# Capacitors, Electric Fields, and Dielectrics

I have a question that is confusing me perhaps one of you can help me.
If I hook up a constant potential difference to a capacitor and place a dielectric inside of it, will the electric field decrease even if the plate separation remains constant?
I think that the capacitance will increase as well as the charge on the plates, meaning that the electric field would have to remain the same, since the potential is constant? or am I wrong? if so why?
Thanks for the help

Doc Al
Mentor
You are correct.

Do you know of anyplace that offers credible proof like in a book or an article?

This was a test question and I argued with my physics teacher about it. He thinks the electric field will decrease.. where I do not and he said I am wrong.

I would like to show him that the electric field doesn't decrease when a contstant potential difference is applied.

Doc Al
Mentor
The electric field in a parallel plate capacitor, as you apparently realize, depends only on the potential difference and the distance between the plates:
[tex]E = \frac{\Delta V}{d}[/itex]

This kind of thing should be in any textbook; here's a web site that discusses it: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c2

That is the formula I explained to him. I don't see why he doesn't get it

Is it because he thought that the electric field is decreased due to the presence of the opposite electric field set up by the dielectric? But he forget that the voltage is still applied after the insertion of dielectric, which will bring the electric field back.

Pls correct me if I'm wrong.

That must be what he is thinking.

DocAl, would this also apply to non parallel plate capacitors?

Doc Al
Mentor
gokugreene said:
DocAl, would this also apply to non parallel plate capacitors?
Sure. The electric field is determined by the applied voltage and the geometry of the conductors, not by the presence of a dielectric.

As Alpha2005 notes, your instructor is probably thinking of this sequence:
(1) Apply a voltage to the (empty) capacitor
(2) Remove voltage source
(3) Insert dielectric​
In this case, she would be correct. The induced polarization charge within the dielectric will reduce the effective electric field.