Re: capacitance of a disk

["John S. Denker" <jsd@MONMOUTH.COM>]

At 10:17 AM 2/9/01 -0500, Eugene Mosca wrote:
> John Denker:
>
> I very much appreciated John Denker's arguments placing an upper and lower
> bound on the capacitance of a disk.

I'm glad somebody liked it.  I wish I had come up with a tighter lower
bound and/or a simpler argument, but John M. kindly provided that for us.

> However, I do have a concern about the
> following assertion:
>
>
> << At surfaces, the electric field strength is proportional to the charge
> per unit area.  For the uniformly-charged disk (case c) this is
>         sigma  = Q / (2 pi R^2)                 (eq 4)
> where the factor of 2 is because the disk has two sides. >>

I stand by that assertion.  Every field line ends on a charge.

> That "the electric field strength is proportional to the charge per unit
> area" is correct for a uniformly charged disk of infinite radius and for a
> conducting disk of finite radius.

That's true, we agree on that.

> However, it seems to me it is not correct
> for a uniformly charged disk of finite radius.

Why not?

> The electric field of such a
> disk is not normal it's the surface, except on the axis.

That's a separate statement.  I never said it was normal.  I did not imply
that it was normal, and assuming normal orientation is not necessary to the
argument I was making.  Frankly I didn't think about orientation at all;  I
made my arguments in terms of field strength, |E|.

In retrospect, I see that I could have made an argument about orientation,
leading to the same result.  Maybe this helps:  remember that I was seeking
an upper bound on the field.  If you want to assume normal orientation,
that only makes the field bigger, so it is allowed by the rules of the
"find a bound" game.

In this context, at 11:37 AM 2/9/01 -0500, Ludwik Kowalski wrote:
> The field is always normal at the surface of a conductor.

That's true, but it doesn't answer Eugene's question about the
uniformly-charged disk.