Potential difference between 2 points in a capacitor circuit

In summary: If you use the charge continuity equation, then the equation becomes ##\dfrac {q_2}{C} = q_1 - \mathcal{E}##.
  • #1
vcsharp2003
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Homework Statement
What is the potential difference between points A and B in the circuit below.
Relevant Equations
##C= \dfrac {q}{V}##
Sum of potential rises = Sum of potential drops in a circuit loop
CamScanner 05-07-2023 19.14_4.jpg

In the given circuit, a transient current will flow and when this current finally stops at equilibrium, the charges ##q_1## and ##q_2## are assumed to deposit at the capacitor plates as shown below. The dashed line indicates an isolated system that will have it's total charge conserved.

CamScanner 05-07-2023 19.14_6.jpg


If I assume ##V_1## and ##V_2## as the potential differences across the 4 ##\mu F##and 2 ##\mu F## at equilibrium state, then the following equations are true.

##2 \times 10^{-6}V_2 = q_2##
##4 \times 10^{-6}V_1 = q_1##

I need two more equations, since there are 4 unknowns. One of the equations can be obtained by applying law of conservation of charge to the isolated system within the dashed curve, for which we can say total charge at start = total charge when circuit reaches equilibrium i.e. ##0 = q_1 + q_2##; but in the isolated system there is the 24 V battery whose terminals will also have charges when equilibrium is reached. My question is whether I can assume the battery terminals to have equal and opposite charges when not delivering current?
 
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  • #2
vcsharp2003 said:
My question is whether I can assume the battery terminals to have equal and opposite charges when not delivering current?
I don't understand what you mean by this. You can assume that the potential difference across each battery is fixed at ##\mathcal{E}_1=##12 V and ##\mathcal{E}_2=2\mathcal{E}_1=##24 V. Then start at point A and go to B adding potential drops/rises clockwise and counterclockwise. That should give you two equations. The charge conservation equation is a third equation. You don't need a fourth. Write something down and see what you get. I would do everything in symbolic form before substituting numbers.
 
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  • #3
kuruman said:
I don't understand what you mean by this. You can assume that the potential difference across each battery is fixed at ##\mathcal{E}_1=##12 V and ##\mathcal{E}_2=2\mathcal{E}_1=##24 V. Then start at point A and go to B adding potential drops/rises clockwise and counterclockwise. That should give you two equations. The charge conservation equation is a third equation. You don't need a fourth. Write something down and see what you get. I would do everything in symbolic form before substituting numbers.
I see. When applying conservation of charge to the isolated system within the dashed line, we will get (charge on lower plate of left capacitor + charge on lower plate of right capacitor + charge on postive terminal of 24 V battery + charge on negative terminal of 24 V battery) = 0.

Equation when going anticlockwise from A to B is ## V_b-V_a = V_2## and when going clockwise ##V_b-V_a = 12 +V_1 -24##. These two equations can be combined into one equation ## V_2 = 12 +V_1 -24##.

We have the following unknowns: ##q_1##, ##q_2##, ##V_1## and ##V_2##. Since, there are 4 unknowns, shouldn't we need a total of 4 equations?
 
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  • #4
There must also be some initial conditions of Vc(t)=0 at t=0 included. As mentioned, there is a Kirchchoff Voltage Law KVL where the loop voltage adds up to 0 in either direction. There must also be some current equations to get there based on the rate of change in voltage to get to steady-state. That's all you need to simulate it.

Here's an interactive simulation after you try the math to check your answer. Click on "switch" and hold ( it's in a slow-motion sampling-rate) and click [ reset ] to try again. You can also plot current, if you want, as I had to add some resistance {1m} to prevent infinite current which aborts the simulation and smokes a few electrons. This is a must in all simulations with ideal caps and batteries have zero resistance. (Real ones do not) https://tinyurl.com/2eslbk3h

The simpler answer is if charge is shared in the entire loop then the change in total charge must be 0 in a lossless loop.
 
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  • #5
And what are ##V_1## and ##V_2## in terms of ##q_1##, ##C_1=2C=~##4 μF, ##q_2## and ##C_2=C=~##2 μF? Please use only symbols for your answers. No numbers.
 
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  • #6
kuruman said:
And what are ##V_1## and ##V_2## in terms of ##q_1##, ##C_1=2C=~##4 μF, ##q_2## and ##C_2=C=~##2 μF?
##V_1= \dfrac {q_1} {2C}##
##V_2=\dfrac {q_2} {C}##
 
  • #7
So what happens when you put all this in your equations in post #3?
 
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  • #8
kuruman said:
So what happens when you put all this in your equations in post #3?
##\dfrac {q_2}{C} = \dfrac {q_1}{2C} -12##

Another equation relating the charges ##q_1## and ##q_2## is needed using law of conservation of charges applied to the isolated system.
 
  • #9
No numbers please. I assume you meant to say
##\dfrac {q_2}{C} = \dfrac {q_1}{2C} -\mathcal{E}##.

OK now remember that you are looking for ##V_b-V_a=\dfrac{q_2}{C}##. What happens if at this point you use the charge continuity equation?
 
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  • #10
kuruman said:
No numbers please. I assume you meant to say
##\dfrac {q_2}{C} = \dfrac {q_1}{2C} -\mathcal{E}##.

OK now remember that you are looking for ##V_b-V_a=\dfrac{q_2}{C}##. What happens if at this point you use the charge continuity equation?
What is charge continuity equation? We haven't been taught that.
 
  • #11
The conservation of charge law is a fundamental principle in physics that states that the total electric charge in a closed system remains constant over time. Thus, if ##C_1V_1+C_2V_2=0## before the loop is closed, it must be the same after and assumes no other losses of charge in the loop. Yet clearly that 12V difference must be shared somehow. Now state what it is.
 
  • #12
vcsharp2003 said:
What is charge continuity equation? We haven't been taught that.
Same as charge conservation. Sorry if I confused you.
vcsharp2003 said:
One of the equations can be obtained by applying law of conservation of charge to the isolated system within the dashed curve, for which we can say total charge at start = total charge when circuit reaches equilibrium i.e. ##0=q_1+q_2##;
 
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  • #13
kuruman said:
Same as charge conservation. Sorry if I confused you.
Shouldn't we also consider the charge on +ve and -ve terminals of the battery when applying charge conservation to the isolated system I showed in OP?
 
  • #14
Yes & No. Let's assume the batteries are both ideal and large so what is the C value you would like to assume for each battery to see if it is relevant when the charge Q circulates until equilibrium to see if their voltage changes using the same law?. For example an 18650 Li-Ion cell is about 10k Farads. Intuitively, consider the 24V battery "charge" ends up in the 12V pack while charging up C1 & C2 which still have a net charge of 0.
vcsharp2003 said:
Shouldn't we also consider the charge on +ve and -ve terminals of the battery when applying charge conservation to the isolated system I showed iny OP?
 
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  • #15
vcsharp2003 said:
Shouldn't we also consider the charge on +ve and -ve terminals of the battery when applying charge conservation to the isolated system I showed in OP?
My answer to that is "No", not for this introductory physics problem.
 
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  • #16
TonyStewart said:
Yes & No
No & No. When considered as an ideal circuit element, No. Anything else far exceeds the level of lumped circuit analysis.
Capacitance of a Lithium Cell?????
 
  • #17
kuruman said:
I don't understand what you mean by this.
In effect, @vcsharp2003 is asking whether the batteries might have initial net charges. If they were to have such, that would invalidate your solution in post #2.
So the short answer is that you can assume they do not.
 
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  • #18
haruspex said:
In effect, @vcsharp2003 is asking whether the batteries might have initial net charges. If they were to have such, that would invalidate your solution in post #2.
So the short answer is that you can assume they do not.
Batteries as we well know have massive capacitance to hold charge , much larger than any Supercap, but also more resistance.
the reason I said “yes and no” which I believe is important to teach is that we assume they have infinite charge from infinite C and thus cannot change in voltage from a charge flowing in this loop . Quickly you will learn dV/dt = Ic/C which we have said is 0 from infinite C & NOT zero charge. Although the net change in charge beteen the 2 batteries we assume is zero because the voltage is assumed to not change.

Then perhaps decades later or when you need to know all battery mAh or watt-hour capacities can be converted to C between initial and final voltage. sorry for too much info, but it’s important to make correct assumptions.
 
  • #19
What is "C" in "infinite C"?
 
  • #20
TonyStewart said:
Then perhaps decades later or when you need to know all battery mAh or watt-hour capacities can be converted to C between initial and final voltage. sorry for too much info, but it’s important to make correct assumption
It is important to teach capacitance using a reasonably conventional definition. Treating a battery as an infinite capacitance is not that, and you may define it as you wish but only for your personal use.
 
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  • #21
haruspex said:
In effect, @vcsharp2003 is asking whether the batteries might have initial net charges. If they were to have such, that would invalidate your solution in post #2.
So the short answer is that you can assume they do not.
But, if the charges on terminals of battery are equal and opposite then that would effectively mean as if the battery terminals carried no charge when the battery stopped delivering current. So, how would the fact that the terminals carry charges invalidate the way conservation of charge is applied?
 
  • #22
kuruman said:
My answer to that is "No", not for this introductory physics problem.
Is it because then the battery would become a capacitor when not delivering current?
 
  • #23
It is because the default state for these ideal circuit problems is that all elements are neutrally charged when the circuit is put together. Of course, if there is an explicit statement that an element, e.g. a capacitor, is initially charged then you have to go by that. Here there is no such statement.
 
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  • #24
vcsharp2003 said:
that would effectively mean as if the battery terminals carried no charge when the battery stopped delivering current.
Not at all. I didn't write that the terminals were uncharged; I wrote that the battery has no net charge.
 
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  • #25
vcsharp2003 said:
Is it because then the battery would become a capacitor when not delivering current?
Just to reiterate
Any isolated chunk of conductor (a car, a human, or a battery) can carry some net excess charge that has been borrowed from the rest of the universe. In that sense everything has a self-capacitence. This is not what one is usually interested in for a circuit and that charge is tacitly assumed zero.
The capacitance that usually matters is mutual capacitance. The capacitance in a capacitor (an object built to have high capacity) is the mutual capacitance between the plates $$C=\frac Q V $$ where Q is the separated charge and V is the potential difference for the plates.
Self-Capacitance can be thought of as the mutual capacitance of an isolated object with a conducting sphere at infinity (representing the rest of the universe)
 
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  • #26
haruspex said:
Not at all. I didn't write that the terminals were uncharged; I wrote that the battery has no net charge.
So, it should be logical to say that when circuit reaches equilibrium i.e. electric current flow has stopped then we could say that the battery's positive terminal carries a charge ##+q_b## and it's negative terminal a charge of ##-q_b##.
 
  • #27
A battery is an ideal circuit element. Its internal charge state does not matter and is very complicated. All one needs to know is that an ideal battery supplies a fixed potential difference and otherwise behaves as a conductor. Why do you think you need to know its "terminal charge", whatever you think that is?
As long as a battery is behaving as an ideal circuit element, it has zero resistance and supplies a fixed voltage. It will supply whatever current (charge) demanded by the rest of the circuit. It is not a capacitor and should not be treated as one.
 
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  • #28
hutchphd said:
It is important to teach capacitance using a reasonably conventional definition. Treating a battery as an infinite capacitance is not that, and you may define it as you wish but only for your personal use.
Two corrections needed:
Please explain how a battery can have no charge yet be able to transfer charges and flow current in either direction and also have an initial voltage. Your personal choice is not only irrational, illogical , but more importantly defines the laws of physics and is not how they are modelled in simulators. In reality, batteries are chemical capacitors.

Also Initial conditions MUST be known as these are the same Conservation of charge conditions that the laws of physics are being applied here. The law states that the total system charge must not change and the initial conditions MUST be assumed to be zero to be valid, otherwise the results are relative and not absolute.
 
  • #29
hutchphd said:
Why do you think you need to know its "terminal charge", whatever you think that is?
As long as a battery is behaving as an ideal circuit element, it has zero resistance and supplies a fixed voltage. It will supply whatever current (charge) demanded by the rest of the circuit. It is not a capacitor and should not be treated as one.
If the battery were allowed to have an unknown net charge the problem could not be solved. Consider the limiting case of the voltage being zero, but the battery having a net charge Q. On connection, potential differences will arise in the capacitors.

This, it seems to me, is the essence of the OP's question in post #1; for my answer see post #17.
 
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  • #30
haruspex said:
This, it seems to me, is the essence of the OP's question in post #1; for my answer see post #17.
It was not clear what the OP was asking an this prompted my discussion of mutual and self capacitance in #25. I agree with your answer obviously.
I have never seen a battery analyzed as an active capacitor nor do I want to.
 
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  • #31
The solution for defining a battery as infinite C implies dV/dt = I/C =0 regardless of the charge transferred or its rate of flow which is current. So we assume a charge is simply transferred in this case from the higher voltage to the lower voltage battery and the conservation of charge equations then only involve the other components. So in the end it doesn’t affect the result but at least it passes the sniff test for reality in physics.

I posted the correct assumption in #11 if if C1V1+C2V2=0.

Also from KVL, V1+V2 +24-12V =0 and if C2=2C1
Thus for C1=2 uF and C2=4 uF

you have 2 equations and 2 unknowns, solve it and what are V1,V2?
 
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Related to Potential difference between 2 points in a capacitor circuit

What is potential difference in a capacitor circuit?

Potential difference in a capacitor circuit refers to the voltage between two points, typically across the terminals of the capacitor. It is the measure of the electric potential energy per unit charge between these two points, and it is what drives the movement of charge within the circuit.

How is potential difference measured in a capacitor circuit?

Potential difference in a capacitor circuit is measured using a voltmeter. The voltmeter is connected across the two points where the potential difference is to be measured, typically across the capacitor's terminals. The reading on the voltmeter gives the voltage or potential difference between those points.

What factors affect the potential difference across a capacitor?

The potential difference across a capacitor is affected by the amount of charge stored on the capacitor and the capacitance of the capacitor. According to the formula V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance, an increase in charge or a decrease in capacitance will result in a higher potential difference.

How does the potential difference change when capacitors are connected in series and parallel?

When capacitors are connected in series, the total potential difference is the sum of the potential differences across each capacitor. In contrast, when capacitors are connected in parallel, the potential difference across each capacitor is the same and equal to the total potential difference applied to the parallel combination.

Why is potential difference important in a capacitor circuit?

Potential difference is crucial in a capacitor circuit because it determines how much energy is stored in the capacitor. The energy stored in a capacitor is given by the formula E = 0.5 * C * V^2, where E is the energy, C is the capacitance, and V is the potential difference. Therefore, understanding and controlling the potential difference is essential for the efficient operation of capacitive devices in various applications.

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