What is the capacitance of a parallel plate capacitor with different areas? 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.
 A: 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 
$$ C=\frac{2\epsilon_0 A_1A_2}{d(A_1+A_2)}. $$
A: The capacitance is a result of the polarization of the medium due to electric field and the attraction of charges on one plate due to the charge on the other (as mediated by the electric field).
When you have two plates facing each other, the electric field is present in their common area (ignoring small fringe effects).
This is why you use the area of overlap to compute the capacitance.
A: I divided every plate in meshes and giving to one plate V=-0.5 and to another one V=0.5V. Then the voltage of every point can be calculated in the following way:
$$Vj=K*\sum \frac{Qi}{r_{ij}}$$
$$\begin{vmatrix}
\\V0 
\\V1 
\\...
\\Vn 
\end{vmatrix}
=K*\begin{vmatrix}
a &1/r01  &...  &1/r0n \\ 
1/r10 &a  &...  &1/r1n \\ 
... &...  &...  &... \\ 
1/rn0 &1/rn1  &...  &1/rnn 
\end{vmatrix}
*\begin{vmatrix}
\\Q0 
\\Q1 
\\...
\\Qn 
\end{vmatrix}$$
Where rij is the distance of the point i to the charge j
And a=1/0.1*mesh size
This can be written in an array form:
[V]=K*[1/r]*[Q]
So making the inverse of the array we have:
$$[Q]=\frac{1}{K}*[1/r]^{-1}*[V]$$
He mesh size was 1mm.
Then as long as C=Q/V we can calculate C in farads:
C= \frac{\sum Q}{V}
Where V=0.5-0.5=1 volt
I added absolute values of charges (if not, a large error is obtained)
In the calculus I had to make inverse of a 3000x3000 array using multithreads
Then obtained following results (SI units):
R1=0.02 R2=0.02 H=1e-3 capacitance=11.637pF (expected 11.13pF) 
R1=0.01 R2=0.02  H=1e-3 capacitance=3.90355pF
The 3.9pF corresponds to a capacitor area for R=0.01185m that is just in the middle between the 2A1A2/(A1+A2) and the minor radius capacitance.
A: Ok, I will try use latex.
Here is the charge distribution that I obtained for same area capacitor and the double radius relation ones:

