If a conductor like copper is placed between two plates of a parallel plate conductor,neither touching any of them, what will happen to the capacitance of the capacitor?
Electrostatics – What Happens if a Conductor is Introduced instead of a Dielectric Medium in Parallel Plate Capacitor?
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Related Solutions
If there is a charge $Q$ and $-Q$ on each plate of the capacitor , when you insert a perfect conductor between the plates (parallel), you simply will have a charge $+Q$ on one side(facing negative plate of capacitor) and $-Q$ on other side of the inserted plate. (Given that the area of plates is large enough to assume constant electric field between them)
Now , because the field between two (effectively) infinite plates doesn't depend on their distance, you will have the same field as before (without the new conductor).So , as you mentioned, the new potential difference is $E(d_0-d_{metal})$ .
Remember that according to uniqueness theorems in electrostatics, when you find an answer for a problem (that is consistent with equations) , this is the correct and only possible answer . So, in electrostatics problems you should try to find a story to create the final form of the system from a simple one , through which you can find the answer more easily.
In theory, provided that the surface area of the electrode remains constant and that the electrode material is conductive, the capacitance will not differ significantly when using different metals. This is just a consequence of the fact that for parallel plates, the capacitance only really depends on plate surface area.
However, for modern electrochemical capacitors this relation is not quite correct; a large part of the capacitance is due to a surface effect called double-layer capacitance (image from http://en.wikipedia.org/wiki/File:EDLC-Charge-Distribution.png):
The capacitance of electrolytic capacitors and supercapacitors is thus largely dependent on the effective surface area of the electrode material used, and thus materials with large surface areas (like activated carbon or other porous materials) are often used. This allows one to construct capacitors with far higher capacitances than would otherwise be expected from a naive application of the parallel plate capacitance formula.
Best Answer
Suppose you have a parallel plate capacitor:
So we have a capacitance:
$$ C = \frac{\varepsilon_0 A}{d} $$
and a charge $Q$ on the plates given by:
$$ Q = CV $$
Suppose we now insert a sheet of copper in between the plates as you describe:
The electrons in the copper plate are free to move, so they flow towards the positive plate and you end up with two capacitors with an increased capacitance:
$$ C' = \frac{\varepsilon_0 A}{d'} $$
The combined capacitance is obtained by using the equation for two capacitors in series:
$$ \frac{1}{C_{tot}} = \frac{1}{C_1} +\frac{1}{C_2} $$
So in this case the new capacitance is:
$$ C_{tot} = \frac{\varepsilon_0 A}{2d'} $$
And since the copper sheet has a thickness greater than zero $2d' \lt d$ and therefore when you increase the copper sheet the capacitance increases.