the integral of a2/2, right? So I would then get:

E * a2x/2

As for the second question, if the above is correct, I would say then you have enough information to find the electric field at points on other areas of the surface of the cube too then.
 
Use the Gauss's law! It directly gives the required integral without any integration!
##\int E.dA=Q/ \epsilon##
Violagirl said:
I would say then you have enough information to find the electric field at points on other areas of the surface of the cube too then.

Note:-If all you have got is the electric flux through the surface you cannot find the electric field at all the points.
To do this you will have to integrate due to the lack of symmetry of the surface.
 
Ok, that makes more sense since the outer surface is not symmetric as you said. You have to integrate at along different points to find each electric field. Thanks for your help with this problem!
 
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