# Understanding Capacitor Equations and Time Constants

• Gourab_chill
In summary: So I'm thinking of the "ground" of the circuit as being the reference point for charge here. So if you add or subtract charge at "ground" it means adding or subtracting charge from the entire circuit.
Gourab_chill
Homework Statement
I attached the question in the attachments.
Relevant Equations
Q=CV; Kirchoff's law of voltage
Well i don't you to solve the question for me but I want you to clarify the concepts pertaining to this question. My question is how do I write a equation for the circuit since the there is same charge on one of the capacitors. While writing the equation should i put the voltage across the charged capacitor as (q+q1)/C2 or (q-q1)/c2? Or what should i do? What is the concept? if there is a way by which the time constant can be calculated and be used to our advantage in computing the charge on the capacitors?

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This question is ill-formulated in my opinion. Specifically, I don't know how to process the statement "C1 is initially uncharged and charge on both plates of C2 is q1." The conductor connecting the two capacitors is an equipotential. One cannot have q1 on the left plate of C2 and zero charge on the right plate of C1. Also this statement seems to assert that each plate of C2 carries charge of the same magnitude and sign.

Maybe the author's intent is that capacitor C2 is separately charged by a battery to charge q1 and then connected to capacitor C1 while the switch open as all this is going on. That's far removed from the posted description.

DaveE and etotheipi
First of all, I'm sorry you have to deal with a question like this; frankly it's crap, and it either means your instructor doesn't really understand the material he's teaching or he isn't paying attention. Matbe he just wants you to solve problems presented in the most confusing way possible? You have my sympathy. Not that that helps, I know.

As an aside I am struck with how often the questions posted by students are the result of poorly constructed homework questions. It simultaneously makes me optimistic about the students and depressed about the quality of some instructors out there.

OK, I'm not sure I can help with the answer, I haven't thought that hard about it. But I can give you some of my impressions:

1) The charge on both sides of C2 is q1. This means that there is no voltage or energy in/across C1.

2) Because C1 is connected to C2 the charge on the right side of C1 MUST be q1 also. According to the schematic they are they same thing. The right side of C1 IS the left side of C2.

3) Because they said "C1 is initially uncharged", then there must be no charge difference across C1. So the charge on the left side if C1 is also q1.

4) OK, now let's recall some basic capacitor stuff: Δq=CV, so if Δq=0 => V=0 across each capacitor. Since you have 2 capacitors each with 0 voltage across them, you can combine them into 1 equivalent capacitor, let's call it C3=C1*C2/(C1+C2)=C/2. C3 also has 0 volts across it. This last step isn't necessary, I would just rather deal with one capacitor than 2.

5) So, to summarize so far: we have a circuit with C3 (=C/2), E, and R all in series with an open switch and there is no initial voltage across C3 (or C1 and C2, if you prefer).

7) When the switch is closed you will have current flowing, limited by R, to charge the capacitor(s) to a total voltage of E eventually. I'm sure you've seen this circuit before, the voltage across the capacitors will be V(t)=E*(1-exp(-t/RC3))=E*(1-exp(-2t/RC)).

At this point, I'm going to stop because I'm losing patience/interest in figuring out what this guy means. I don't really understand about "matching columns" or how the initial charge on all of the capacitor plates q1=CE matters. You can probably figure something out, like doing a crossword puzzle; there are probably only a few (one, hopefully?) ways this question can make sense.

DaveE said:
3) Because they said "C1 is initially uncharged", then there must be no charge difference across C1. So the charge on the left side if C1 is also q1.

4) OK, now let's recall some basic capacitor stuff: Δq=CV, so if Δq=0 => V=0 across each capacitor. Since you have 2 capacitors each with 0 voltage across them, you can combine them into 1 equivalent capacitor, let's call it C3=C1*C2/(C1+C2)=C/2. C3 also has 0 volts across it. This last step isn't necessary, I would just rather deal with one capacitor than 2.

Totally agree the question is senseless. I'm not sure the formula Δq=CV is correct though, since in the normal case that would imply something like 2Q=CV. I've normally seen it written Q=CV with Q as the magnitude of the transferred charge.

In this case though equal charges on each plate, although strange, would imply zero voltage. Though I don't think the standard capacitor equation makes sense in scenarios where the plates are not oppositely charged.

Last edited by a moderator:
One might argue that no matter how the iniital charge is distributed, once the switch is closed and one waits long enough relative to RC, the series combination will be fully charged and the initial charge q1 is irrelevant. However, the question is asking about some intermediate time t in which case it is not clear that the capacitor combination is fully charged. Perhaps we have to assume that this is the case because none of the answers shows explicit time dependence.

Since the initial charge q1 is the same at all capacitor plates, it should be irrelevant. What matters is charge difference. According to me, you can view q1 as a static charge applied to the entire circuit in reference to some mystery definition of what/where zero charge is. Concepts like this are often glossed over. In a similar way it really doesn't make sense to talk about the voltage on the positive side of the battery without some zero voltage reference. It's the voltage across the battery that matters. Same with charge.

Edit: This is why I always like to define a "ground" point in my schematics, even if there's only one. It represents a definition of where voltage and charge are zero, for reference.

etotheipi and Merlin3189
DaveE said:
Since the initial charge q1 is the same at all capacitor plates, it should be irrelevant. What matters is charge difference. According to me, you can view q1 as a static charge applied to the entire circuit in reference to some mystery definition of what/where zero charge is. Concepts like this are often glossed over. In a similar way it really doesn't make sense to talk about the voltage on the positive side of the battery without some zero voltage reference. It's the voltage across the battery that matters. Same with charge.

Sure, but ##C = \frac{Q}{V}## is the ratio of the charge transferred to the voltage that results across the capacitor. If a capacitor has charges, relative to a suitable reference, of ##+3C## and ##-3C## then it is still ##V = \frac{3}{C}## and not ##V = \frac{3-(-3)}{C}##. Just resolving any possible ambiguity

etotheipi said:
Sure, but ##C = \frac{Q}{V}## is the ratio of the charge transferred to the voltage that results across the capacitor. If a capacitor has charges, relative to a suitable reference, of ##+3C## and ##-3C## then it is still ##V = \frac{3}{C}## and not ##V = \frac{3-(-3)}{C}##. Just resolving any possible ambiguity
Yes, because when you start talking about the charge transferred, or the charge across the capacitor, you have redefined what/where zero charge is (either before the transfer = 0, or the other side of the cap = 0).

Remember, voltmeters and coulometers, have 2 probes. It can be confusing figuring out from text what people mean sometimes, but there is always a context for where each probe goes. Whenever someone refers to the voltage (or charge) "at" some point, they are assuming you agree about some zero reference. For time based differences measurements, there is always a before and after.

etotheipi
DaveE said:
Yes, because when you start talking about the charge transferred, or the charge across the capacitor, you have redefined what/where zero charge is (either before the transfer = 0, or the other side of the cap = 0).

Remember, voltmeters and coulometers, have 2 probes. It can be confusing figuring out from text what people mean sometimes, but there is always a context for where each probe goes. Whenever someone refers to the voltage (or charge) "at" some point, they are assuming you agree about some zero reference. For time based differences measurements, there is always a before and after.

Yes the potential at a chosen point in the circuit must be quoted w.r.t. some chosen zero of potential (this difference is the voltage you speak of).

As a question (I don't know the answer ), do we usually consider complete circuits to be electrically neutral? I.e. we could say that the entire circuit has a charge of +7C, so that if ##Q = 1C## then the charges on either plate of the capacitor are ##8C## and ##6C##, however in practice mightn't we just let the total charge inside a surface enclosing the circuit be zero?

etotheipi said:
Yes the potential at a chosen point in the circuit must be quoted w.r.t. some chosen zero of potential (this difference is the voltage you speak of).

As a question (I don't know the answer ), do we usually consider complete circuits to be electrically neutral? I.e. we could say that the entire circuit has a charge of +7C, so that if ##Q = 1C## then the charges on either plate of the capacitor are ##8C## and ##6C##, however in practice mightn't we just let the total charge inside a surface enclosing the circuit be zero?
It doesn't matter to the circuit if you add charge, or voltage everywhere. You can define the zero where ever you like. You might, however, confuse people, since we will all pick a convenient location within the circuit to call zero.

etotheipi
I didn't quite get much of the question as well; so can we say this question is absolutely wrong?
However the solution of this question is interesting, just putting it up for you guys to see, i literally can't find any logic how they derived it.

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I don't understand this. The solution presents answers to 3 questions all of which are some kind of charge. The statement of the problem does not list 3 questions but asks us to "appropriately (match) the information given in the three columns of the following table."

I wouldn't say that this question is absolutely wrong, but that it is not at all formulated for people who cannot read minds. I am sure it makes sense to the person who wrote it and put out a solution. However, even with the solution in front of me, I cannot figure out what that sense is.

I have spent enough time trying to puzzle this out. I'm outta here.

Delta2

## 1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which stores energy until the capacitor is discharged.

## 2. What are the different types of capacitors?

There are many different types of capacitors, including ceramic, electrolytic, film, and variable capacitors. Each type has its own unique properties and is suitable for different applications. For example, ceramic capacitors are commonly used in high-frequency circuits, while electrolytic capacitors are often used in power supply circuits.

## 3. How do you calculate the capacitance of a capacitor?

The capacitance of a capacitor is determined by the surface area of the plates, the distance between the plates, and the type of dielectric material used. It can be calculated using the formula C = εA/d, where C is capacitance, ε is the permittivity of the dielectric material, A is the surface area of the plates, and d is the distance between the plates.

## 4. Can a capacitor store an unlimited amount of charge?

No, a capacitor has a maximum charge it can hold, known as its capacitance. Once the capacitor reaches its maximum charge, it cannot store any more energy. Attempting to overcharge a capacitor can cause it to fail or even explode.

## 5. How do you discharge a capacitor?

To discharge a capacitor, you can either short-circuit the two plates together or connect a resistor to the plates. Short-circuiting is the quickest method, but it can be dangerous if the capacitor is holding a large charge. Using a resistor allows the capacitor to discharge slowly and safely.

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