# Conducting slab between a parallel-plate-capacitor

1. Oct 11, 2008

### Tufts

This is a problem I found while studying for an exam I have next thursday, and conceptually it doesn't make any sense for me!

NOTE - This is not a dieletrics problem.

1. The problem statement, all variables and given/known data
A conducting slab with a thickness d and area A is inserted into the space between the plates of a parallel-plate-capacitor with spacing s and area A. What is the value of the capacitance of the system?

Diagram of the system:

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

//////////////////////////// < Conducter Slab (d wide, with area A)

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

2. Relevant equations

$$C = \frac{\epsilon_{o}A}{d}$$ (1)

Capacitors in parallel combination: $$C_{eq} = C_{1} + C_{2}$$ (2)
Capacitors in series combination: $$\frac{1}{C_{eq}}= \frac{1}{C_{1}}+\frac{1}{C_{2}}$$ (3)
3. The attempt at a solution
At first sight, I immediately identified this problem as a simple series capacitor combination problem. Why? The conducter slab inbetween the plates will divide the existing capacitor into two after reaching eletrostatic equilibrium, as in the diagram:

+Q
--------------------
< First capacitor
-Q
--------------------
///////////////////////
--------------------
+Q
< Second capacitor
--------------------
-Q

Since the slab is a conductor, it acts as a bridge or a conducting wire between these two capacitors. One capacitor after the other, seems to be series for me. Knowing that, I used equation 1 (where $$d$$ = 2(s - d)) to find the individual values of the capacitors, and used it along with equation 3 to find the equivalent capacitor.

My answer: $$C_{eq} = \frac{\epsilon_{o}A}{4(s-d)}$$

Unfortunately that wasn't correct. The real answer was $$\frac{\epsilon_{o}A}{s-d}$$ which after reverse engineering, used parallel combination instead of series. Like I said before, this doesn't make sense at all for me.

Can anyone explain this? I would appreciate any help.

2. Oct 11, 2008

### tiny-tim

Welcome to PF!

Hi Tufts! Welcome to PF!

Isn't $$d$$ = (s - d)/2?

3. Oct 11, 2008

### Tufts

Oh no! Silly me!!

I guess I was thinking too hard on this question, and blatantly ignoring an obvious mistake. Thanks for noticing. I'm glad that at least my analysis of the system was correct.

4. Oct 11, 2008

### nasu

The slab does not have to be in the middle. The nice thing is that the result does not depend on the position of the slab.
It would be interesting (and quite easy) to try to solve the problem in the general case.
I mean, take s1 and s2 as spacing for the two capacitors, with s1+s2=d-s

5. Oct 11, 2008

### Tufts

Well it makes sense why it would work for any position. If one of the distances get larger, the other one has to get smaller in the same proportion. This way the sum is always the same.

6. May 7, 2009

### mcgonia

Now I'm doing some excercises on capacitors and I found a pretty similar one to this, and I this the answer given here is not correct (if I'm wrong, please tell me, because on this depends my exam result :tongue: ).

I think that we have to remember about the the permittivity of the slab (εr), because the electric field in the slab is different than in the capacitor.
So the capacitance will be:

C=$$\frac{\epsilon S}{d-b+\frac{b}{\epsilon r }}$$

where d is the distance between plates of capacitor and b is the thickness of the slab.

If anyone could tell me is it correct, I would really appreciate it. :)

Last edited: May 7, 2009