I searched for this question but I cannot get satisfactory answer. Say if we have a parallel plate or cylindrical capacitor and if I supply unequal charge to it how could I find its Capacitance. We know $Q=CV$. I have the potential difference (i can calculate that). So what should I take the value of Q to find its capacitance?
[Physics] What would be the capacitance of an unequal charged capacitor
capacitanceelectric-fieldselectricityelectrostaticspotential
Related Solutions
First of: the energy W does not increase exponentially it increases quadratic!
Second:
What do you want?
Energy density!
So to get good high energy density you need high C because you can (as you already stated) only go to a certain amount of voltage until you get a breakdown. The problem is, C does not change easily. If you take a simple Plate - Plate capacitor $C \propto \frac{A}{d}$ where A is the Area of the plates and d the distance between them. The smaller the distance the higher the electric field gets between the plates gets relative to Voltage ($|\vec E| \propto \frac{U}{d}$) the faster you get a spark. So the only thing you can actually increase safely is the area. That of course increases the wight/size of you capacitor, reducing the Energy density. This generally applies to all capacitors
Wikipedia does have an excellent article about capacitors:
http://en.wikipedia.org/wiki/Capacitor
Third:
About the capacitance:
A spring is a very good analog to your problem:
A newtonian spring has a certain flexibility, describing the force you have to pull it relative to the expansion of the spring:
$F=k \times x$
Where F is the Force k is the flexibility and x the way. So the capacitance is actually k in this analog. But F is not the Energy W.
$E = \int_0 ^x F(x) dx$
So the more you pull the harder it gets to pull because the force from the spring increases. So to get the energy you can not just calculate $E = F \times x$ but you rather have to integrate the way.
That of course gets you to
$E = \frac{1}{2} k x^2$
Back to the Capacitor:
The more electrons you put in it, the stronger becomes the electric field between the plates of the capacitor. So you have to increase the voltage. The capacitance just tells you how high your voltage has to be (how hard you have to push) to put more electrons on the plate.
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
$$ C=\frac{2\epsilon_0 A_1A_2}{d(A_1+A_2)}. $$
Best Answer
Remember that the word "capacitance" is just that, a word. The standard definition of capacitance given in introductory textbooks only applies when the two objects have charges $Q$ and $-Q$, in which case we define $$C = \frac{Q}{V}$$ where $V$ is the potential difference between the objects. There is no inherently "correct" way to generalize this definition to unequal charges; there are just more and less useful ways. Furthermore, you never need the idea of capacitance. You can just solve for everything directly using Coulomb's and Gauss's laws. Capacitance is merely a definition, not a law.
That said, there can certainly be bad definitions. When you have unequal charges $Q_1$ and $Q_2$, it seems many people in this thread and others would like to define capacitance so that it stays the same as before, which they claim happens if you define $C = |Q_1 - Q_2| / 2 V$. However, this claim is wrong. As a simple example, consider two distant spheres of different radii $r_i$. Their pairwise capacitance is certainly nonzero, but if you give the spheres the same charge, $Q_1 = Q_2 = Q$, then $V_1 \approx k Q_1 / r_1$ and $V_2 \approx k Q_2 / r_2$. As a result, their claimed definition of capacitance yields zero!
The problem is that you can't summarize the response of two conductors to general charges with a single number; instead, you need three, which are roughly the original pairwise capacitance and the self-capacitances of each one alone. Only by using all three of these quantities can you compute the voltages. Similarly, for $n$ conductors you need to specify $\binom{n}{2} + n$ numbers.
In electrical engineering courses, this information is summarized in an elegant way. To motivate it, suppose you had $n$ conductors, which had charges $Q_i$. The charge on any one conductor affects the potentials and charge distributions on all of the other conductors in a complicated way. However, the equations governing the system are still linear, which means the superposition principle works. In particular, that means the potentials $V_i$ of the conductors in the general case can be found by adding together the potentials you get by only charging the first conductor, and then only charging the second conductor, and so on, assuming the potential is set to zero at infinity.
Therefore, the $Q_i$ and $V_i$ are always related by a linear transformation, and we define $$Q_i = \sum_j C_{ij} V_j.$$ The matrix of $C_{ij}$ is called the Maxwell capacitance matrix. For example, in the special case where there are only two conductors with opposite charges, you can show that the "usual" simple definition of pairwise capacitance is related to these coefficients by $$C = \frac{C_{11} C_{22} - C_{12}^2}{C_{11} + C_{22} + 2 C_{12}}.$$ Furthermore, it can be shown that $C_{ij} = C_{ji}$, and that the usual self-capacitances are the $C_{ii}$. For more about this, see section 3.6 of Purcell and Morin, Electricity and Magnetism.
Once again, you never need this idea to solve problems, because it is derived from more basic things, such as Gauss's law. But it's the most useful way to parametrize things in certain contexts.