Capacitor Units are Wierd

29 May, 2025

Warning: The ideas below are purely my own speculation, and are likely wrong.

Post-mortem: This is in every physics textbook.

In physics, most quantities represent more of something. More mass means greater inertia, more force means greater acceleration, velocity means more distance. But capacitance is not like that, its units counteract each other.

The equation for capacitance is

C = \frac{Q}{V}

Capacitance is measured in Farads, which are $\frac{C o l u m b s}{V o l t s}$ which in base units is $\frac{s^{2} C^{2}}{k g m^{2}}$ .

This is weird, because high voltage and high charge is rated by Farads to be of the same importance as low voltage and low charge. I will give my explanation on what is going on.

Now. Here are a few more equations related to capacitance.

This one is in terms of a capacitor's physical properties.

C = \frac{ϵ_{0} A}{w}

where $A$ is area of the plates, $w$ is the width of the gap between the plates, and epsilon is the electrons' opinion of the charges on the other side. (This is the traditional 2-plate capacitor.)

And because a capacitor is a circuit component, here are relevant circuit behavior equations:

I (t) = \frac{ε}{R} e^{- t / R C}

As a reminder on how capacitors work. Capacitors are two very large plates with a bit of space between them. The large plates allow capacitors to store large amounts of static electricity. After one plate of the capacitor is charged by placing lots of electrons on one of the plates, there will be a voltage (potential difference) between the negatively charged and positively charged plates, allowing the capacitor to act like a battery.

Since capacitors aren't batteries, their voltage begins to drop quickly as they start releasing charge. The equation above models the current leaving the capacitor. Notice that current decays over time. This happens exactly because the voltage drops as higher current quickly extracts charge from the capacitor.

One aspect of $C = \frac{Q}{V}$ now becomes evident. Since both charge $Q$ and voltage $V$ are decreasing when a capacitor is discharging, the capacitance will remain the same. The ratio of the top and bottom will be maintained even as both values decrease.

So the capacitance of a capacitor remains the same regardless of its charge. This is great, because the capacitance of a capacitor should hopefully depend only on the capacitor's physical properties. Such as plate area $A$ and distance between the plates $w$ . $C = \frac{ϵ_{0} A}{w}$

Now, let's return to $I (t) = \frac{ε}{R} e^{- t / R C}$ . One sad thing about this reality, is that current will drop the fastest at the start. $P o w e r = I^{2} R$ . For the circuit to do something useful using the capacitor, we would need a high power output for as long as possible. But physics are against us, and our current and voltage will drop quickly, before they can get done anything useful.

Turns out, the time for which the capacitor can deliver consistent current is far more important than the actual voltage numbers. So long as the capacitor works for longer, we will be able to get more useful stuff done.

Coincidentally, there is a value $τ = R C$ which represents the time it takes the capacitor to drop to $\frac{1}{e}$ (37%) of its initial current. And it is in fact in the current equation.

I (t) = \frac{ε}{R} e^{- t / R C} = \frac{ε}{R} e^{- t / τ}

As you might notice, $τ$ depends on $C$ - the capacitance. In fact, it is directly proportional. $τ \propto C$

More capacitance means more time. Farads were actually buying us time all along. Suddenly, physics quantities make sense again, and everyone can live happily ever after.

Also, capacitors don't have a voltage limit (at least in ideal theory). You can always stuff in more charges and get a higher voltage, so long as you have a strong enough battery to charge the capacitor. This is a bit counterintuitive, because you would think capacitors are about storing charge. Yet, capacitors care very little about charge, and even the smallest ones can hold however much you stuff into them.

C = \frac{Q}{V} = \frac{ϵ_{0} A}{d}

So what if we get a better capacitor which has larger plates and a smaller gap? $Q = \frac{C}{V}$ means that we will be able to store more charges in the capacitor at each voltage ( $Q \propto C$ if we keep V constant). If we have more charge, it will take the strong initial current longer to discharge them, while maintaining the voltage and current.

Capacitors with higher capacitance are better because they give more useful time. Long live capacitors!

- very qualified electrical engineer