# Some questions about capacitor discharging

Milotic
1. why does the voltage of the capacitor eventually go to 0 when discharging the capacitor? I heard that's because "current starts flowing when discharging", but how exactly does that lead to V going down? I know that I = C * dV/dt, but that doesn't seem to help me understand why V goes down (which is equivalent to dV/dt < 0).
2. Alternatively, I heard this is due to Q = VC; discharging capacitor means reducing the charge on capacitor, so V must decrease as well. Okay, in that case, where does the charge go?
3. should the capacitor be connected to the ground node in both directions if it's to be discharged?I saw a diagram/circuit with a short circuit and pre-charged capacitor only (), which would apparently lead to V decreasing, but the circuit didn't have any ground nodes labelled. Is it possible to discharge a capacitor without having a ground node?

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2022 Award
why does the voltage of the capacitor eventually go to 0 when discharging the capacitor? I heard that's because "current starts flowing when discharging", but how exactly does that lead to V going down? I know that I = C * dV/dt, but that doesn't seem to help me understand why V goes down (which is equivalent to dV/dt < 0).

Because there are opposite charges on each side of the capacitor, and discharging allows the charges to flow and cancel each other out.

Alternatively, I heard this is due to Q = VC; discharging capacitor means reducing the charge on capacitor, so V must decrease as well. Okay, in that case, where does the charge go?

Electrons flow from one plate of the capacitor to the other plate. Specifically they flow from the negatively charged plate to the positively charged plate and the charges cancel out, making the voltage zero.

should the capacitor be connected to the ground node in both directions if it's to be discharged?

No, as there are equal but opposite charges on each side of the capacitor. It is sufficient to let them neutralize each other.

Note that the circuit as a whole isn't charged. So there's no need to connect it to ground to get rid of excess charge. We're only moving charges around within the circuit (specifically the capacitor).

• etotheipi and vanhees71
Gold Member
2022 Award
You can make this excellent answer also quantitative with what you already wrote. The wires of your circuit have some resistance ##R##, and that implies from Kirchhoff's rule that
$$U(t)=R I(t).$$
On the other hand you know that ##I=-\mathrm{d} Q/\mathrm{d} t=-C \mathrm{d} U/\mathrm{d} t## (where ##Q## is the charge on the positively charged plate; think about why there's a - sign!). So you get a differerntial equation for the voltage:
$$R C \dot{U}(t)=-U(t)$$
or
$$\frac{\dot{U}}{U}=-\frac{1}{RC}.$$
Integrating this from ##t'=0## to ##t'=t## you get
$$\ln \left (\frac{U(t)}{U_0} \right)=-\frac{t}{R C} \; \Rightarrow \; U(t)=U_0 \exp \left (-\frac{t}{R C} \right).$$
This means the voltage decreases exponentially to 0. The current indeed is
$$I(t)=-C \dot{U}(t)=\frac{U_0}{R} \exp \left (-\frac{t}{R C} \right)=\frac{U(t)}{R}.$$

Milotic
@ Drakkith thank you for your explanations! I was wondering if you could still explain
(1) if or how the equation I = C * dV/dt relates to discharging
and
(2) you said that "electrons flow from the negatively charged plate to the positively charged plate and the charges cancel out". Since no current is allowed to flow in between the capacitor plates, would that mean the electrons go the "long way" around the rest of the circuit to balance out the charges? And is the reason why the electrons start flowing again (in the absence of voltage source) because the electrons are attracted to the positive charge on the + plate / repelled by the many other electrons on the - plate?

Milotic
@vanhees71 I'm following through your explanations right now—just to confirm, are all those "U(t)"'s referring to voltages?

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2022 Award
Yes, ##U(t)## is the voltage across the capacitor. To understand the relation with the current, just remember that the current is the charge ##\mathrm{d} Q## flowing through the wire from the positively charged capacitor plate in a time interval ##\mathrm{d} t## devided by this time interval. There's a negative sign, because the decrease of the charge on the positively charged plate, i.e., ##\dot{Q}<0## corresponds to a positive current, i.e., you have ##I(t)=-\dot{Q}(t)=-C \dot{U}(t)## (where finally I used ##Q(t)=CU(t)##).

Gold Member
why does the voltage of the capacitor eventually go to 0
It can be a big help if, when you want a deeper understanding of circuit 'situations', you use the term "Potential Difference" instead of Voltage. That will constantly remind you that there are two terminals connected to your mental Voltmeter. Also PD is only two characters!

Staff Emeritus
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