Polarities of capacitor plates in a complex circuit

In summary: No, each capacitor has a specific potential. It is not a general rule that a plate with - charge will have a -ve potential and a plate with a + charge will have a +ve potential.
  • #1
vcsharp2003
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Homework Statement
I am trying to understand why the polarities for the plates of capacitors are as shown in the complex circuit below after equilibrium state is reached i.e. transient currents have ceased. I have two questions related to this circuit.

(a) Could the top plate of capacitor C₅ end up having either a + or - charge? (the diagram shows it as a + charge)

(b) Will the plates of capacitors C₁, C₅ and C₃ that are connected to each other via a wire be at the same potential once the circuit reaches an equilibrium state i.e. when transient current flow ceases?
Relevant Equations
None
CamScanner 05-04-2023 08.18_1.jpg

(a) I think the top plate of C5 could end up with either + or - charge, and not necessarily + charge as shown. This is because the connected plates of C1, C5 and C3 form an isolated system to which we can apply the law of conservation of charge i.e. Total charge just before transient currents between the connected plates starts = Total charge just after transient currents between connected plates stops and this gives us ##-q_1 + q_5 + q_3 = 0 \implies q_5 = q_1-q_3##. Clearly ##q_5 > 0 ## or ## q_5 = 0 ## or ##q_5 < 0##, and so the charge on the top plate of C5 could be + or - or 0.

(b) I am confused on this one.
  1. One logic says that at equilibrium when transient currents have ceased, then there should be no potential difference between the connected plates since electric current only flows when there is a potential difference.
  2. But then I get confused when I see different polarities for connected plates and think that the plate with a - charge will always be at a lower potential than a plate with a + charge; one of the connected plates is shown with - charge and the others are with a + charge, so it makes me conclude that potentials of connected plates could not be the same when equilibrium is reached (even though the transient electric currents have ceased). Currents should not stop if potential difference exists.
I don't know which of my above two arguments are correct
 
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  • #2
Let mathematics think the problem.
vcsharp2003 said:
Homework Statement: I am trying to understand why the polarities for the plates of capacitors are as shown in the complex circuit below after equilibrium state is reached i.e. transient currents have ceased. I have two questions related to this circuit.

(a) Could the top plate of capacitor C₅ end up having either a + or - charge? (the diagram shows it as a + charge)

(b) Will the plates of capacitors C₁, C₅ and C₃ that are connected to each other via a wire be at the same potential once the circuit reaches an equilibrium state i.e. when transient current flow ceases?
Relevant Equations: None

Total charge just after transient currents between connected plates stops and this gives us −q1+q5+q3=0⟹q5=q1−q3. Clearly q5>0 or q5=0 or q5<0, and so the charge on the top plate of C5 could be + or - or 0.
You are right. We have no intuition what signature does q_5, plate charge of condenser 5 at M side, has.
Conservation of charge on M piece, D piece and Outer circuit are
[tex]-q_1+q_5+q_3=0[/tex]
[tex]-q_2-q_5+q_4=0[/tex]
[tex]q_1+q_2-q_3-q_4=0[/tex]
Potential of M and D are
[tex]V_M=\frac{q_3}{C_3}=V-\frac{q_1}{C_1}[/tex]
[tex]V_D=\frac{q_4}{C_4}=V-\frac{q_2}{C_2}[/tex]
[tex]V_M=V_D+\frac{q_5}{C_5}[/tex]
where I made ##V_E=0## for simplicity.
I hope solution of these relations would show you the results to your question.
 
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  • #3
anuttarasammyak said:
Let mathematics think the problem.

You are right. We have no intuition what signature does q_5, plate charge of condenser 5 at M side, has.
Conservation of charge on M piece, D piece and Outer circuit are
[tex]-q_1+q_5+q_3=0[/tex]
[tex]-q_2-q_5+q_4=0[/tex]
[tex]q_1+q_2-q_3-q_4=0[/tex]
Potential of M and D are
[tex]V_M=\frac{q_3}{C_3}=V-\frac{q_1}{C_1}[/tex]
[tex]V_M=\frac{q_4}{C_4}=V-\frac{q_2}{C_2}[/tex]
[tex]V_M=V_D+\frac{q_5}{C_5}[/tex]
I hope solution of these relations would show you the results to your question.
Thanks for the detailed answer. However, I still have one doubt and that is whether the right plate of C1, the left plate of C3 and the top plate of C5 are at the same potential when transient currents have stopped?
 
  • #4
The diagram is not necessarily claiming the charges will be a particular way around. The analyst has to choose a convention at each capacitor in order to define the variables. The result may be a negative value for q5.

vcsharp2003 said:
I see different polarities for connected plates and think that the plate with a - charge will always be at a lower potential than a plate with a + charge
No, knowing which is + and which is - within a capacitor (when we do know) only says which of the two is at the higher potential. It tells you nothing about how those potentials compare with potentials at other capacitors.
 
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  • #5
vcsharp2003 said:
Thanks for the detailed answer. However, I still have one doubt and that is whether the right plate of C1, the left plate of C3 and the top plate of C5 are at the same potential when transient currents have stopped?
Yes, they are. In static state a metal piece has same electric potential everywhere because if not current flows in it.
 
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  • #6
anuttarasammyak said:
Yes. In stattic state a metal piece has same electric potential everywhere because if not current flows in it.
Ok. Can we not say that a plate with - charge will have a -ve potential and a plate with a + charge will have a +ve potential?
 
  • #7
haruspex said:
No, knowing which is + and which is - within a capacitor (when we do know) only says which of the two is at the higher potential. It tells you nothing about how those potentials compare with potentials at other capacitors.
Doesn't the + or - sign on a plate of a parallel plate capacitor tell whether the plate is +vely or -vely charged rather than which is at a higher potential? I understand the + or - sign on a plate as the polarity of the charge on it, but I could be wrong.
 
  • #8
vcsharp2003 said:
Ok. Can we not say that a plate with - charge will have a -ve potential and a plate with a + charge will have a +ve potential?
No, we cannot. You can carry charges to any place where V has any value, though work needs to be done sometime.
 
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  • #9
vcsharp2003 said:
Doesn't the + or - sign on a plate of a parallel plate capacitor tell whether the plate is +vely or -vely charged rather than which is at a higher potential? I understand the + or - sign on a plate as the polarity of the charge on it, but I could be wrong.
Consider two identical capacitors in series, no charges anywhere. Now a potential difference V is applied across the pair. Each capacitor will now be charged positively on one side, negatively on the other. There is a potential drop V/2 across each.
Start again, but now there is a substantial net charge on the wire connecting the capacitors. After applying the external p.d., there is still a V/2 drop across each, but it could be that three of the four plates are positively charged.
 
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  • #10
haruspex said:
Consider two identical capacitors in series, no charges anywhere. Now a potential difference V is applied across the pair. Each capacitor will now be charged positively on one side, negatively on the other. There is a potential drop V/2 across each.
Start again, but now there is a substantial net charge on the wire connecting the capacitors. After applying the external p.d., there is still a V/2 drop across each, but it could be that three of the four plates are positively charged.
Ok. I am trying to figure out the scenario you described but am stuck at Fig 3 in diagram below. I am unable to decide what direction the transient current will flow in Fig 3 and whether current will flow at all. Could you please help me with Fig 3?

CamScanner 05-04-2023 08.18_2.jpg
 
  • #11
Are you referring to my precharged case? The transient currents will flow in the same way as in the uncharged case.
 
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  • #12
haruspex said:
Are you referring to my precharged case? The transient currents will flow in the same way as in the uncharged case.
Yes, I am trying to follow what you described in post#9. Fig 3 is the last step where the charged capacitors are connected to the same battery as in Fig 1. The voltages shown in Fig 3 are the initial voltages when connected to battery. If transient currents flow in same direction as in Fig 1, then the final charges on the charged capacitors could change and also the voltages. Is that right?
 
  • #13
vcsharp2003 said:
Yes, I am trying to follow what you described in post#9. Fig 3 is the last step where the charged capacitors are connected to the same battery as in Fig 1. The voltages shown in Fig 3 are the initial voltages when connected to battery. If transient currents flow in same direction as in Fig 1, then the final charges on the charged capacitors could change and also the voltages. Is that right?
If a region would end up with charge q in the case where there are no initial charges, and starts with charge q' in the case where there are initial charges, then in the final state of the latter case it will have charge q+q'.
 
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  • #14
haruspex said:
If a region would end up with charge q in the case where there are no initial charges, and starts with charge q' in the case where there are initial charges, then in the final state of the latter case it will have charge q+q'.
But if the charged capacitors did change their charges in Fig 3, then even the potential difference across each capacitor would change since capacitance is unchanged. If potential differences changed from existing 50 V for each capacitor, then they would still need to add 100V and also be equal. In other words, potential difference cannot change from the initial value of 50 V and therefore, even the charge on each should not change in Fig 3 from their charged states. So, it's amounting to no effective transient current in Fig 3. Not sure if I'm wrong in above reasoning.
 
  • #15
vcsharp2003 said:
Homework Statement: I am trying to understand why the polarities for the plates of capacitors are as shown in the complex circuit below after equilibrium state is reached i.e. transient currents have ceased. I have two questions related to this circuit.

(a) Could the top plate of capacitor C₅ end up having either a + or - charge? (the diagram shows it as a + charge)

(b) Will the plates of capacitors C₁, C₅ and C₃ that are connected to each other via a wire be at the same potential once the circuit reaches an equilibrium state i.e. when transient current flow ceases?
Relevant Equations: None

But then I get confused when I see different polarities for connected plates and think that the plate with a - charge will always be at a lower potential than a plate with a + charge; one of the connected plates is shown with - charge and the others are with a + charge, so it makes me conclude that potentials of connected plates could not be the same when equilibrium is reached (even though the transient electric currents have ceased). Currents should not stop if potential difference exists.
I think I am understanding this paradoxical scenario where the plate with a - charge can have the same potential as another connected plate with a + charge. A - charge on a plate does not make its potential negative or a plate with + charge doesn't mean its potential is positive, which was the source of confusion for me. My explanation is as below. Of course, I am assuming all potentials are with respect to potential at infinity.

I am using the diagram to explain this paradoxical situation of how potentials at points 1 and 2 can be the same even though they are on oppositely charged plates.

CamScanner 05-04-2023 08.18_4.jpg

Firstly, when transient currents stop in a circuit then we know from basic electricity that the potential difference between any two points on the wire will be zero as per Ohm's law i.e. any two points on the wire will have the same potential. Thus, all points on the wire connecting capacitors C1 and C2 will be at the same potential, which includes the points 1 and 2 that are on oppositely charged plates of C1 and C2.

Secondly, since electric potential obeys the Superposition Principle so we can say that the electric potential at point 1 is a sum of potentials created by the + charge plate of C1, the - charge plate of C2 and the + charge plate of C2. Similarly, the electric potential at point 2 is a sum of potentials created by the + charge plate of C1, the - charge plate of C1 and the - charge plate of C2. As a result, the sum of potentials at 1 equals the sum of potentials at 2 i.e. V1 = V2.

So, clearly the sign of charge on a plate tells us nothing about its potential except that its potential is lower (if its a - plate) or higher (if its a + plate) than the other plate of the same capacitor. This means the potential on a - charge plate could be positive like +2 V and not a negative value.
 
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