Electric Fields and Flux

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1. Jan 19, 2017

WWCY

1. The problem statement, all variables and given/known data
Not so much of homework questions but more to do with clarifying concepts:

1. How do i assess whether or not field lines emitted by charged objects (i.e uniformly charged cube) are both perpendicular to their surfaces uniform? This is in relation to taking the E of ∮φ = EdA as a constant.

2. What is the difference between having a charged conductor and charged insulator in a Gaussian surface? Do I have to take in any particular considerations when calculating flux and fields in either case? E.g: Is calculating the flux of a uniformly charged spherical conductor and insulator the same thing?

3. What does symmetry actually entail? Is a symmetrical object one that has uniform electric fields passing through its surface?

Thanks for any help!

2. Relevant equations
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3. The attempt at a (re)solution
1. I was told that I could use coulomb's law for point charges to work out whether or not fields were uniform over the surface of a uniformly charged cube, however i'm not quite sure how to apply this. Does this have to do with integration (extended charged body) and summing up the fields experienced over various points on the surface and making comparisons?

2. Again I was told that the main difference was that net charges reside on conductors (due to mobile charges) while insulators don't have mobile charges. Again i'm not sure how this would affect calculating flux and field.

2. Jan 19, 2017

BvU

For an ideal conductor there can be no component parallel to the surface: the charges would simply move.
In an insulator the charge density may be irregular and then Gauss isn't very useful. It is useful if there is symmetry enough (e.g. dependent on $\ |\vec r| \$ only in the case of spherical symmetry)
Yes, as long as you are talking outside the charge distribution.
Symmetry occurs when physically nothing changes under some form of coordinate(*) transformation.

(*) other kinds of symmetries occur too: check here - if really curious
Do the integration and find out ... pick a few points: above the center of a side, above the center of an edge at identical distance, on a long diagonal at the same distance from a corner point... (lot of work...)
charges reside on the surface of conductors. In an insulator they can be anywhere. Makes a difference.

3. Jan 19, 2017

WWCY

Does this mean that fields above a conductor surface are uniform?

Gosh, could i do it just by imagining a test charge interacting with the many particles above a finite squarish surface and concluding that the fields experienced by the test charge are non-uniform because 1/r^2 with these particles vary across different points above the surface?

Would I be right in saying that this only makes a difference only when we measure charge (or lack thereof) inside a charged conductor or insulator?

Thanks BvU, you've been of great help!

4. Jan 19, 2017

Cutter Ketch

No, not at all. Charges are free to move within a conductor, but they cannot leave the conductor. They are repelled by each other and spread out until the field everywhere in the conductor is zero. This is because, as WWC points out, if the field wasn't zero the charges would experience a force and would keep moving until the field was zero. So this is why there can be no field component parallel to the surface. But just because the field is perpendicular to the surface does not mean that the field strength has to be the same everywhere.

For a sphere the most the charge can spread out is a uniform even distribution on the surface. This however is an unusual case of high symmetry. For conductors of just about any other shape the charge distribution and the field won't be uniform across the surface. For example in a rod the charge (and so the field) concentrates at the ends. With a football shape the points would have more charge. In fact with any shape the charge and field will concentrate at points and edges.

5. Jan 20, 2017

WWCY

That cleared things up a lot, thank you. Would I then be right in saying that flux cannot be calculated for conductors of arbitrary shapes just by using symmetry alone?

6. Jan 20, 2017

BvU

Basically you try to explore any symmetry there is. A good way to become familiar is by working through exercises and examples

7. Jan 20, 2017

WWCY

Got it, thanks a lot!

8. Jan 20, 2017

BvU

Yes, that would be enough to disprove uniformity. wise approach
No. Inside a conductor the field is 0. The charge distribution on the surface doesn't have to be uniform. It can be influenced by the shape or by other charges outside the conductor (see 'method of mirror images') .
For the field outside an insulating object, it makes a difference if e.g. all charge is on top as opposed to all charge being concentrated at the bottom.