Capacitors and charge?

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Homework Statement



A 2.70µF capacitor is charged to 475 V and a 4.00µF capacitor is charged to 525 V. (a) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

Homework Equations



C=Q/V

The Attempt at a Solution



For part (a) I tried the following:

I used this formula: C=Q/V ---> Q=CV

For C, I added the individual capacitances (2.70µF + 4.00µF) and got a total capacitance of 6.70µF.
C = 6.70µF

For Q, I added the individual charges (475 V + 525 V) and got a total charge of 1000 V.
Q = 1000 V

I then applied the previously stated formula.
V=Q/C
V=1000 / 6.70µF
V≈149.25 V
[Are those units correct? Did I even do the problem right?]

For part (b), the first thing that came to mind was to divide my answer for part (a) in half.
V≈149.25 V/µF / 2
V≈74.63 V/µF

Is all that correct?
 
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Answers and Replies

  • #2
gneill
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Volts are not charge; Coulombs are charge. For a capacitor of value C farads, the charge on it, in coulombs, is Q = CV.
 
  • #3
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Volts are not charge; Coulombs are charge. For a capacitor of value C farads, the charge on it, in coulombs, is Q = CV.
Ok, I used Q=CV and got
Q1 = (2.70 µF)(475 V) = 1282.5 µC
Q2 = (4.00 µF)(525 V) = 2100 µC

I suppose I would add these charges together to find the total charge?
Q = Q1+Q2 = 2100 µC + 1282.5 µC = 3382.5 µC

However, what does the problem mean by they switched connections? How do I incorporate that change mathematically?
 
  • #4
gneill
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Ok, I used Q=CV and got
Q1 = (2.70 µF)(475 V) = 1282.5 µC
Q2 = (4.00 µF)(525 V) = 2100 µC

I suppose I would add these charges together to find the total charge?
Q = Q1+Q2 = 2100 µC + 1282.5 µC = 3382.5 µC

However, what does the problem mean by they switched connections? How do I incorporate that change mathematically?
They mean that the capacitors' connections were physically rearranged, thus posing an interesting (according to some) problem for students to figure out what happens to the charges that were put on each capacitor in the beginning. :devil:

First there's a bit of problem in electrostatics, figuring out how the charges combine for a given connection arrangement. Then there's figuring out what the resulting capacitance and voltage will be. As for the mathematics to describe the situation... that's what the problem's all about.
 
  • #5
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Okay, so for part (a) now I have the following work:
V=Q/C
Q1+Q2 = 475 µC + 525 µC = 1000 µC
C1+C2 = 2.70 µF + 4.00 µF = 6.70 µF
V = 1000 µC / 6.70 µF
V = 504.851 V

And for part (b):
V=Q/C
Q1 = 475 µC
Q2 = 525 µC
I switched the C values here because in the problem it switches the capacitors.
C2 = 2.70 µF
C1 = 4.00 µF
V1 = 475 µC / 4.00 µF
V1 = 118.75 V
V2 = 525 µC / 2.70 µF
V2 = 194.44 V

The wording of the problem confuses me (or perhaps its my lack of understanding what its really talking about), but it talks about things being switched a lot. Did I do it correctly?
 
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  • #6
gneill
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For part (a) they give you the capacities and voltages for the capacitors. You've taken the voltages as charges. You'll have to first find out what the actual charges are on the capacitors.

For part (b), the capacitors are again being connected in parallel, but this time one of them is turned around so that its plate holding the negative charges is matched with the other capacitor's plate hold the positive charges. You have to determine what is going to happen to those charges when they meet.
 
  • #7
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(a)
Q=CV
Q1 = 475 V * 2.70 µF = 1282.5 µC
Q2 = 252 V * 4.00 µF = 1008 µC
V1+V2 = 475 V + 525 V = 1000 V
C1+C2 = 2.70 µF + 4.00 µF = 6.70 µF
Q = 1000 V * 6.70 µF
Q = 6700 µC

And for part (b):
V=Q/C
Q1 = 1282.5 µC
Q2 = 1008 µC
I switched the C values here because in the problem it switches the capacitors.
C2 = 2.70 µF
C1 = 4.00 µF
V1 = 1282.5 µC / 4.00 µF
V1 = 320.625 V
V2 = 1008 µC / 2.70 µF
V2 = 373.333 V

How does that look?
 
  • #8
gneill
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(a)
Q=CV
Q1 = 475 V * 2.70 µF = 1282.5 µC
Q2 = 252 V * 4.00 µF = 1008 µC
V1+V2 = 475 V + 525 V = 1000 V
C1+C2 = 2.70 µF + 4.00 µF = 6.70 µF
Q = 1000 V * 6.70 µF
Q = 6700 µC

And for part (b):
V=Q/C
Q1 = 1282.5 µC
Q2 = 1008 µC
I switched the C values here because in the problem it switches the capacitors.
C2 = 2.70 µF
C1 = 4.00 µF
V1 = 1282.5 µC / 4.00 µF
V1 = 320.625 V
V2 = 1008 µC / 2.70 µF
V2 = 373.333 V

How does that look?
The initial voltage given for the second capacitor I thought was 525V. It became 252V in your workings. I suspect that one of them is a typo...

Can you explain why you're adding the voltages in part a? It should be the charges that add together (in part (a) positive charge meets positive charge, and negative charge meets negative charge, so they sum and spread themselves across the available plate areas.)

For part (b) you should have a similar situation, only in this case there are positive charges meeting up with negative ones. Some mutual cancellation will occur as they combine.
 
  • #9
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Ugh, yes, I meant to type 525 V.

(a)
Q=CV
C = C1+C2 = 2.70 µF + 4.00 µF = 6.70 µF
Q1 = 475 V * 6.70 µF = 3182.5 µC
Q2 = 525 V * 6.70 µF = 3517.5 µC
Q1+Q2 = 6700 µC
Q = 6700 µC

(b)
C2 = 2.70 µF
C1 = 4.00 µF
Q1 = 3182.5 µC
Q2 = 3517.5 µC
V = Q/C
V1 = 795.63 V
V2 = 1382.78 V
 
  • #10
gneill
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You need to figure out the charges on the individual capacitors *before* they are connected together. Once they are connected, they cannot have different voltages.

When they are connected, the charges on each plate that connect to each other will sum. The resulting charges will spread across the two connected plates.
 
  • #11
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(a)
Right... so... find the individual charges without changing anything first?

Q1 = (2.70 µF)(475 V) = 1282.5 µC
Q2 = (4.00 µF)(525 V) = 2100 µC

Now add them together.

Q = Q1+Q2 = 1282.5 µC + 2100 µC = 3382.5 µC

(b)
So I "turn around" the plate by switching C1 and C2, right? You never commented on this.
C1 = 4.00 µF
C2 = 2.70 µF

Now using the total charge and each capacitor, I can find the voltages.

V=Q/C
V1 = 3382.5 µC / 4.00 µF
V1 = 845.63 V
V2 = 3382.5 µC / 2.70 µF
V2 = 1252.78 V

The question now asks to find the charges... again.

Q=VC
Q1= V1*C1 = 3382.52 µC
Q2= V2*C2 = 3382.51 µC

They have the same charges, different only through rounding error.



If this isn't right, please just show me how you would do it.
 
  • #12
gneill
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(a)
Right... so... find the individual charges without changing anything first?

Q1 = (2.70 µF)(475 V) = 1282.5 µC
Q2 = (4.00 µF)(525 V) = 2100 µC

Now add them together.

Q = Q1+Q2 = 1282.5 µC + 2100 µC = 3382.5 µC

(b)
So I "turn around" the plate by switching C1 and C2, right? You never commented on this.
C1 = 4.00 µF
C2 = 2.70 µF

Now using the total charge and each capacitor, I can find the voltages.

V=Q/C
V1 = 3382.5 µC / 4.00 µF
V1 = 845.63 V
V2 = 3382.5 µC / 2.70 µF
V2 = 1252.78 V

The question now asks to find the charges... again.

Q=VC
Q1= V1*C1 = 3382.52 µC
Q2= V2*C2 = 3382.51 µC

They have the same charges, different only through rounding error.



If this isn't right, please just show me how you would do it.
For (a), you're mostly on target. You have the total charge which is the sum of the charges which were on the individual capacitors before they were connected together. Now, after they are connected in parallel, they are sharing that total charge. They are now two capacitors in parallel with a total charge of Q = 3382.5 µC.

The capacitance of the parallel pair is Ceq = C1+C2 = 6.7 µF. The voltage across Ceq is given by V1 = Q/Ceq, or 504.9 V. So that's the voltage that the paralleled capacitors have on them. Both of the individual capacitors of that parallel pair have that voltage across them -- they must have the same voltage because they are wired in parallel.

To determine the individual charges that the capacitors now have on them (remember that the total charge from before the connection of the capacitors in parallel is now spread over the two parallel-connected capacitors), you use the V = Q/C formula for each of the capacitors, where V is now the new shared voltage of the two capacitors, V1, that we just determined.

For part (b), There is no "turning around" a plate. The problem states that the capacitors are connected so that the negative and positive plates of the opposing capacitors are wired together, rather than positive and positive, negative negative as in part (a). Shuffling capacitor variable names is not going to accomplish this. Here is what is happening:

Take a 2.70 µF capacitor in your right hand. It has two leads. Let's paint one lead red and the other lead black, strictly for purposes of identification. Now, the red lead is connected to the positive terminal of a 475V supply. The black lead is connected to the negative terminal of the same supply. After a short time, you remove the capacitor's connections from the supply. The capacitor is now charged to 475V and holds 1283.5 µC of charge, which it retains after disconnection from the supply. Now you take a 4.00 µF capacitor in your left hand. You give it the same treatment as the other capacitor, only this time you connect it to a 525V supply. It ends up with 2100 µC stored and 525V across its (painted) leads.

The plot so far:

C1: 2.70 µF ; 475V ; 1283.5 µC ; red lead connects to plate with +++ charges ; black lead, --- charges
C2: 4.00 µF ; 525V ; 2100.0 µC ; red lead connects to plate with +++ charges ; black lead, --- charges

Now you connect the capacitors in parallel. The "switch" that's described in the problem statement involves how the leads of the capacitors are connected. In this case, the positive (red) lead from each capacitor is connected to the negative (black) lead of the other. The capacitors are in parallel, but they have been wired together in such a way as to connect their oppositely charged plates via their leads. Red to black, black to red.

What happens? The opposite charges residing on the plates of the individual capacitors suddenly "see" a path to opposite charges via the newly connected leads. They jump at the chance to meet, and promptly race together, joining and canceling each other out. This happens on both plates of each capacitor simultaneously; the charges on both plates of each capacitor have access to the charges on the plates of the other capacitor, and they are charges of the opposite sign.

When all is said and done (in a few nanoseconds or so), after the mad rush of charges between capacitors, all the charges that can cancel have done so. What remains is whatever charges did not find an opposite charge to cancel with. So whichever capacitor initially had the largest charge "wins"; It's remaining charges determine the whole charge for the parallel combination. Those remaining charges spread themselves across the parallel capacitors just as in part (a). With the remaining charge you can determine the voltage across the equivalent capacitance of the paralleled capacitors. With the voltage, you can determine how much of the remaining charge resides on the individual capacitors of the pair.

So in this case C2 contained the greater charge (2100 µC) before the capacitors were connected. The remaining charge after the mutual annihilation will be Q2 - Q1. Because C2 had the greater charge, it will determine the voltage polarity of the paralleled result; C2's red lead will be connected to the positively charged plates of the paralleled pair.

So to sum up for part (b). Capacitors C1 and C2 with charges 1283.5 and 2100 µC and 475 and 525 volt potential differences are connected with opposing plate polarity. The charges on their plates meet, greet, and annihilate, leaving Q2 - Q1 = 817.5 µC to populate the plates of the now parallel pair. Their parallel capacitance of 6.7 µF then sports a voltage of (Q2 - Q1)/(C1 + C2) = 122V. Both of the capacitors, because they are in parallel, now share this voltage. The charges that reside on each capacitor can be found via the Q = C*V formula. The sum of the charges will equal the "remains", Q2 - Q1.

How's that?
 
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  • #13
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That makes much more sense, thank you very much!
 

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