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I am confused as to whether/how capacitance changes when each plate has a different charge. For example, consider a coaxial cable and put $20Q$ on the outer cable, and $-Q$ on the inner. Or how about concentric spheres, grounding either the inner or the outer?

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This is a duplicate of…. You have to consider the ground as an additional terminal. Then everything makes sense. Especially, charging always means separation of charges under supply of some form of work to the system which becomes electrical energy of the system. – Tobias Mar 1 '14 at 5:41
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Systems of plates are not typically considered capacitors unless they are globally neutral. Nevertheless, capacitance is a geometric property that is to do with the system more than the actual voltages and charges you apply to it, so that your question still makes sense: the capacitance is the same as it would be with symmetric charges.

More specifically, the (mutual) capacitance of two conducting surfaces is defined as the charge that must be applied to one surface so that the potential in the other one will rise by one unit; by energy considerations it must be symmetric.

If the charges on both surfaces are antisymmetric (i.e. $+Q$ and $-Q$) then there will be a potential difference between the plates of $V=Q/C$. If they are asymmetric, a similar statement holds: if plate 1 has charge $Q_1$ and plate 2 has charge $Q_2$, then there will still be a potential difference between them of $V=(Q_1-Q_2)/2C$.

The problem with this, though, is that you're no longer seeing the whole picture, and you'll also have to deal with the self capacitance of the plates, which wasn't a problem before: if you put 100 C of charge on one plate and 99 C on the other, there will still be some potential difference between the plates, but they are also at a very high potential with respect to anything else you might consider, and you will have a host of other problems. This is why the situation is hardly ever considered.

In the general case, then, you have not one but two independent voltages to consider; that is, the mean and the difference, or the two voltages of the plates separately. To deal with this appropriately, you need to use a general capacitance matrix as Suresh describes in his answer.

Finally, you also need to worry about what other charged systems you must consider, and where they are. You can't just conjure a large charge on one plate without taking it from somewhere, and depending on where you're grounding there may also be some significant energy of interaction there.

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I think this answer is incorrect. For the coaxial capacitor in the question, the electric field between the two conductors is determined by the inner conductor's charge only, which then determines the potential difference. The "excess" charge of the outer conductor produces an external field which must terminate somewhere else. – Art Brown Feb 28 '14 at 0:49
This was already in my final paragraph but I've emphasized it a bit. – Emilio Pisanty Mar 5 '14 at 14:51
Thanks, but my objection is to the formula at the end of your third paragraph, which is true for a parallel plate configuration but is not true in general, and in particular is not true for the geometries mentioned in the question. I've edited my answer to make this point more explicit. – Art Brown Mar 6 '14 at 6:57
While the capacitors we use in simple circuits are indeed globally neutral, it is no trouble to define either the capacitance of single pole object nor the energy stored by charging it up (this finds a use in touch switched lamps). We just don't usually bother the undergrads with these cases. – dmckee Apr 26 '14 at 0:03

Suppose you have two conductors kept at voltages $V_1$ and $V_2$ and they have charges $Q_1$ and $Q_2$ respectively. Then, one has the relation $$ Q_1 = C_{11}\ V_1 + C_{12}\ V_2\quad \textrm{and}\quad Q_2 = C_{12}\ V_1 + C_{22} \ V_2 \ , $$ which defines a (symmetric) Capacitance matrix that is determined by the geometries of the two conductors. This generalizes the case that one uses for capacitors. See Purcell's Electricity and Magnetism in the Berkeley Series in Physics for a discussion.

Note added: The assumption in the above formulation is that the conductors have finite size. Then, when the charges $Q_1$ and $Q_2$ are finite, one assumes that the potential at spatial infinity defines the zero of the potential.

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You should add that $V_1$ and $V_2$ are voltage drops w.r.t. the ground. You model the ground abstractly as infinity in the comment to Art Brown. This is a good approximation if the system in question is far away from the ground. Certainly there are other systems such as interstellar winds of charged particles but those also have their counterparts. – Tobias Mar 1 '14 at 7:50

Suresh's answer gives the correct general formalism.

(1) For the specific case of a coaxial cable, the electric field between the two conductors is determined by the charge $-Q$ on the inner conductor, which terminates on $+Q$ worth of charge on the outer conductor. (There can't be any field inside the inner conductor, so all the field generated by its charge extends outward.) The potential difference between the two conductors will just be $Q/C$, where $C$ is the capacitance of the coax.

The "excess" $19Q$ on the outer coax conductor produces a field extending outwards from the coax. If this field extends to infinity the potential drop will be infinite. Instead, one can terminate the field on $-19Q$ worth of new charge situated somewhere outside the coax. If that new charge resides on a new third conductor, that new topology is characterized by a new capacitance $C_{new}$, which will set the potential difference between the outer coax and the new third conductors: $19Q/C_{new}$.

(2) It's interesting to generate the capacitance matrix for this configuration. (See suresh's answer for the capacitance matrix equation formulation.) Because the coax cable length is infinite, one must consider the "specific" charges and capacitances, that is, charge and capacitance per unit length.

First consider the infinite coax in isolation. Evaluating the matrix elements (with the inner conductor denoted "1") gives: $$C_{11}=-C_{12}=-C_{21}=C_{22}=C=\frac{2 \pi \epsilon}{\ln\left(r_2/r_1 \right)}$$

This matrix is degenerate, forcing $Q_1=-Q_2$.

Now add a third, grounded cylindrical conductor at radius $r_3$, $r_3>r_2$. $C_{11}$, $C_{12}$, and $C_{21}$ are unchanged, but $C_{22}$ acquires a new term:

$$ C_{22}= 2 \pi \epsilon \left( \frac{1}{\ln\left(r_2/r_1 \right)} + \frac{1}{\ln\left(r_3/r_2 \right)} \right) = C + C_{new}$$

This addition removes the degeneracy, allowing arbitrary charges to be assigned to the two coax's conductors, with the resulting voltages measured with respect to the grounded conductor. As $r_3$ increases, $C_{new}$ decreases, reaching the degenerate case in the limit.

Inverting this matrix equation and solving for the potential difference, one finds: $$ V_1 - V_2 = \frac{Q_1}{C} $$ agreeing with the first analysis result.

In general, the potential difference in terms of the capacitance matrix elements and charges is: $$ V_1 - V_2 = \frac{(C_{22}+C_{21}) Q_1 - (C_{11} + C_{12}) Q_2 }{C_{11} C_{22} - C_{12} C_{21} } $$

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The net charge of the system need not be zero -- it suffices that the total charge be finite. Then, the potential at infinity can be chosen to be zero as is assumed in the formula that I wrote. – suresh Mar 1 '14 at 3:13
@suresh: got it, thanks! I've edited my answer. – Art Brown Mar 1 '14 at 5:41

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