capacitance

**StatusX** · Dec13-04, 04:12 PM

the general definition of capacitance is that if you have an arrangement of two conductors, and you put a charge +Q on one and -Q on the other, and a potential difference of V results between them, the capacitance is Q/V. But what if the charges on each conductor don't have the same magnitude, say +Q and -Q/2? Is there any way to measure the capacitance from this setup? Or, given the capacitance for the normal case, could you derive the potential difference that will result here?

**pack_rat2** · Dec14-04, 10:13 PM

It's the potential DIFFERENCE, and the DIFFERENCE in the number of charged particles that matters. Suppose one plate is +5Q and the other is +3Q. The difference of 2Q is what is important. Also, C = Q/V may be a definition of capacitance, but it's not used in measuring it (not directly, at least). Capacitance is measured either by determining its reactance to a particular frequency AC voltage, or the time-constant of an R-C circuit it's part of.

**spacetime** · Dec15-04, 07:51 AM

Say you have two conducting plates. You put a charge of +Q and -Q/2 respectively on them. Then you place them facing each other as the plates of a parralel plate conductor.

Now, the charges on the plates will be redistributed in such a way that the sides of the two plates facing each other will have equal but opposite charges. The capacitance is calculated taking only these charges into account.

You can verify this if you apply the theorems that electric field inside a conductor is zero in static conditions and take the superposition of electric fields due to charges on all four faces of the two plates.( after assuming suitable variables for them )

spacetime
www.geocities.com/physics_all/

**StatusX** · Dec19-04, 11:30 PM

Thanks for your replies, but I'm still not sure.

pack rat: If that's the case, then there should always be a potential difference of 0 between any two conductors with the same charge, regardless of shape or position. I don't see how this could be, since if you brought one in from infinity, it would seem they would repel for most of the trip, and thus you must be establishing a potential difference.

spacetime: I don't think that's necessarily true (what if you have two plates with equal charge: on which one will the charges move?), and anyway, that only works for parellel plates. I'm talking about the general case.

**Gokul43201** · Dec20-04, 12:56 AM

You want to first ask yourself what you mean by "capacitance of a pair of conductors with different charges". Capacitance is actually defined only for a single conductor. It has become common to extend this definition (to the capacitance of a pair of conductors with equal and opposite charges on them) to the popular version of "charge per unit voltage".

For a general system of conductors, one has :

$LaTeX Code: Q_i = \\sum _j {C_{ij}V_j$
where

is the capacitance of the i'th conductor.

This comes from solving the Poisson Equation with the Green's Function for the system of conductors, and inverting it (since that would give you $LaTeX Code: V_i = \\sum {Csingle-quote_{ij}\\rho _j}$ )

So really, the capacitance of a conductor is the charge on it when it is maintained at 1 volt, and all the other conductors are grounded.

~~rayjohn01~~ · Dec20-04, 09:05 PM

The definition of capacitance is as you already said -- it does not change just because you charge two plates differently and then bring them into proximity
which would change the potential difference between them.
The definition is fixed ..
Ray.