Hey.
I am having major trouble understanding why the energy under the V against Q graph formed when a capacitor is charging up is equal to the energy stored in the capacitor.
I think that I do not understand this because I do not truly appreciate what the existence of a potential difference across the capacitor means. What does it mean? (Because electrons are unable to flow across this electric field due to the presence of the dielectric).
The problem I have is that the electrons do not flow through the capacitor like in most components and so I cannot explain the fact that W = Integral of V with respect to q using the perhaps basic definition "V is energy lost per unit charge as it flows through the component."
Please help me.

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 29032011 19:28
Last edited by Magu1re; 29032011 at 19:49. 
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 29032011 20:54
In the case of a capacitor, it's better to think of the energy stored in it as the work done charging it up.
You have to do work against the repulsion of the charge that's already on it. The more charge on there, the more work you have to do.
When charged, the energy is stored in the electric field, and when you connect the plates together in a circuit, the capacitor does work pushing the charges around the circuit as it discharges.
The energy is released.
Although a "direct" current doesn't flow through a capacitor, charge can flow onto and off the plates. This constitutes a current. If you have a circuit with alternating current, the capacitor conducts a current with no problem at all, first pushing charge onto one plate, and then onto the other.
The larger the value of the capacitance, the more charge can be stored on it (at the same V) so the bigger the alternating current can be for the same volrtage. This is why the "resistance" (correctly called reactance) of a capacitor to AC goes down as you increase its capacitance.Last edited by Stonebridge; 29032011 at 20:55. 
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 29032011 21:03
(Original post by Stonebridge)
In the case of a capacitor, it's better to think of the energy stored in it as the work done charging it up.
You have to do work against the repulsion of the charge that's already on it. The more charge on there, the more work you have to do.
When charged, the energy is stored in the electric field, and when you connect the plates together in a circuit, the capacitor does work pushing the charges around the circuit as it discharges.
The energy is released.
Although a "direct" current doesn't flow through a capacitor, charge can flow onto and off the plates. This constitutes a current. If you have a circuit with alternating current, the capacitor conducts a current with no problem at all, first pushing charge onto one plate, and then onto the other.
The larger the value of the capacitance, the more charge can be stored on it (at the same V) so the bigger the alternating current can be for the same volrtage. This is why the "resistance" (correctly called reactance) of a capacitor to AC goes down as you increase its capacitance.
Why does a greater force of repulsion mean more work must be done by the supply? Surely the electromotive force is constant?Last edited by Magu1re; 29032011 at 21:09. 
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 29032011 23:04
(Original post by Magu1re)
Okay. A lot of that makes sense to me. But why is the area under the graph of V against Q equal to the work done charging the capacitor up? What physical understanding/picture underpins the reason for doing this integration?
2. The energy stored in a capacitor is the work done charging it up. The work is done in pushing charges onto the plates against the repulsion of the charges already on them.
3. The work done moving a charge through a potential of one volt is by definition, one joule. So to move a charge onto a plate which is at a potential of, say, 1 volt needs 1 joule of work. This is then stored in the capacitor.
4. As you store more charge on it, the capacitors voltage rises. This means it needs more work now to get a charge of one coulomb onto the plates.
5. The total work done is found mathematically by integration, because the voltage does not remain constant as you add more charge.
6. You could think of it very simplistically by saying that the voltage rises from zero to V. The average voltage was ½V
The work done moving a charge Q through a voltage V is QV (by definition), but the pd on the capacitor was not a constant V, it went from 0 to V giving an average of ½V.
Work done is ½QV
[ A bit like distance travelled is average velocity times time. In that case, it's the area under a vt graph. If the velocity goes from 0 to v, the average is ½v and the area under the vt graph is a triangle of area ½vt, half base times height]
Why does a greater force of repulsion mean more work must be done by the supply? Surely the electromotive force is constant?
And the formula? ½Fx where F is the final tension in the spring and x the compression. (There's that half again. Here it comes from average force because the force opposing the compression increases as you compress the spring, from 0 to F. The area under the graph is a triangle and its area is half base times height.)
The spring example is a good analogy to help trying to visualise whats going on in a capacitor regarding stored energy.
Finally, if you do the actual integration you finish up integrating VdQ from 0 to V by considering the work done moving a small charge dQ onto a plate with potential V
As V=Q/C the integral becomes (Q/C) dQ
Integrating this wrt Q gives Q²/2C
as C=Q/V this becomes ½QVLast edited by Stonebridge; 30032011 at 09:50. 
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 30032011 15:30
(Original post by Stonebridge)
1. The area under the graph is a triangle and is equal to ½QV
2. The energy stored in a capacitor is the work done charging it up. The work is done in pushing charges onto the plates against the repulsion of the charges already on them.
3. The work done moving a charge through a potential of one volt is by definition, one joule. So to move a charge onto a plate which is at a potential of, say, 1 volt needs 1 joule of work. This is then stored in the capacitor.
4. As you store more charge on it, the capacitors voltage rises. This means it needs more work now to get a charge of one coulomb onto the plates.
5. The total work done is found mathematically by integration, because the voltage does not remain constant as you add more charge.
6. You could think of it very simplistically by saying that the voltage rises from zero to V. The average voltage was ½V
The work done moving a charge Q through a voltage V is QV (by definition), but the pd on the capacitor was not a constant V, it went from 0 to V giving an average of ½V.
Work done is ½QV
[ A bit like distance travelled is average velocity times time. In that case, it's the area under a vt graph. If the velocity goes from 0 to v, the average is ½v and the area under the vt graph is a triangle of area ½vt, half base times height]
Because work done against a force is force times distance, and if the force increases the work done increases. It's the force of repulsion from the charge already on the plates that has to be overcome. As more charge builds up, you need more force to overcome it. This is the physical meaning of energy stored in a capacitor. Just like energy stored in a spring is the work done compressing it. The more you compress it the more force is needed. And the energy stored in the spring is the area under the forceextension graph. Again it's mathematically an integration, and is represented by the are under the appropriate graph.
And the formula? ½Fx where F is the final tension in the spring and x the compression. (There's that half again. Here it comes from average force because the force opposing the compression increases as you compress the spring, from 0 to F. The area under the graph is a triangle and its area is half base times height.)
The spring example is a good analogy to help trying to visualise whats going on in a capacitor regarding stored energy.
Finally, if you do the actual integration you finish up integrating VdQ from 0 to V by considering the work done moving a small charge dQ onto a plate with potential V
As V=Q/C the integral becomes (Q/C) dQ
Integrating this wrt Q gives Q²/2C
as C=Q/V this becomes ½QV 
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 30032011 17:16
(Original post by Magu1re)
So when you say the energy required to move a charge from one position to another do you literally mean a quantity of charge (not an actual electron which moves from one position to another). Because my visualisation of electrons in a wire is like smarties in a tube and I can see an electron vacating a position causing another at the end of the tube to enter a position and as such "charge" has been moved from one position to another but the actual electron has not moved between those positions. Is this the case? If it is then I understand what you have written.
Current is defined as quantity of charge flowing per second.
I'm not sure the smarty analogy works too well.
To move a charge you need an electric field. If you place a charge in an electric field it experiences a force. (This defines an electric field) In an electric circuit charge actually does move from place to place. Electrons do move. They move because there is force on them from the applied electric field, and they are held loosely enough to their parent atoms. In a circuit the force is the emf from the battery. This is an electric field applied to the wire. Charges can also move, as a spark, between electrical terminals at high voltage. (Or clouds as lightning).
So electrons do move in a circuit. I don't see how the smarty analogy quite works. How can a charge "be moved from one position to another" but yet "not move between those positions". The fact that more electrons (these are the charges in the circuit) come in to take the place of those that have moved away, does not invalidate the notion of charge movement any more than the fact that you get more water coming along in a river to replace that which has flowed away invalidates the idea of a river being a flow of water..
To go back to your original question because we have drifted away a bit;
When you connect a capacitor to a battery, there is an electric field applied to the wires and the capacitor. (This field comes from the emf of the battery)
Electrons in the conducting wires experience a force and move in response.
They are pushed onto one plate of the capacitor and move off the other.
This continues until the electric field (pd) that builds up on the capacitor, is equal to the applied field. Then the pd across the capacitor is equal to the emf of the cell and no more charge flows. (The electric fields, and forces, are balanced)
We say the capacitor is charged. In order to charge it, the battery had to do work moving the electrons onto the plate (against repulsion) and also pushing them around the rest of the circuit against resistance. As I explained in my other post, the energy stored in the capacitor is ½QV where Q is the charge and V the voltage on the plates. Not all of the energy the battery expended will be stored in the capacitor, some is lost as heat in the rest of the circuit. 
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 30032011 21:27
(Original post by Stonebridge)
The theory does talk about quantities of charge being moved from place to place. If you move a quantity of charge then you have a current.
Current is defined as quantity of charge flowing per second.
I'm not sure the smarty analogy works too well.
To move a charge you need an electric field. If you place a charge in an electric field it experiences a force. (This defines an electric field) In an electric circuit charge actually does move from place to place. Electrons do move. They move because there is force on them from the applied electric field, and they are held loosely enough to their parent atoms. In a circuit the force is the emf from the battery. This is an electric field applied to the wire. Charges can also move, as a spark, between electrical terminals at high voltage. (Or clouds as lightning).
So electrons do move in a circuit. I don't see how the smarty analogy quite works. How can a charge "be moved from one position to another" but yet "not move between those positions". The fact that more electrons (these are the charges in the circuit) come in to take the place of those that have moved away, does not invalidate the notion of charge movement any more than the fact that you get more water coming along in a river to replace that which has flowed away invalidates the idea of a river being a flow of water..
To go back to your original question because we have drifted away a bit;
When you connect a capacitor to a battery, there is an electric field applied to the wires and the capacitor. (This field comes from the emf of the battery)
Electrons in the conducting wires experience a force and move in response.
They are pushed onto one plate of the capacitor and move off the other.
This continues until the electric field (pd) that builds up on the capacitor, is equal to the applied field. Then the pd across the capacitor is equal to the emf of the cell and no more charge flows. (The electric fields, and forces, are balanced)
We say the capacitor is charged. In order to charge it, the battery had to do work moving the electrons onto the plate (against repulsion) and also pushing them around the rest of the circuit against resistance. As I explained in my other post, the energy stored in the capacitor is ½QV where Q is the charge and V the voltage on the plates. Not all of the energy the battery expended will be stored in the capacitor, some is lost as heat in the rest of the circuit.
Is this the case? 
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 30032011 22:39
(Original post by Magu1re)
What I was trying to check was that to the "movement of charge through a potential difference" i.e. removing 1.6EXP19 C of charge from the soon to be positive plate and depositing it on the now negative plate does not actually involve an electron moving from the positive plate to the negative plate. Rather it involves an electron flowing of the positive plate and another, different electron flowing on to the negative plate. In a sense, charge has moved through the potential difference. I was trying to check this was correct and what is meant by "the energy transferred moving a charge through a potential difference".
Is this the case?
Charge moves off the one plate and charge moves onto the other, but via the battery, not directly, of course. The drift speed of the electrons in the circuit is very slow (typically around a mm per second) so it won't be the same electrons.
I think the confusion lies in the way it's described in electrostatics when talking about charging a capacitor.
After all, once it's charged up, the end result is that charge has been "transferred" from one plate to the other. The amount of positive charge on the one plate is exactly the same as the amount of negative on the other. It's as if the charge was just taken off the one and placed on the other.
It's not what actually happens but the phrase is still often used.
So long as you understand what is really going on it shouldn't cause a problem.
In the end it's just a matter of the words used. It's just traditional with capacitors to talk about "the transfer of charge from one plate to another". Confusing, I agree. 
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 30032011 23:29
(Original post by Stonebridge)
Yes.
Charge moves off the one plate and charge moves onto the other, but via the battery, not directly, of course. The drift speed of the electrons in the circuit is very slow (typically around a mm per second) so it won't be the same electrons.
I think the confusion lies in the way it's described in electrostatics when talking about charging a capacitor.
After all, once it's charged up, the end result is that charge has been "transferred" from one plate to the other. The amount of positive charge on the one plate is exactly the same as the amount of negative on the other. It's as if the charge was just taken off the one and placed on the other.
It's not what actually happens but the phrase is still often used.
So long as you understand what is really going on it shouldn't cause a problem.
In the end it's just a matter of the words used. It's just traditional with capacitors to talk about "the transfer of charge from one plate to another". Confusing, I agree.
When the negative plate gains electrons the positive plate loses an equal number of electrons so that the net charge on both plates is equal in magnitude but opposite in sign. Why is this?
At first I understood it to be the "smarties effect" leaving a hole near the positve plate which attracts an electron off it. However, this picture doesn't seem to explain it in the case where you have two capacitors in parallel and an isolated section between them. Could you shed any light on this? 
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 31032011 11:23
(Original post by Magu1re)
When the negative plate gains electrons the positive plate loses an equal number of electrons so that the net charge on both plates is equal in magnitude but opposite in sign. Why is this?
At first I understood it to be the "smarties effect" leaving a hole near the positive plate which attracts an electron off it. However, this picture doesn't seem to explain it in the case where you have two capacitors in parallel and an isolated section between them. Could you shed any light on this? 
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 31032011 13:35
(Original post by Stonebridge)
Conservation of charge. If one plate gains Q and the other plate doesn't lose Q you have some explaining to do about where your charge has gone or how charge has appeared from nowhere.
You will need to elaborate here as I'm not really sure what it is you are asking me to shed light on.
However, I cannot use this "argument" in the case of two capacitors in series due to the fact that the section of wire between the two capacitors is isolated. If the lefthand plate of the capacitor on the left gains negative charge then this encourages the same amount of charge to be pulled off the righthand plate of the capacitor on the right (using this argument with two capacitors in series (one to the left and one to the right)). However, I can see electrons in the right handplate of the capacitor on the left are going to be repelled (forced to the right) and electrons in the lefthand plate of the capacitor on the right are going to be attracted (forced to the right) and this will result in positive charge on the righthand plate of the capacitor on the left and negative charge on the lefthand plate of the capacitor on the right. However, I do not see why the amount of charge on these two plates separately must be equal in magnitude to the amount of charge on the other two plates that make up the system. 
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 31032011 14:43
The isolated bit in the middle (ringed) has no charge initially.(no net charge = neutral)
The battery pushes negative charge onto the right of B.
This negative charge repels electrons off the left plate of B, which becomes positive. Each negative charge on the left induces a positive charge on the right.* As the isolated bit is neutral, there must be an equal amount of negative charge that has moved to the right plate of A. (Conservation of charge)
This means there is an equal amount of positive charge on the left plate of A, and this perfectly balances the negative charge on the right plate of B.
* Induced charges are always equal and opposite.
Do you not have a decent text book that explains this?

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 31032011 19:35
(Original post by Stonebridge)
The isolated bit in the middle (ringed) has no charge initially.(no net charge = neutral)
The battery pushes negative charge onto the right of B.
This negative charge repels electrons off the left plate of B, which becomes positive. Each negative charge on the left induces a positive charge on the right.* As the isolated bit is neutral, there must be an equal amount of negative charge that has moved to the right plate of A. (Conservation of charge)
This means there is an equal amount of positive charge on the left plate of A, and this perfectly balances the negative charge on the right plate of B.
* Induced charges are always equal and opposite.
Do you not have a decent text book that explains this?

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 31032011 21:23
(Original post by Magu1re)
Nope. It sort of treats the reader like a minor idiot and explains everything in what I would describe as "GCSE terms". What does it mean to induce charge on the other plate? And why must it always be an equal amount induced?
In the books it's usually drawn such that there is a line of force going from a negative charge on the right plate, straight across to the left plate. Where the line strikes the left plate, a positive charge is "induced". So for "4" negative charges on the right, you get 4 positive charges on the left. I know it's a bit of a hand waving explanation, but unfortunately, if you want a more rigorous explanation, it gets buried in a lot of complex maths. But read on...
{The idea of "induction" also appears in magnetism where you induce a north pole in a lump of iron by bringing a south pole near it; and in electromagnetic induction where you generate a current from a changing magnetic field}
It originates from the old idea of "action at a distance" and is a result of a force field. If you bring a negatively charged object near to an insulated conductor, there is a positive charge induced on the surface of the conductor nearest the object, and an equal negative charge on the side of the conductor furthest away from the object.
It may help you understand this if you can find something on "Faraday's Ice Pale Experiment". It was first performed in 1843 and demonstrated that the induced charge is equal and opposite to the inducing charge. (Under certain circumstances). You may also have heard of the Faraday Cage. It's related to this theory.
I have a link here
http://en.wikipedia.org/wiki/Faraday...ail_experiment
but I'm sure you can find others.
The wiki is not "idiot level" but there are better sites if you take the time to look.
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