# Capacitor asymmetrically charged

Can a capacitor be asymmetrically charged? I mean, not to have equal and opposite charges on each plate, -Q and +Q, but two different charges of any sign, say, q1 and q2. I ask firstly if it's possible, because maybe it can be done by putting each plate in contact with different sources, but upon disconnection, charges maybe find an equilibrium by polarization of one end. Secondly, if it's possible, must I rewrite the famous capacitance definition $$C=\frac{Q}{V}$$ as $$C=\frac{q_1-q_2}{2V}$$?

• Note that such an 'asymmetrically charged' capacitor would not be electrically neutral which implies that there is more than one capacitance to consider. – Alfred Centauri Oct 13 '18 at 21:11
• This question tickled a memory of an answer I wrote some time ago here. Your question is related if not a duplicate of Charge On parallel plate capacitor? – Alfred Centauri Oct 13 '18 at 21:24

Can a capacitor be asymmetrically charged?

Sure. For instance, you can apply $$10$$V to one of the capacitor plates, while the other plate is floating. As a result, one plate will receive some charge, while the other will won't.

Secondly, if it's possible, must I rewrite the famous capacitance definition $$C=\frac Q V$$ as $$C=\frac {q1−q2} {2V}$$?

We can look at a capacitor from two different perspectives: first, as as a conventional two plate capacitor, which is charged by moving some charge from one plate to the other, second, as two pieces of metal held together, where each piece (plate) has some self-capacitance, which could be charged independently of the other.

Both capacitors are defined by the same formula, $$C=\frac Q V$$, but the meaning of $$Q$$ and $$V$$ is different. For a conventional two plate capacitor, $$Q$$ is the charge moved from one plate to the other and $$V$$ is the voltage between the plates. For the self-capacitance of a plate, charge $$Q$$ is delivered from somewhere else and voltage $$V$$ is measured relative to some reference point, like infinity or ground.

The self-capacitance of a plate could be many orders of magnitude smaller than the capacitance of the two plate capacitor. As a result, the charge required to raise the voltage of a plate, say, relative to ground is many orders of magnitude smaller than the charge required to charge the two plate capacitor to the same voltage. For instance, the charge required to raise the voltage of a plate of a $$1\mu F$$ capacitor by $$10V$$ relative to ground, would change the voltage between the plates by a few microvolts, which would hardly be detectable. This is because the self-capacitance of a plate could be a fraction of a picofarad.

Based on the above, we can also conclude that, in order to put any consequential charge on one of the plates of that capacitor, the applied voltage should be on the order of tens or hundreds of kilovolts.

In summary, we are dealing here with two very different types of capacitance and we should treat them separately, using the same universal definition of capacitance for both with different meaning of charges and voltages.

• Excellent answer. One thing that I would add is that one of the assumptions of circuit theory is that there is no net charge on any component. So to properly represent this scenario in circuit theory you would need to do what you describe here as far as treating it as a pair of self capacitors to ground. – Dale Oct 16 '18 at 1:36

It's possible to have asymmetric charges on a capacitor but there will tend to be a huge energy cost associated with an asymmetrically charged capacitor with any substantial charge and it won't want to stay asymmetrically charged. Since the capacitor as a whole will have a net static charge, I think that - depending on the surrounding environment - either the charged plate will tend to lose its charge or the opposite plate will tend to gain an opposite charge.

What makes a capacitor so easy to charge without a huge energy cost in normal operation is that the two plates are oppositely charged so that the capacitor remains electrically neutral.