The capacitance of a parallel plate capacitor after inserting a dielectric material

  • Thread starter moderate
  • Start date
  • #1
21
0
A derivation of the capacitance for a parallel plate capacitor is as follows:

[ assume uniform field, plate area is infinite in relation to the separation, the conductors are perfect conductors ]

So, V=Ed

where
V=voltage difference between plates
E= electric field (uniform)
d=separation distance

Also, apply Gauss' law to a surface that is a rectangular prism, with one plane passing through the conductive plate and another plane passing perpendicularly through the field E. Dotted lines: planes of Gaussian surface. Solid line: plate. Plus signs: positive charge.

--------------
+++++++++++
____________


--------------

result of Gauss' law:
EA=q/[tex]\epsilon[/tex]o
q=EA[tex]\epsilon[/tex]o

dividing:

C=q/V
C=EA[tex]\epsilon[/tex]o / Ed
C=A[tex]\epsilon[/tex]o / d

Supposing a dielectric material is inserted. The insulator's molecules become somewhat polarized, and an induced negative charge appears next to the plate containing the positive charge (and vice versa).

The net result is the reduction in the electric field E. As a consequence, the voltage across the capacitor drops (for an isolated capacitor), and the capacitance (C') increases (from C=Q/V).

I am having difficulty reconciling this explanation with the above derivation for the capacitance.

As can be seen from the red equation, the strength of the electric field cancels out, and does not affect the capacitance! But, the capacitance is clearly different after the dielectric is inserted! What is going wrong?

I am thinking that maybe Gauss' law is no longer applicable, but I can't figure out if this is true/why.
 

Answers and Replies

  • #2
Pythagorean
Gold Member
4,214
272


As a first thought, [tex]\epsilon[/tex]o is the permittivity of air. Once you insert the dielectric, you have to replace that with [tex]\epsilon [/tex] which will reflect the permittivity of the dielectric itself.

This is also assuming it's a linear dielectric.
 
  • #3
21
0


I've figured out why I was wrong (by looking in a textbook).

What was wrong was that for the second case, the induced charge needed to be taken into account.

So, the charge enclosed by the volume was no longer q, but was q-q', where q' is the induced charge of the opposite sign.

Once that's considered, the capacitance is no longer the same as without a dielectric. :tongue2:

(maybe using a different permittivity would lead to the same end result? I don't know)
 
Last edited:
Top