Why Does a Charged Disk Generate a Non-Zero Electric Field at Its Center?

[hell-hawk]

I have two questions.
The first concerns a charged disk. Consider the center of the surface of the charged disk, O. Mathematically, the value of E at O comes out to be a non-zero finite value. However i am having difficulty linking this with the physical sense. The disk can be thought to be composed of an infinite number of rings; i think that each ring will set up zero electric field at O since for each charge element on the ring, the diametrically opposite charge element of the charge element cancels its effect. So how come does E come out to be non-zero at O? Please explain the flaw in my reasoning in physical terms, not mathematical.
The second question involves two charges. Consider one charge to be fixed, while the other moves past it in a straight line. What fields will be set up?
What force, if any, will act on each of the charges?
Please help. I'll be highly grateful.

 

[ZapperZ]

I have two questions.
The first concerns a charged disk. Consider the center of the surface of the charged disk, O. Mathematically, the value of E at O comes out to be a non-zero finite value. However i am having difficulty linking this with the physical sense. The disk can be thought to be composed of an infinite number of rings; i think that each ring will set up zero electric field at O since for each charge element on the ring, the diametrically opposite charge element of the charge element cancels its effect. So how come does E come out to be non-zero at O? Please explain the flaw in my reasoning in physical terms, not mathematical.

What if I put a point charge at O? Do you still expect to measure a non-zero E-field there? If you do, then no matter how many of these rings you add, even if the fields for these rings cancel, you will have a net field from this point charge at O. So you should think of a disk as being assembled by all of these rings, except that the ring of infinitesimal diameter at the origin approaches that of a point charge.

Zz.