# Capacitance of bodies with different charge [duplicate]

How does one calculate the capacitance of two bodies with different charges? I was looking at coefficients of potential, but they don't seem helpful.

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## marked as duplicate by Emilio Pisanty, Qmechanic♦Mar 5 '14 at 15:13

This question was marked as an exact duplicate of an existing question.

possible duplicate of Capacitor with different charges on each plate – Emilio Pisanty Mar 5 '14 at 14:49

How does one calculate the capacitance of two bodies with different charges?

To be clear, capacitance doesn't depend on the amount of charge; the capacitance is determined by the geometry of the bodies.

If you have two conductors, there are actually three capacitances to consider, the self-capacitance of each and mutual capacitance of the two conductors.

In electrical circuits, the term capacitance is usually a shorthand for the mutual capacitance between two adjacent conductors, such as the two plates of a capacitor. However, for an isolated conductor there also exists a property called self-capacitance, which is the amount of electrical charge that must be added to an isolated conductor to raise its electrical potential by one unit (i.e. one volt, in most measurement systems).[20] The reference point for this potential is a theoretical hollow conducting sphere, of infinite radius, centered on the conductor.

Let $Q_1$ be the charge on conductor 1 and $Q_2$ the charge on conductor 2.

Further let $C_1$ be the self-capacitance of conductor 1, $C_2$ the self-capacitance of conductor 2, and $C_{12}$ the mutual capacitance.

We can then write:

$$Q_1 = C_1 V_1 + C_{12}(V_1 - V_2)$$

$$Q_2 = C_2 V_2 + C_{12}(V_2 - V_1)$$

For an intentional capacitor, the self-capacitance of the conductors is insignificant, i.e., the mutual capacitance dwarfs the self-capacitance of the conductors and we speak of the capacitance of the capacitor which is understood to be the mutual capacitance.

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Is there a proof of correctness of coefficients of potential somewhere? Also, why did you take the difference of the potentials? – user24082 Mar 5 '14 at 17:43
@Anthony, because the voltage between (or across) the two conductors is precisely the potential difference: $V_{12} = V_1 - V_2$. – Alfred Centauri Mar 5 '14 at 17:54