Potential Difference. across capacitors

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When a fully charged capacitor is connected to an uncharged capacitor with different capacitance, would the p.d. across each of them be the same?

I thought that only the charge stored on each would be the same so p.d. across each would depend on their capacitance? Does that mean that they will not have the same p.d.? If they do not have the same p.d., will the sum of their p.d. add up to be the voltage stored in the originally uncharged capacitor?
 

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  • #2
phinds
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How could parallel capacitors possibly not have the same PD? For that matter how could ANY elements in parallel not have the same PD?

Now, whether or not the PD is the same as that on the original capacitor is a different matter. If you add two caps in parallel, what happens to the total capacitance?
 
  • #3
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How could parallel capacitors possibly not have the same PD? For that matter how could ANY elements in parallel not have the same PD?

Now, whether or not the PD is the same as that on the original capacitor is a different matter. If you add two caps in parallel, what happens to the total capacitance?

I should have stated it more clearly. The fully charged capacitor is only connected with the uncharged capacitor, so shouldn't they be in series in this case? This why I am confused because technically the p.d. should be "split" between the two capacitors when they are connected in series, but my textbook says they have the same p.d.
 
  • #4
Chandra Prayaga
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I suggest that you make a circuit diagram.
 
  • #6
berkeman
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but my textbook says they have the same p.d.
Does your textbook have drawings or figures for this topic?
 
  • #7
Chandra Prayaga
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If you are connecting just two components together, automatically, they are parallel. The potential across each is the same. The charge distributes proportional to the capacitances capacitances
 
  • #8
berkeman
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If you are connecting just two components together, automatically, they are parallel. The potential across each is the same. The charge distributes proportional to the capacitances capacitances
That is not true. The OP has mentioned series connection, so we need to wait to see what the textbook is asking about.
 
  • #9
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When the caps are connected together in series, the stored electric charge redistributes itself. The voltage on the charged cap will go down, and the voltage on the uncharged cap will go up, until the voltage across both caps will (after some amount of time) be the same.
 
  • #10
berkeman
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When the caps are connected together in series, the stored electric charge redistributes itself. The voltage on the charged cap will go down, and the voltage on the uncharged cap will go up, until the voltage across both caps will (after some amount of time) be the same.
Not unless the circuit is completed somehow. Let's let the OP respond with the actual question out of their textbook please.
 
  • #11
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Not unless the circuit is completed somehow. Let's let the OP respond with the actual question out of their textbook please.

Hi, this is the exact question on my book. It doesn't come with a diagram however. But I think the meaning is quite clear.
 

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  • #12
davenn
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But I think the meaning is quite clear.

totally unclear
we still don't know how the capacitors are connected
 
  • #13
Merlin3189
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I thought it is clear enough. They are connected in series and parallel! If you have only two components this is always the case.
You may have thought that only one wire of each capacitor is connected, but that would be a fairly trivial & pointless exercise. You could join all 4 leads to the same point and there are other trivial possibilities, but the only non-trivial way for 2 elements is in series and parallel at the same time.
And although your mind, like mine, probably immediately thought electrolytics when you saw microfarads, the question does not say that. So it does not matter which end connects to which.
It does not matter whether you say parallel so the voltages are the same, or series so that the sum of the voltages round the loop is zero. If there are only two components, the voltages must be equal in magnitude.
Similarly the amount of charge transferred comes out the same whether series or parallel.
It's like skinning cats.
 
  • #14
phinds
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I thought it is clear enough. They are connected in series and parallel! If you have only two components this is always the case.
Exactly.
 
  • #15
jbriggs444
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I should have stated it more clearly. The fully charged capacitor is only connected with the uncharged capacitor, so shouldn't they be in series in this case? This why I am confused because technically the p.d. should be "split" between the two capacitors when they are connected in series, but my textbook says they have the same p.d.
If these two capacitors have no other connections then they are both in parallel and in series. However, there is a catch.

Make that circuit drawing. Label the two ends on each capacitor. A and B on the charged capacitor and a and b on the uncharged capacitor. If you join A to a and B to b and consider the two capacitors as being in parallel then you have:

pd(AB) = pd(ab).

If you join A to a and B to b and consider the two capacitors as being in series then you have:

pd(AB) + pd(ba) = 0.

When you keep track carefully, you see that you have to flip conventions for the potential difference across one of the two capacitors. The result is that both equations say the same thing.
 
  • #16
Merlin3189
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I'm not sure what you're saying about conventions here?
I think your series equation is exactly what Kirchoff would say. I can't see where any change in convention is required.
My labels are the reverse of yours, but I get P(BA) + P(ab)=0, ∴ P(BA) - P(ba) =0 [since P(xy)=-P(yx) ], ∴ P(BA) = P(ba) or P(AB)=P(ab)

But what I'd like to know is, what happens to the energy? I used to think it was dissipated in the resistance (of the wires and the capacitors themselves), but I was once told that this does not account for enough. (I've been trying for a while this afternoon to calculate this myself, but my maths is failing me.) The explanation I was given was that some is dissipated in EM radiation. While I thought that possible at the time, I've come to doubt it, so I'd like to see a proof that it can all be dissipated as I2R. If so, any links?
capacitors_2.png
 

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  • #17
jbriggs444
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I'm not sure what you're saying about conventions here?
You have the same capacitor with with the same endpoints and the same potential difference. This is transcribed into equations once as +4V and once as -4V. [Edit, nothing wrong with that. The equations, are, of course, correct]. There must have been some reasoning or some standard behind that assignment. That is the sort of convention that I am referring to.

But what I'd like to know is, what happens to the energy? I used to think it was dissipated in the resistance (of the wires and the capacitors themselves), but I was once told that this does not account for enough.
If the resistance is very high, it does account for essentially everything. If the resistance is low then the circuit will resonate due to the finite inductance in the wires. It amounts to an RLC circuit. Ultimately, the energy in the resonance will damp due to both resistance and EM radiation. [A glance at the equations for radiated power from dipole and monopole antennae makes it less than obvious how to calculate the radiated power for a given current, frequency and wire length]
 
  • #18
cnh1995
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But what I'd like to know is, what happens to the energy? I used to think it was dissipated in the resistance (of the wires and the capacitors themselves), but I was once told that this does not account for enough. (I've been trying for a while this afternoon to calculate this myself, but my maths is failing me.) The explanation I was given was that some is dissipated in EM radiation. While I thought that possible at the time, I've come to doubt it, so I'd like to see a proof that it can all be dissipated as I2R. If so, any links?
I am not sure if this is what you are looking for:
In a simple RC series circuit, the energy stored in a capacitor after charging is ½CV2. It can be shown mathematically that the energy lost in the resistor is also ½CV2 (I don't know where EM radiation comes into picture. Edit: Reading jbriggs444's reply gave me some idea:wink:).
The expression will be E=∫i(t)2R dt from 0 to ∞, where i(t) is ioe-t/RC. Solving this, we get E=½CV2.
Similarly, in case of two capacitors connected in series and parallel at the same time, I believe it is possible to show that the energy can all be dissipated as I2R.
 
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  • #19
Merlin3189
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Thanks for those comments, esp. cnh and the maths. No wonder I was struggling, as I'd got i(t) wrong and couldn't find the integral. I just need to check that this is still ok for C1RC2 circuit, but I'm already feeling reassured!
 
  • #20
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When charge travels from one capacitor to another, if we neglect the resistance of the connecting wires and the dielectric loss, the change in energy is zero.
 
  • #21
Merlin3189
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I don't think so.
Say the capacitors are 1 F and one is charged to 10V
The energy is 0.5 CV2 = 50 J and charge is CV=10 Cb
When joined, total capacitance is 2F with 5Cb on each capacitor
then V=Q/C = 10/2 = 5V for pair, or 5/1 = 5V for each
Energy = 0.5CV2 = 0.5 x 2 x 25 = 25J for the pair, or =0.5 x 1 x 25 = 12.5 J on each.
"Lost" energy = 50 - 25 = 25J

I haven't time to show it algebraically at the moment, but this 50% loss is general. See Hyperphysics for more.
Edit: Sorry, I think the generality of the 50% may not be accurate - I'll redo the algebra after rereading Hyperphys.
Edit2: I think the energy after joining charged C1 to uncharged C2 is
= C1 / (C1 + C2 ) of the original energy.
So it's only 50% for equal capacitors. But always less than 100% unless C2 is 0.
 
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  • #22
davenn
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I thought it is clear enough. They are connected in series and parallel!

there is another possibility
and that's why I said it was unclear without a diagram
( it's an incredibly vague Q)
 
  • #23
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It takes energy to charge a capacitor, but you get it all back when it discharges.
 
  • #24
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Correction: When one capacitor charges another capacitor, half the energy that was stored in the first capacitor is lost. I apologize for the error.
 
  • #25
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totally unclear
we still don't know how the capacitors are connected
I beg to differ.
A battery is connected to a capacitor. The battery is removed and replaced by an additional capacitor.
How many possibilities are there to experienced physicists.
 
  • #26
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Correction: When one capacitor charges another capacitor, half the energy that was stored in the first capacitor is lost. I apologize for the error.
This is incorrect.
The energy 'lost' depends on the value of the capacitors.
Also energy is always lost by electro magnetic radiation from the wires. Electro magnetic radiation always occurs when a current changes. It can be detected, usually, as radio waves.
 
  • #27
berkeman
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This is incorrect.
The energy 'lost' depends on the value of the capacitors.
That may be true. Can you show us the math?
 
  • #28
jbriggs444
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That may be true. Can you show us the math?
See Merlin3189's response in post #21.
 
  • #29
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That may be true. Can you show us the math?
merlin post 21 has shown his example.
This is a trivial calculation, there must be thousands of examples in basic text books.
I am sure this is not the first time this question about capacitors has appeared here
More than 20 posts litttered with mistakes, corrections and misunderstanding. More confusion has been caused by this thread than enlightenment.
I am surprised posts like this do not have some sort of moderation regarding validity of posts. Too much subjective debate.
 
  • #30
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I'm sticking with Merlin3189. The 50% loss is general for any size capacitor. It also doesn't make any difference whether the capacitor is charged by a battery or power supply. Only half the energy drawn from the power supply will be stored in the capacitor. The other half will be dissipated. The EM radiation, on the other hand, can be made small depending on how the circuit is arranged.
 
  • #31
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I'm sticking with Merlin3189. The 50% loss is general for any size capacitor. It also doesn't make any difference whether the capacitor is charged by a battery or power supply. Only half the energy drawn from the power supply will be stored in the capacitor. The other half will be dissipated. The EM radiation, on the other hand, can be made small depending on how the circuit is arranged.

perhaps I misunderstood you. I thought this thread was about connecting capacitors together (merlins post is about this!) In that case the energy lost is only 1/2 when the capacitors are equal and one charged capacitor is connected to an uncharged capacitor.
What you say about energy stored on a charged capacitor is absolutely correct. E = 1/2CV2 or 1/2QV or 1/2Q2/C
the lost energy can be lost by series resistance, sparking at the switch or electro magnetic radiation. There will always be electro magnetic radiation.
If resistance = 0 and there is no sparking at the switch then all of the lost energy is by electro magnetic radiation.
I hope this clears up what I had posted. Sorry for any misunderstanding
 
  • #32
berkeman
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I am surprised posts like this do not have some sort of moderation regarding validity of posts.
If you see something incorrect in a post, please click the Report link on the post and let the Mentors know about it. We do not read every post in every thread... :smile:
 
  • #33
Merlin3189
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Strange this thread should resurrect itself just as I'm watching another thread about more or less the same issue. I was preparing a post for that making a 'good old fashioned' water analogy! Perhaps I can offer it here for your diversion.

Imagine (because I haven't drawn the diagrams yet!) two tanks for containing liquid: for simplicity lets say they are right prisms (such as a cuboid or a cylinder,) standing on a level plane, with a pipe (of negligible size) connected to the base.
Start with one full of liquid and the other empty, then join the two tanks with a small pipe and let the water flow. The flow will end when the level is equal in both tanks and, if the tanks are the same size, the depth will be half the original depth of the full tank. The pressure at the base of the tanks will obviously be half the original pressure. But as for the potential energy of the water, this will be halved. The total mass of water is the same, but its depth is halved. Half the energy has been dissipated in turbulence.

If the volume of liquid in a tank is v and the area of the tank (base) is A
Pressure is proportional to the depth of liquid in the tank, ##P = ρgh. = \frac{ρgv}{A}## or ## v\frac{ρg}{A}## [Cf ##V= \frac{Q}{C}## ]
[So ##\frac{ A}{ρg}## would be the analogue of C]
PE is proportional to half the height, E= weight * height of CG =## ρgv\times \frac{h}{2} = \frac{v^2 ρg}{2A}## [ Cf ##E= \frac{Q^2}{2C} ## ]
Or ##E= vρg\times \frac{h}{2} = \frac{Pv}{2}\ \ \ ## [ Cf ##E = \frac{VQ}{2}## ]
Or ##E= \frac{Pv}{2} = \frac{PAh}{2} = \frac{P^2A}{2ρg}\ \ \ ## [ Cf ## E= \frac{V^2 C}{2}## ]

So for the two equal tanks, one filled with V volume of liquid,
Initial PE = V2 ρg / 2A
Final PE = 2 x (V/2)2 ρg / 2A = V2 ρg / 4A

If we don't have equal tanks, say the second tank is half the area of the first (full) tank, equilibrium will be reached when a third of the volume has been transferred. The second tank then has half the volume of the first, but both have the same depth = 2/3 of the original depth.
Initial PE = V2 ρg / 2A as before
Final PE = (2V/3)2 ρg / 2A + (V/3)2 ρg / A = [ 4/18 + 1/9 ] V2 ρg / A = V2 ρg / 3A
So final energy is 2/3 of initial (matching the C1/(C1 + C2) ratio) and dissipated energy is 1/3 (matching the C2 / (C1 + C2) ratio )

If we fill the second tank from a pump or enormous reservoir so that the input pressure is constant, like charging a capacitor from a battery, the work input is simply the pressure x volume. Of course the back pressure due to the height of water in the second tank is less than this until the second tank has filled to the necessary depth. So the rest of the pressure must be used overcoming resistance to flow: and flow will be higher at the beginning, tapering off as the other tank fills.
When V volume has flowed and pressure has reached P, the PE of the liquid in the tank is now VP/2 (from the second line of PE expressions above.)
As with the capacitor being charged from a battery, as everyone has said, half the energy supplied is lost. (Here in turbulence and heat, or maybe radiated as sound?)
Of course the proportion of energy lost is tiny, because the reservoir (or C1) is very much larger than the tank (C2)

I don't claim this proves anything, just maybe people struggling with capacitors and their invisible energy, might find tangible water more understandable. For me they key point is to realise that the CG of the water rises at half the rate that the surface level rises.
 

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