why is knowing the total charge on the conductors enough?

alemsalem

how do you prove that the electric field is determined uniquely from knowing the total charge on a conductor (just the outline of the proof).

Thanks!!

Simon Bridge

You don't - you also need the charge distribution.
That will be determined by the properties of the setup.
See Maxwell's equations.

Meir Achuz

The net charge on the conductor is enough.
That is the 'uniqueness theorem' tha is proved in most EM texts.
You start with the volume integral of phi grad phi, where phi is the difference of two possible potentials for the same rho (so delsquared phi=0.
Then use the divergence theorem.

Simon Bridge

... and here's me thinking that the electric field is stronger near the pointy bits of a charged conductor... requiring knowledge of the shape of the conductor as well as the net charge.

Perhaps there is a context I'm missing?
No doubt you have the right of this question though.

LastOneStanding

Meir Achuz is correct—with the caveat that the OP meant to say "in electrostatics". That may be the context you are missing. That is the context in which the uniqueness theorem is proved, though it is easy to see intuitively: on a conductor, charges are free to move to be moved around by any electric field. Hence, due to their mutual repulsion, they will arrange themselves until they all lie on the conductor's surface and the electric field is everywhere perpendicular to the surface—at which point they can move no further. Of course, the situation is very different in electrodynamics since we don't require the charge distribution to ever have shuffled itself into its lowest energy arrangement. OP, if you want a more detailed explanation, I suggest p.118 of Griffith's "Introduction to Electrodynamics" (3rd ed.)

alemsalem

thanks i found the proof in Griffiths, I've seen it along time ago and didn't remember where.

the theorem doesn't say that the electrostatic field doesn't depend on the shape of the conductor, it just says given the total elecrtic charge on the conductors there is only one solution.

DocZaius

Just to be clear I hope everyone agrees that in electrostatics one can know the electric field around a conductor by knowing:

1) Its shape and
2) Its total charge

If not, then I am missing something very important!

Simon Bridge

The proposition under consideration was:...
consider: the field inside a conductor is zero
we identify the inside from our knowledge of the conductor's shape
if all we know is the total charge, we do not know it's shape
therefore, knowledge of the total charge is not sufficient to determine the electric field everywhere.

Perhaps if we modify the proposition:
the electric field, outside the conductor, is determined uniquely from knowing the total charge on it

But the charges could be moving ... lets try again:
the electrostatic field, outside the conductor, is determined uniquely from knowing the total charge on it

... now we are getting somewhere - as noted the charges are free to move, and will spread themselves over the surface as far apart as they can from each other. This means they will tend to cluster about ridges and corners - so the field lines about a corner will be denser than the field lines elsewhere.
i.e. http://physics.bu.edu/py106/notes/Conductors.html

So the electric field outside a needle of charge Q is not going to be, everywhere, the same as the electric field outside a ball-bearing of charge Q ... or is it?

I think a clear statement about what this particular "uniqueness theorum" is saying would be useful. BTW: it is known by a different name?

LastOneStanding

Yes, sorry, the shape does matter of course. I was responding to your original statement which didn't say anything about not knowing the shape, and it hadn't registered for me that Simon said something different.

LastOneStanding

The shape does matter. That registered implicitly for me in the OP since I know the theorem, and I didn't realize you had said otherwise. My mistake.

The point is that given the total charge and the shape of the conductor, there is a unique electrostatic solution—you don't need to be told the charge distribution, since there is only one possible.

Simon Bridge

No worries.
The OP wording could just have been relying on the context or it could have been due to a misunderstanding. I didn't want to assume ;) Hopefully the question is now answered.

alemsalem

Yup! thanks!