capacitanceelectrostatics
Suppose there is a parallel-plate capacitor but the two plates have different areas, $A_1$ and $A_2$. How will we derive an expression for its capacitance?
I have been told to take the area common between the two, but from where does this follow?
Ideally we start off by assuming that one plate has $+Q$ charge and the other has $-Q$. Then we find that the potential difference $V$ is directly proportional to $Q$, from which we find capacitace $C$.
I am unable to find the potential difference.
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.
The solution is incorrect. Capacitance depends on geometry, and a parallel plate with a diagonal connector is plainly a different geometry than a parallel plate so the statement "effective distance between the plates becomes zero" is irrelevant.
Capacitance is not a very useful concept to apply to something that is not a capacitor. If you insist on assigning a capacitance to a simple wire, though, the reasonable value is actually infinity. An ideal wire has 0 reactance at all frequencies $\omega$ and reactance is given by $X=\omega L-\frac{1}{\omega C}$. It's clear that this can only be zero for all frequencies for $L=0$ and $C=\infty$.
There is the separate concept of "self capacitance" or "capacitance to ground" which can be computed and is useful in some cases, but it's not infinite in this case and isn't directly related to the capacitance of a parallel plate capacitor to begin with.
Best Answer
In general, the areas of the plates add like $(\frac{1}{A_1}+\frac{1}{A_2})$, as I will prove. The capacitance reducing to "taking common area" only in the limit of one of the plates being much larger than the other.
$$C=\frac{Q}{|\Delta V|}$$
WLOG, let $A_1\geq A_2$, let the plates be separated by a distance $d$, and suppose that $|Q|$ is on both plates. Now,
$$\Delta V=-\int_0^d \mathbf{E}\cdot d\mathbf{l}.$$
Where $\mathbf{E}$ is defined to be the electric field between the plates. Now, the electric field between two finite plates can get quite complicated if we are not armed with the assumption $$d<<A_1,A_2.$$
Given that, the electric field between the plates can be taken to be uniform. Now, use Gauss' law (using the "pillbox" surface). Since $A_1$ is bigger by assumption, the electric field between the plates, then, is determined by
$$E_1\cdot A_{pillbox}=E_1\cdot 2 A_{circle} =\frac{1}{\epsilon_0}\sigma_1 A_{circle} $$
Therefore,
$$E_1=\frac{\sigma_1}{2\epsilon_0}=\frac{Q}{2A_1\epsilon_0}. $$
For, the other plate we have that
$$E_2=\frac{\sigma_2}{2\epsilon_0}=\frac{Q}{2A_2\epsilon_0},$$
Therefore, the integral becomes
$$\Delta V =\frac{Q}{2\epsilon_0}\frac{A_1 + A_2}{A_1A_2}d$$
in magnitude, and the capacitance is
$$ C=\frac{2\epsilon_0 A_1A_2}{d(A_1+A_2)}. $$