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    Hi all,

    I'm struggling my way through the capacitors chapter in my OCR Physics and have a query with this question:

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    Part of the answer is that the total charge remains constant. Which, as I understand, is from Kirchoff's Charge law that charge is conserved.

    I have done previous questions exactly like this but without the resistors and normally I use the conservation of charge rule which makes sense.

    But in this question there is a resistor in the 2nd loop when the switch is thrown and I can't understand how the total charge remains constant. As surely the resistor drops the voltage and uses up energy in the process, and as Q=VC; if V drops so should the charge right?

    If someone can enlighten me I'd be very grateful.
    Malc.
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    Thing is, the equation you quote is for one capacitor. Once you flip the switch you have two in the circuit and charge can flow from one to the other.

    The charge on C1 drops because half of its charge is transferred to C2. The total charge remains the same but V = Q/C for each capacitor is now half what it was for C1 before the switch was flipped.
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    (Original post by rsk)
    Thing is, the equation you quote is for one capacitor. Once you flip the switch you have two in the circuit and charge can flow from one to the other.

    The charge on C1 drops because half of its charge is transferred to C2. The total charge remains the same but V = Q/C for each capacitor is now half what it was for C1 before the switch was flipped.
    Thanks for the reply, but maybe I didn't explain my question properly.

    What I'm asking is once the switch is flipped, why doesn't the resistor in the parallel capacitor setup lower the charge or voltage stored?

    I would have expected the resistor to lower the voltage and prevent the capacitors getting an even split of the 10V.

    It seems as if the resistor being there has no effect, beside the charge/discharge time.
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    (Original post by Sir_Malc)
    Thanks for the reply, but maybe I didn't explain my question properly.

    What I'm asking is once the switch is flipped, why doesn't the resistor in the parallel capacitor setup lower the charge or voltage stored?

    I would have expected the resistor to lower the voltage and prevent the capacitors getting an even split of the 10V.



    Voltage acts like a pressure pushing charge around a circuit. Only when enough charge is transferred to equalise the p.d.'s between the two capacitors (V = Q/C), will the current cease to flow.

    A potential difference will only appear across the resistor when a current flows through it. V = IR

    At the start, when the switch is first closed, there is an imbalance of charge between the two capacitors. Like-charges repel and so electrons from the from the first capacitor will readily rush onto the second capacitor because the only thing limiting the flow at the start is the resistor.

    As the charge on the second builds, the charge on the first must deplete by the same quantity. (Total charge is conserved). However, the building charge on the second capacitor, will now also start to increasingly repel charges flowing in from the first. The p.d. across the second capacitor builds while the p.d. across the first falls. Which means that the current flowing through the resistor when the switch is closed, starts off as [IR = Vc1 / R] (all of the p.d. across the first capacitor is across the resistor), but as the charge on the second capacitor (and hence p.d.) builds, the current flowing through the resistor tapers off. [IR = {Vc1 - Vc2} / R]

    Current will only stop flowing when the p.d's acroos the two capacitors reach equilibrium. i.e. when Vc1 = Vc2

    At that time, there is no difference in voltage pressure and hence all charge forces are in equilibrium and no current will flow form one to the other.

    It seems as if the resistor being there has no effect, beside the charge/discharge time.
    In essence, you are correct. The resistor regulates the rate at which charge equalises (di = dq / dt) between the two capacitors. A p.d. only appears across the resistor when charge flows: When the capacitor p.d.'s equalise, the voltage on the two capacitors must be the same, no further charge can flow, no p.d. is developed across the resistor.
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    (Original post by uberteknik)
    Voltage acts like a pressure pushing charge around a circuit. Only when enough charge is transferred to equalise the p.d.'s between the two capacitors (V = Q/C), will the current cease to flow.

    A potential difference will only appear across the resistor when a current flows through it. V = IR

    At the start, when the switch is first closed, there is an imbalance of charge between the two capacitors. Like-charges repel and so electrons from the from the first capacitor will readily rush onto the second capacitor because the only thing limiting the flow at the start is the resistor.

    As the charge on the second builds, the charge on the first must deplete by the same quantity. (Total charge is conserved). However, the building charge on the second capacitor, will now also start to increasingly repel charges flowing in from the first. The p.d. across the second capacitor builds while the p.d. across the first falls. Which means that the current flowing through the resistor when the switch is closed, starts off as [IR = Vc1 / R] (all of the p.d. across the first capacitor is across the resistor), but as the charge on the second capacitor (and hence p.d.) builds, the current flowing through the resistor tapers off. [IR = {Vc1 - Vc2} / R]

    Current will only stop flowing when the p.d's acroos the two capacitors reach equilibrium. i.e. when Vc1 = Vc2

    At that time, there is no difference in voltage pressure and hence all charge forces are in equilibrium and no current will flow form one to the other.



    In essence, you are correct. The resistor regulates the rate at which charge equalises (di = dq / dt) between the two capacitors. A p.d. only appears across the resistor when charge flows: When the capacitor p.d.'s equalise, the voltage on the two capacitors must be the same, no further charge can flow, no p.d. is developed across the resistor.
    Thanks for the reply.

    When current is flowing through the resistor during discharge/charging from C1 to C2, would the resistor heat up a bit; losing a bit of energy and therefore decreasing the total Voltage stored between the two capacitors before they equalise?

    This is my main problem. To me the resistor should be taking some energy out of the system, even if it's a tiny amount in this case.
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    (Original post by Sir_Malc)
    Thanks for the reply.

    When current is flowing through the resistor during discharge/charging from C1 to C2, would the resistor heat up a bit; losing a bit of energy and therefore decreasing the total Voltage stored between the two capacitors before they equalise?

    This is my main problem. To me the resistor should be taking some energy out of the system, even if it's a tiny amount in this case.
    You are correct to say that work is done when the charges move through the resistor.

    It's important to recognise the definition of Voltage, which is Joules per Coulomb of charge. It's also important to realise that voltage is referred to as 'potential' for a reason.

    When the voltage potential falls as the first capacitor discharges, it's another way of saying the potential energy stored in the first capacitor has fallen. i.e. its started out as say, 10 Joules per Coulomb of charge, (10V) potential and falls to say 5 Joules per Coulomb (5V) as the capacitors reach equilibrium. (These are completely arbitrary numbers to illustrate only)

    The quantity of charge moving through the resistance is how that potential is expended.

    Thing is, work had to be performed to move the charge onto the first capacitor. That came from the power supply. The energy form the power supply came from somewhere else (chemical energy from a battery or dynamo, or steam turbine etc etc)

    An analogy is gravitational potential. There has to be a difference in potential when an object moves form a higher potential to a lower potential. The potential is converted to kinetic in motion. Potential is zero when the object comes to rest.

    Gravitational potential energy is always with reference to position (height). So it is with voltage potential.

    In the case of voltage potential, as long as there is a 'potential difference' then there is a potential to perform work and expend energy. i.e. voltage is always with reference to some other energy level.
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    (Original post by uberteknik)
    You are correct to say that work is done when the charges move through the resistor.

    In the case of voltage potential, as long as there is a 'potential difference' then there is a potential to perform work and expend energy. i.e. voltage is always with reference to some other energy level.
    So in the question I'm doing; how come some (or all) of the 10V potential energy that C1 has, is not used up going through the resistor to reach C2.
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    (Original post by Sir_Malc)
    So in the question I'm doing; how come some (or all) of the 10V potential energy that C1 has, is not used up going through the resistor to reach C2.
    The energy potential stored by the capacitor is most definitely used. Voltage potential is analogous to pressure. The potential energy stored on the capacitor is E = \frac{1}{2}QV. i.e. The energy is a conserved quantity and is equivalent to the work done moving the total collective charge quantity of trillions of electrons onto the capacitor against the electric field generated as the charge builds.

    In the original question, some of the stored charge moved from one capacitor to the other. The proportion of the charge that transferred via the resistor did the work that heated the resistor.

    Let's be clear about conservation of charge:

    The electric force exerted by charge carriers - electrons and protons - is a fundamental property of matter. The same as saying energy can nether be created nor destroyed. It's the limit of current scientific understanding to know why any of the four fundamental forces exist; we don't know why any more than science knows why the universe exists. All we know is that they do because we observe them to exist.

    We must simply, at present, be content (until science discovers something as yet undiscovered) to accept that electrons carry the force described as electromagnetic. Each and every electron exhibits that property and will repel each and every other electron, but also will attract each and every oppositely charged proton, ad inifinitum.

    So far, nothing is observed to contradict the fact that the electric force never decays or that electrons themselves are not made of smaller particles.

    That the electric force is a fundamental property of the universe, is an inescapable observed fact, true for each and every charge carrier. Electrons have never been observed to decay and are postulated to have a lifetime of at least 6.6x1028 years. Putting that into context, any given electron will repel like charges and attract unlike charges for longer than 1 billion trillion times the current age of the universe since the big bang.

    The electric force stays with the electron forever (or at least 6.6x1028 years at any rate). It does not decay, get smaller, or can be separate from the particle. The force exerted between charge carrying particles is there forever.

    The fundamental property of charge is conserved. It's only the motion of charge that causes useful work to be done. That only happens when enough charge can flow from one place to another.
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    (Original post by uberteknik)
    The fundamental property of charge is conserved. It's only the motion of charge that causes useful work to be done. That only happens when enough charge can flow from one place to another.
    This last part makes more sense to me now.

    The electrons are doing work on the resistor when they are flowing between capacitors, but once the charge on the capacitors reach equilibrium the flow of current stops and work ceases to be done, but the charge carriers (electrons) still have the potential to do the same amount of work if they were to get moving again. But to get them moving again requires an input of energy like the flicking of the switch. Otherwise you'd have a free energy machine just moving current back and forth between capacitors.

    Thanks for your detailed responses.
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    (Original post by Sir_Malc)
    This last part makes more sense to me now.

    The electrons are doing work on the resistor when they are flowing between capacitors, but once the charge on the capacitors reach equilibrium the flow of current stops and work ceases to be done, but the charge carriers (electrons) still have the potential to do the same amount of work if they were to get moving again. But to get them moving again requires an input of energy from an external source like a power supply to create a new potential difference. Otherwise you'd have a free energy machine just moving current back and forth between capacitors.

    Thanks for your detailed responses.
    Good summary. But note the amendment in red to your reply in the quote box above.


    Voltage (Joules per Coulomb) potential can only be stated between two points - one is the reference (usually ground) and the other is the energy potential per quantity of charge.

    If two places are at the same voltage potential when referenced to the same place (point in a circuit), then no potential difference exists between them. there is no potential for work because the electric charge force is in equilibrium and no flow of charge can occur.

    It's only when externally supplied work adds charge so that the two measured places (both referenced to a single point) becomes unbalanced, will there exist a potential difference, which can then be used to perform useful work.
 
 
 
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