# Does mutual capacitance increase for two positively charged plates when brought close?

My understanding is that any two charge carriers will exhibit what is known as mutual capacitance. That is, there is an amount of charge that can be stored between any two objects, which is called mutual capacitance (typically just called capacitance).

As answered here, it seems very clear to me that this capacitance will increase when two objects of opposite charge come closer. (You generate a stronger electric field which draws more holes to the positively charged object and more electrons to the negatively charged object).

As I see it, mutual capacitance exists between any two positively (or negatively) charged objects. According to the Wikipedia stub for Mutual Capactiance, no distinction is made about the charges, and it plainly says:

All objects in the universe, conducting or non-conducting, that hold charge with respect to another exhibit capacitance. An object's capacitance increases when another object is brought closer to it.

The following diagram demonstrates what my question is about:

Since the right side has a larger (magnitude) charge than the left side, I'd expect the net field to go from right to left. (If I threw a proton in the field, it would repel more from the right and end up on or close to the left).

But as I bring these two objects closer together I would expect that the capacitance would decrease because the right side would attract electrons from the left side, forcing out the holes in the left to some other part of the system.

Conversely, if the objects were moved further apart, I'd expect the field to diminish to the point where both objects have no effect on each other. This would cause the capacitance to increase to its maximum.

So, what is true? Does the mutual capacitance increase or decrease for these two objects as they come closer? Is it even possible for these two objects to have a capacitance if they have like charges (both positive in this case)?

I may need to open up another question for this, but I'm trying to see how mutual capacitance is used for proximity/touch sensing. As it seems right now, I can't tell how this mutual capacitance can be monitored if only one object is connected to the sensor system, while the target (a human hand for example) is not connected.

Mutual capacitance is independent of the actual charges that may exist. It is only a function of the geometry of the capacitor and the dielectric constants of the materials involved. In general, for a parallel plate capacitor, the capacitance is inversely proportional to the distance between the plates. So therefore if the plates are brought closer, the capacitance will increase. However, the value of this capacitance depends only on the size of the plates and the dielectric properties of the material between the plates. The value of capacitance is not affected by whatever charges may be present.

• Hm... but unit capacitance is the amount of charge (be it Coloumbs or electrons) that can be stored between two points with 1 volt difference (i.e. C = Q/V). Surely the charge affects the capacitance between two objects? If V is held constant, and Q increases, C must increase as well. Or if Q is held constant, and V decreases, then C must increase. The latter can be understood as lowering the work done needed to inject charges into the capacitor, hence it can store more. Commented Apr 15, 2016 at 13:26

Q = CV.

If the two objects were dragged away from each other, capacitance decreases (as per the definition of what physically capacitance is) but voltage increases thus Q remains constant.

But as I bring these two objects closer together I would expect that the capacitance would decrease because the right side would attract electrons from the left side

How are these electrons to cross the barrier of the insulation?

• I've just read on the Wiki that Q=CV (or C=Q/V) does not apply when there is a non-zero net charge or when there are more than 2 objects. So is C=(ε0×εr×S) / d the true definition of capacitance? Commented Apr 15, 2016 at 14:09
• Maybe you have misinterpreted wiki so please provide a link and where I can find it. The picture is capacitance between two plates and is a true definition. OK I found it - you have to fiddle factor things but the formula still applies after the fiddling due the complexity of 2+ plates. Commented Apr 15, 2016 at 14:11
• Also, the electrons never cross the barrier. They would be attracted to the end of the left plate (next to the dielectric). The Capacitance section in the Wiki describes qualitatively what happens when to oppositely charged objects are put near each other. It would appear in my scenario that more work is needed to add charge because two positively charged objects brought together has a higher voltage across the dielectric. If the voltage increase, C=Q/V indicates capacitance decreases. Commented Apr 15, 2016 at 15:21
• Correct me if I'm wrong, but this means that changing the distance doesn't always increase capacitance. It depends on the initial charge on the objects when the two objects are moved closer. Perhaps in most scenarios it indeed increases capacitance (because of oppositely charged objects), but I do not think this is always the case. Another example would be to look at the effects of two neutrally charged objects, which would result in no change of capacitance when brought together. Commented Apr 15, 2016 at 15:23
• Take this fact onboard..... Capacitance is independent of charge or applied voltage just as inductance is nothing todo with current flow despite the definition of inductance being total flux produced per amp. Take on board. Commented Apr 15, 2016 at 20:21

First let me try to interpret what you are saying: if there are two objects that are very far apart, therefore the field between them are very weak for a given net charge. If an extra charge is brought to one of the object, since the field is so weak, the capacitance must change by a lot.

But it does not work this way.

Forget about capacitance for a moment. When an extra charge is brought to one of the two objects, the field has changed. And the potential between the two objects has changed. If I measure the potentials or integrate using Coulomb's law to get the potentials and then do a little calculations: $$\frac{initial.net.difference.of.charge}{initial.potential} = C_1$$ With final numbers: $$\frac{slightly.larger.net.difference.of.charge}{slightly.larger.potential} = C_2$$ It turns out $C_1 = C_2$ and it is conveniently called Capacitance.

You need to take $C = \frac{Q}{V}$ similar to $R = \frac{V}{I}$. If I don't apply a voltage to a resistor, does it mean the resistor has no resistance?