# Why exactly do capacitors charge and discharge exponentially?

Hi this is my first time posting so please feel free to correct me on any formatting errors I may make.

I understand that as a capacitor charges, the amount of electrons that are deposited on one plate increases, thereby the overall voltage across the capacitor increases. And I kind of understand that because of that, the rate at which 1 coulomb of charge flows in the circuit starts to fall because of this. But what I don't understand is why this decrease in current is exponential, or how any relationship between these variables are exponential. I don't know if I was doing this wrong but I couldn't find a true explanation with a google search, so I was hoping my question could be answered here, thanks in advance.

• Have you ever took a lesson on elementary circuit analysis? Furthermore, do you know anything about differential equations? Answers those questions in order to seek a way to answer your question. Feb 14 at 19:24
• That is not an inherent property of a capacitor. What you are describing is the transient response of a circuit in which a capacitor that initially was charged to voltage, $V_i$, suddenly is connected to a constant voltage source, $V_s$, through a resistor $R$. In other circuits, the voltage on a capacitor could be a linear function of time, or it could be a periodic function of time, (e.g., a sine wave) etc. Feb 14 at 19:25
• @SolomonSlow I think it's possible to answer OP's question in the spirit of it, which appears to be about simple decay of a capacitor's charge over something passive like a simple resistor.
– Gert
Feb 14 at 19:39
• @Gert, sure enough, but I'm too lazy to show how the differential equation describing that circuit can be derived from the law that describes a capacitor, from Ohm's Law (describes the resistor), and from Kirchoff's Laws. Never the less, I thought that the OP should know that it's not just the capacitor that is responsible for the behavior that they described. Feb 14 at 19:45

As a capacitor charges, electrons are pulled from the positive plate and pushed onto the negative plate by the battery that is doing the charging. Looking just at the negative plate, note that electrons repel each other, so they will spread out evenly on the negative plate as they accumulate. Since electrons repel each other, as more electrons accumulate on the negative plate, it becomes increasingly more difficult to add the next electron to that plate because the repulsive force on the next electron that you are trying to add to the negative plate is directly proportional to the number of electrons that are already on that plate. The differential equation that describes this effect is formulated such that the rate of charging is negatively affected by the charge that is already on the negative plate, and is proportional to the charge that is already on that plate. The solution of the differential equation results in an exponential mathematical form with a negative exponent.

A similar argument can be made for discharging the negative plate, as the repulsive force on an electron leaving the negative plate is proportional to the charge on that plate. Likewise, a similar argument can be made for the positive plate regarding how easy it is to either remove or add electrons to that plate as the capacitor is charging or discharging.

Note that there are many instances in nature of a rate depending on how much of some substance or energy already exists (e.g., Newton's Law of cooling), and for that reason, the exponential mathematical form noted above occurs frequently.

• OK, so THIS is why we developed calculus: to avoid long tracts explaining simple things! :-)
– Gert
Feb 14 at 19:48
• @Gert, you make an excellent point, but I interpreted the OP's question as meaning that he was looking for the concept behind the math. Feb 15 at 1:23
• I really was just joking, Sort of. DW. Good answer! +1 from me.
– Gert
Feb 15 at 3:59

When a capacitor discharges through a simple resistor, the current is proportional to the voltage (Ohm's law).

That current means a decreasing charge in the capacitor, so a decreasing voltage.

Which makes that the current is smaller.

One could write this up as a differential equation, but that is calculus.

One can also reason that when half of the charge is gone after a certain time $$\tau_{1/2}$$, the current is half as large, so it again it will take the same $$\tau_{1/2}$$ to halve again.

This is exponential decay, in the same way as radioactive decay.

(But it is not true when you connect an LED to the capacitor.)