# Why doesn't capacitance increase if voltage decreases?

I was asked to determine how to increase a parallel-plate's capacitor, and I isolated two ways:

• decreasing the distance between the plates
• decreasing the voltage

The first method is based off the formula $$C=\frac{\epsilon_{0}A}{d},$$ where $C$ is the capacitance, $A$ is the plate's area, $d$ the distance between the two. But as the same time, the formula $$C=\frac{Q}{V},$$ where $Q$ is the plates' charge in Coulombs and $V$ is the plates' voltage, also mathematically fits. My book indicates the correct answer was to decrease $d$. Thus my question is why does decreasing $V$ conceptually not decrease $C$?

• How are you going to decrease the voltage while holding the charge Q fixed?
– user93237
Jul 23 '17 at 5:51
• In Newton's second law, $F=ma$, why does the mass not decrease when you decrease the force $F$? Or in the expression for kinetic friction, $f_k=\mu n$, why does the roughnesses and thus $\mu$ not decrease when the friction decreases? Jul 23 '17 at 7:47

In most capacitors (including the simple parallel plate capacitor, which is the one you refer to), changing the applied voltage simply results in more charge being accumulated on the capacitor plates, and has no effect on the capacitance.

A capacitor is nothing more than two conductors which are separated from each other by a dielectric material of some kind. When a voltage is applied across a capacitor, a certain amount of charge builds up on the two conductors. The amount of charge which accumulates is a function of the voltage applied - $Q=Q(V)$. The capacitance is then defined by $C = \frac{Q}{V}$.

Under most conditions, the capacitance ends up being purely a function of geometry. In the case of the parallel plate capacitor, one finds that

$$Q(V) = \frac{\epsilon_0 A V}{d}$$ so $$C \equiv \frac{Q}{V} = \frac{\epsilon_0 A}{d}$$

The capacitance therefore depends on the area of the plates and the distance between them - nothing else.

It's worth noting that one could construct a device which has a more complicated, voltage-dependent capacitance. For example, one could consider a parallel plate capacitor where the plates are held apart by electrically insulated springs. In that case, the distance separating the plates would depend on the electrostatic attraction between the plates, which is determined by the applied voltage. Increasing the voltage would result in a larger charge buildup, which would cause greater attraction between the plates, which would further compress the springs, which would increase the capacitance. Working this out explicitly is an interesting exercise.

But again, for a simple case like the one under consideration in your example, the capacitance is merely a geometrical constant.

• There are devices whose capacitance depend on voltage but it is again a matter of changing the geometry (separation of "plates") of the capacitor en.m.wikipedia.org/wiki/Varicap Jul 23 '17 at 13:35