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Electrostatic charge and law of conservation of energy

  1. Dec 25, 2009 #1
    Take a capacitor arrangement...the capacitor having 2 plates which are square shaped. Assuming one of the plates as A and the other as B.

    If A has been given a static charge, it can be said that the all energy possessed by A can be computed when A is discharged...overall discharging will release all the energy possessed by A...let this energy be x.

    Now, since B is in front of A and A has been given a charge, if B is earthed it will gain a charge equal and opposite to A (let's assume it be equal, it will be a bit less actually cause of E.F losses). Cause of this gained charge, a charge has to flow from B to earth; now energy can be harvested from this charge flow and it will be almost equal to x (assume it to be equal to x for simplicity, the difference will be very less if the right dielectric is used); thus it can be said that energy used for charging A has been recovered though this discharge.

    But still there can be one more discharge which will result in an energy output of 2x...joining of plates A and B; this energy is extra...where did it come from?

    There's another way by which this can be explained.
    Have a look at the 2 attached images.
    There are 2 curves, a red one and a blue one; the red one is fixed while the blue one has the ability to rotate along the surface of the yellow transparent cylinder (or the blue curve forms the locus of the yellow transparent cylinder.).
    Both the curves are metallic and the red curve is charged while the blue one is grounded all the time. If the blue curve is made to rotate continuously, it will charge and discharge cause of the red charged conductor depending on it's location on the yellow cylinder; energy can be harvested though these charge/discharge cycle.

    Now, assuming the blue curve has an angular momentum (large enough), under ideal conditions it will continue to rotate forever since the force of attraction between both the curves will be conservative, i.e at certain location of the blue curve, it's motion will get aided while at the remaining places it will get a resistance; overall, both equal out and it won't matter in a complete rotation of the blue curve...there will be ups and downs in it's angular velocity, that's all.

    But still we can harvest energy from the blue curve...where did this energy come from?

    This thing has screwed me up for ~2 years and I'm not getting any answers.
     

    Attached Files:

  2. jcsd
  3. Dec 25, 2009 #2

    Dale

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    This is a good example. Net energy cannot be harvested through this cycle. If the change in the kinetic energy of the rotor is negligible (i.e. spinning at a steady rate) then it takes the exact same amount of work to go from the "near" to the "far" configuration as the work that you can extract by going from the "far" to the "near" configuration.
     
  4. Dec 25, 2009 #3
    In the above arrangement, energy is not harvested using the rotation, it's harvested from the charge flow between the blue curve and earth.

    Since the direction of charge flow does not matter when it comes to harvesting energy, energy is being harvested yet the rotation does not stop. I expect it to stop with the number of rotations, but that will not happen here. Even the M.F cannot stop this since there's just one charged rotating body, so the M.F will not interact with anything. Anyway, I'm not sure about this. I would appreciate if someone puts some light on the M.F interaction since I absolutely do not know E.M.
     
  5. Dec 25, 2009 #4

    Dale

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    The direction of the current does matter. When current goes from a higher potential to a lower potential energy can be extracted, but when current goes from a lower potential to a higher one then energy is required.

    I will re-read your description to see if I am misunderstanding, but unless I misunderstood your device each half-cycle just gives you back the energy you put in the previous half cycle.
     
  6. Dec 26, 2009 #5
    Humm...ok. But this will actually produce a square wave AC...if that was so, why does AC has the ability to do work?


    If we have 2 charged plates, one negatively charged and the other positive, and if both are grounded, the the work done by the charge flow will be the same for both the cases.

    In one case, current flows form the plate to earth, and in the other from earth to the plate. The energy generated can be harvested thought the heat generated in the wires; which will be the same in both the cases. This is what I mean by the direction of current.
     
  7. Dec 26, 2009 #6

    Dale

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    I have looked at the description in your OP again in detatil. Let's say that the static element is positively charged. The rotating element begins far away and essentially uncharged. As it approaches the static element Coulombic attraction causes negative charge to accumulate. This has two effects, first, the voltage of the rotating element becomes negative and, second, current leaves the ground. Since the ground is the more positive terminal the current leaving can do work. During this half the energy in the electrostatic field of the static element is reduced.

    Now, on the second half the rotating element begins nearby and negatively charged. As it leaves it becomes less negatively charged. This is a current going in to the ground which is still the more positive terminal, so it requires energy from some other source. During this half the energy in the electrostatic field is increasing.
    AC does work on both half cycles because both the polarity of the voltage and the direction of the current are switched every half cycle, so you always have current leaving the more positive terminal.
     
  8. Dec 26, 2009 #7
    I think that earth should behave like a negative body, since charge will flow from the rotating element to earth...the direction of current will be from the rotating plate to earth; that's the reason why the electrons will be drawn from earth to give the blue conductor a net negative charge.

    Why will it require a net work to be done to make the negative charge flow to earth? If we have a negatively charged conductor, the current flows from earth to the conductor doing work in the process, same is being done here. The rotation of the blue conductor will be resisted -- yes, but in the other half cycle it will be aided by the same amount, so it should not make a difference.

    If in both the half cycles, charge is drawn from earth, the arrangement will gain a net negative charge indefinitely with time...that's not happening here.
     
  9. Dec 26, 2009 #8

    Born2bwire

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    You do a certain amount of work to charge up plate A. This is an easy exercise of bringing each charge in from infinity to its designated position, one charge at a time. But this is in the absence of our grounded plate B. Since B is in the picture here, you are doing more work, because when you bring in your charge from infinity to A, you also have to repel a charge from B to the ground. Remember, plate B was originally neutral, as you bring in an electron from infinity to A, you must also move an electron from A to ground in response to your induction.

    EDIT: It would probably be enlightening to work it out on paper for the first charge. Essentially, you will be bringing in a charge to plate A and inducing a growing "dipole moment" in B as you move an electron away in response and leave behind a positive charge. This seems conceptually similar to the polarization of a dielectric. In a dielectric, an applied field induces dipole moments in the dielectric. These dipole moments store energy and the equivalent energy stored is taken out of the net electric field by the opposing field induced by the dipole moment.
     
    Last edited: Dec 26, 2009
  10. Dec 26, 2009 #9

    Dale

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    No, earth is always a neutral body, that is why it is traditionally assigned a 0 voltage.
    Again, power is voltage times current. If you have a component like this:
    (-) --> (+)
    then that component is like a battery that provides power. If you have a component like this:
    (-) <-- (+)
    then it is like a resistor that consumes power.

    In your device the rotating plate is always the terminal at a lower potential, so it provides power when current goes from plate to ground and consumes power when current goes from ground to plate.
     
  11. Dec 26, 2009 #10
    Assume A was made to gain charge when B was not around...after it did gain the desired charge, B was put in the picture. It was bought close unearthed, then at a certain distance, B was earthed.
     
  12. Dec 26, 2009 #11
    Yup!...that's the final answer.

    The thing which you have stated above is strictly a convention used in a closed circuit, suppose in the second case, when the current passes though a resistor, it lowers in potential (-)<---(+), but if there's another resistor in series, then it will further lower the potential (-)--->(--); thus this is all related to the circuit and is a potential difference relative to the voltage source (as it approaches the potential of negative terminal); here, in case of electrostatics, the circuit is open...it's like an AC wire grounded to earth...no matter what the polarity (negative to positive) it will do work.

    In case of (-)--->(+) it means that the component is increasing the current (and thus the potential)...it does not have to do anything with static charge. BTW you cannot apply this to a charged place since a plate does not take in charge...it has charge stored in it, this is against KCL since this is not a circuit. Also the negative in closed circuits is strictly different form the negative in electrostatics, in closed circuits, negative is defined to be the terminal which has lower charge relative to the positive terminal...this is not an absolute measure...it might be that relative to earth the negative terminal has potential of 5 volts and the positive has 10 volts...thus both are charged positively yet one of them is called negative...just cause it's in relation to the other 10v terminal. In electrostatics, the potential is absolute, it measured as the deficiency/excess of electrons.

    Actually, in the arrangement, work is actually being done by the angular momentum on the field to pull the current (earth)(+)---->(-)(plate) in the other half cycle...but work done by the angular momentum can be harvested tough a resistor which is bidirectional; in the next half cycle energy can again be harvested through the resistor but the angular momentum has also increased (earth)(-)<----(+)(plate); thus one can say work is being done on the system by the field. If we state that the red plate is negatively charged, the direction of current will flip but the work still will be done by the angular momentum to draw current (earth)(-)<----(+)(plate) despite the current traveling from the plate to earth.

    Actually, there's a bit of confusion here, work can be done in 2 ways by the whole system -

    1) By the stored electrostatic charge.
    2) By the angular momentum.


    I think it can be made more clear if you define though which medium work is being done by the system or on the system like I defined above.
     
    Last edited: Dec 26, 2009
  13. Dec 26, 2009 #12

    Born2bwire

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    Same thing occurs by virtue of superposition. If you brought B in from infinity grounded, it would be the same as bringing in all the charges from infinity to A while B was static. If you bring B in ungrounded and then ground it, you still are bringing in a ground line from somewhere. On the ground line you would be moving charges as you moved it into the picture to connect it to B. Just before it is connected you would have a potential difference that would facilitate the moving of the charges from B to ground.
     
  14. Dec 27, 2009 #13
    Yes, but I sorta didn't get your point; I do not understand this -

    The charges will move automatically, there will be no work done on the charges so as to move them, instead work should be done by the charges.

    Also notice, after charging A, it has been completely isolated. As we bring B close to A, there will be no force on B, if we ground it, then there will be a force, but it wont do any work cause B has been fixed in place.
     
  15. Dec 27, 2009 #14

    Born2bwire

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    You are still doing work when you physically bringing in the ground line. You cannot just magically ground your plate. You need to bring in a physical wire to short it to ground.

    Let me ask you a different question. Say I have an infinite slab of perfect conductor. It goes from z=0 to infinity say in terms of thickness. What is the work that is done if you bring in a charge from infinity to say z=-a if there is no slab, if the slab is present, and if the slab is an imperfect conductor?
     
  16. Dec 27, 2009 #15
    Suppose we have a terminal which's supposed to be earth...we have to connect the wire emerging form plate B to this terminal...on doing so, at most the 2 open terminals (the ground one and the one emerging though plate B) will attract each other doing work; I would have expected them to have repelled, but that's not the case.

    1) If there is no slab, the work done will be 0.
    Now I'm not sure about the last 2 cause problems still persist in that area.
    2) There will be a work done if there is a conductor...or actually work will be done on the charge.
    3) Still there will be a work done in the same manner as in the previous case.

    I think you're telling me about the charge separation within the conductor...yes there will be forces acting on a very thick but neutral conductor, but it will be aiding the motion of plate B towards A doing work...here again I would have expected a repulsion if law of conservation of energy is to be see.

    If plate B is sufficiently thin, we can ignore this force.
     
  17. Dec 27, 2009 #16

    Dale

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    Sorry, I was not clear. Above (+) means "higher potential" and (-) means "lower potential". They are not intended to indicate positive and negative voltages wrt ground. So for your second example (-) would be the higher potential (+) and (--) would be the lower potential (-).
    This is your key error. This is simply wrong. I can derive this rigorously later if the following explanation is insufficient:

    Remember that the change in voltage is proportional to the E-field and that the E-field is proportional to the force. Now, remember also the basic definition of work, W=f.d, if the displacement is in the same direction as the field force then the field is transfering energy to the charge which in turn can be transfered to the environment, but if the displacement is in the opposite direction as the field force then the charge is losing energy to the field which must be supplied by the environment. The current is proportional to the displacement, so if you reverse the current in relation to the change in potential then you switch between energy going in or out. You simply cannot harvest energy both ways.

    Note that none of the above has anything to do with Kirchoff's laws or circuit theory. It is all simply due to the definitions of the various quantities, and it applies to a differential element in an electrostatics problem like this one just as well as to a lumped element in a circuit problem.
     
  18. Dec 27, 2009 #17
    Lightning is negative in polarity WRT earth...i.e current flows from earth to the atmosphere...but this results in work being done by the lightning.

    We have positive lightning also with the same effect.

    Comparing this fact to the rotation, the angular momentum is doing work on the blue curve...which is analogous to the charge here.

    Current is proportional to displacement per unit time.

    Reverse the current...I didn't understand. Which current are you referring to?

    IMO, in the other half cycle, we do not have to do additional work (it will be done by the angular momentum) to make the current flow to earth from the rotating conductor.
     
  19. Dec 27, 2009 #18

    Born2bwire

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    This is a trivial problem. In the first case no work is done since there is no external field exerting a force. In the second case, for a grounded conducting sheet or slab (the thickness does not matter at all), the amount of work to bring a charge a distance a above the sheet is,

    [tex] W = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{4a} [/tex]

    This can be found by use of an image charge. In the second case, we suddenly find that work occurs to arrange a charge near a grounded conductor due to the field induced on the conductor. In fact, the work done is half that if we had placed a true charge at the location of the image and brought our other charge in from infinity. This is because moving charges in a conductor takes no work, the potential difference inside a conductor is always zero. That is, we only worked on the charge we brought in from infinity, the charge distribution inside the conductor was a free lunch. Thus, if we allow for a loss in our conductor then the work done will be something like,

    [tex] W = -\frac{1}{4\pi\epsilon_0}\frac{q^2}{2a} [/tex]

    Because now with a load, we have to do work to move our charges around the conductor.

    This is one of the points I believe that you are not considering in all of this. You keep talking about doing work but your problems here are not expending any work, you have yet to actually consider the consequences of hooking up a load in your problems.

    So, when you have your loads, you do double the work because you have to move your charges through the load as you bring your charges on to plate A. It doesn't matter how you convoluted you try to connect the ground to plate B, as long as you can easily show the physics with one grounded configuration it will apply to any other scenario that you can think of when it comes to hooking up a ground. That is:

    In the absence of any loads, you do not actually do any work to rearrange the charges on your grounded plate, so the doubling of your power is ficticious because you end up double-counting. However, if you were to consider a load and thus have to expend work in rearranging the charges on the grounded plate, that work is done when you bring the charge into position.
     
  20. Dec 28, 2009 #19
    I did not know it was grounded.

    Why will work have to be done to bring the charge from infinity?...work will be done on the charge since the infinite conductor will attract it.

    I do not know the image charge method, so it will be good if that is out of the picture.

    A load was put after equilibrium was attained.

    art. Actual case -
    The actual case that I am putting forward is that the point charge is brought from infinity to a neutral ungrounded infinite conductor and made to stop at a distance...no work will be done in this case (or will be?).

    Now, what will be the work done to ground the conductor? IMO, no work will be done. If we bring a wire (which is grounded) to this infinite place, the wire will itself will get inducted with charge and actually get attracted towards the plate, so no work will be done on the wire, work will be done by the wire.

    Yes, that's the problem. With my understanding work does not have to be done but work is being done...but actually work will have to be done, but I do not understand why...at most work will be done by the charge, not on the charge.

    The potential scopes of doing work considering the "actual case" article, is bringing the charge to the conductor (since it is a neutral unearthed conductor, it cannot apply force on the charge)...that is one of the 3 scopes of doing work (the other 2 being joining of the point charge and the conductor, and the last being joining of the wire to the infinite plate, which will again will not cause any work to be done) , but since there is no force on the charge how can work be done to bring the charge to the conductor?

    After everything has it's static position (the charge, and the plate), then the load will come into action.

    The load will be applied...of suppose x ohms in series to the grounding wire. This will not change the nature of the wire to get attracted towards the infinite plate (thus no work will be done on the wire to bring it tough the plate; again). As the charge flows work will be done on the resistor, generating heat. With time all the charge will flow.

    After this, the point charge needs to flow into the the plate (right now, it has started experiencing force)...this will be done by a wire which will connect the point charge to the plate.

    What will be the work done on putting the wire between the 2?...I think there will be no work done...at most the wire will be attracted, it cannot be repelled cause it's neutral; so there will be no force application on it (or might be force application by it) so no work will be done on it.

    So, question here is, where how is work being done in any form?
     
  21. Dec 28, 2009 #20

    Dale

    Staff: Mentor

    Are you claiming that there is lightning where the current goes from a negatively charged region to a positively charged region? If so, please back it up with a reputable source. If not, do you now understand that electrical power depends on both the voltage and the current and that the direction of the current in relation to the change in voltage does matter?

    Sure, that is fine. The energy can be supplied by the environment either mechanically or electrically. The point is that if you do no net mechanical work then there is no net electrical power generated.

    Correct, which is why current times voltage gives power instead of energy.

    The current from the plate to earth.

    This is fine, and in the process of letting the angular momentum do the work the kinetic energy will decrease, thus conserving energy.

    http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" [Broken], in your device there are 4 possible sources, electrical power (current times voltage), KE from the rotating mass, the electrical field, and mechanical work. So this is a fairly complicated device and it is easy to lose track of something. I would recommend simplification in order to avoid confusion. Think about which energy transfer mechanism you really want to understand and eliminate the others by some simplification.
     
    Last edited by a moderator: May 4, 2017
  22. Dec 28, 2009 #21

    Born2bwire

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    If you do not know about image charges then you need to start studying more electrostatics; this is first year material. Your original problem stated the following:

    It is obvious from this that you had the load placed between the grounded plate and the ground from the outset, otherwise there would be no energy harvested from the current on B's ground line. As I explained above, once you place the load there, you have to do work because now there is a potential difference between B and the ground that you have to move the charges through. This work is done when you bring in the charges to charge A. Without the load, you would only need to do work upon the charges that you bring in, the induced charges on A would not require any work. And yes, you do have to expend work to bring in charges onto A regardless of the grounded plate B. It should be an easy exercise to calculate the work expended to bring in a second charge above an infinite conductor in the example I gave previously.

    In truth, your example is a very close analog to a parallel plate capacitor.
     
  23. Dec 29, 2009 #22
    No, I claimed that that in negative lightning current goes from the ground (positive) to the negatively charged region, doing work...i.e work does not have to be done to make the charge flow...work is done by lightning...the negative charge.

    Yes, that is the issue. Following the current arrangement, by my understanding, you do not have to do work. The angular momentum lost by the other half cycle is gained back by the first half cycle.

    In the first half cycle, work is done on the system by the field, resulting in 2 changes -

    1) Increase in angular momentum.
    2) Flow of current.

    In the other half cycle, there is a flow of current but by virtue of the stored angular momentum -

    3) Angular momentum does work on the current to make it flow...reducing the angular momentum by the same amount as it was increased in the first half cycle.
    4) Energy can be harvested though this current flow (and actually you will be harvesting the energy possessed by the angular momentum stored though this flow)

    Now, go back to step 1.

    There is a resistance on the wire which connects the ground and the plate...if the resistance is very high, it will require increased time for the charge to flow to earth, as a result the the plate will not completely get charged as it completes the first half cycle; thus as compared to a situation where there was no resistance in the wire, the increase in angular momentum will be less.

    Proceeding to the other half cycle with this increased resistance, since the rotating conductor has not completely discharged, there will be a reduction in the force which resists angular momentum (by the same amount as there was reduction in the attracting force in the first half cycle). All this converts to lower reduction in angular momentum (by the same amount as there was a reduction in the increase of the angular momentum in the first half cycle.)...thus high resistance too should not matter.

    But there's on major factor missing in the above analysis. It happens that in a conductor, a field x falls such that it has the ability to induce the charge y on the conductor, if the conductor's grounding connection has a high resistance in between, it will take considerable amount of time for the conductor to gain the charge y. Suppose time t has passed and the conductor has gained a charge y/10 in the time t...at time time t, if the E.F x is reduced by a very small amount, dx, then the conductor will still continue to charge since the E.F x - dx still has the ability to induce a charge y - dy (where dy is an infinity small quantity) on the conductor. I've assumed the current charge on the conductor (after passing though time t) to be less than y - dy.

    This might be the crack to this concept. After just finishing the the first half cycle (still assuming a very high resistance), and entering the other half, the E.F will start decreasing on the rotating plate, but since the resistance is very high, it (the rotating plate) will continue to charge...let's take the worst case scenario, the resistance is so high that throughout the second cycle, the rotating plate continues to charge; this will result in higher reduction of the angular momentum in the second half cycle but as the machine enters the first half cycle (again), this time the increase in angular momentum will be higher cause of the increase in initial charge...thus it is really making no difference...the extra reduction in the angular momentum is compensated by this extra increase in the angular momentum.

    If the plate starts to discharge while it is somewhere in the second half cycle, again by the time this cycle ends, the plate will discharge less as compared to when the resistance was low, thus in the next cycle, the attractive forces will increase the angular momentum by a higher amount as compared to when the machine started in the first half cycle.

    So this theory will not prove the thing wrong..

    But the K.E will increase equally in the other half cycle cause of the attractive forces making no difference.

    I've even tried to analyze after putting high resistance in the wires...but as I've stated above, it won't matter.

    You're referring to the capacitor arrangement that I'm discussing with born2bwire...if that is proved wrong somehow, this will automatically get proved wrong.
     
  24. Dec 29, 2009 #23

    Vanadium 50

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    Whoa...whoa...whoa!

    This is starting to sound less and less like a request for knowledge and more and more like a defense of a perpetual motion machine. Please note that this is ahttps://www.physicsforums.com/showthread.php?t=5374".

    You have been given two very good pieces of advice that you shouldn't ignore.

    One is to simplify the system: by making it more and more complex, all you do is increase the likelihood of confusing yourself and making a mistake. There's a reason why perpetual motion machines have all the bells and whistles they do: they serve to confuse the inventor, or the potential investors, depending on whether the inventor is malign or simply befuddled.

    The other is to learn how to use tools you are unfamiliar with - you pooh-poohed the image charge method because you hadn't seen it before. If you want to understand something, you need to put some work into it yourself. If you're not willing to do that work, why should everybody else do the work to explain it to you.

    If this thread continues in this vein, it will have to be closed.
     
    Last edited by a moderator: Apr 24, 2017
  25. Dec 29, 2009 #24

    Dale

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    It seems like you understand the idea of electrical drag, but just don't apply it correctly. As long as you are supplying power to the environment there is a drag force which resists the motion. It always acts to slow you down. So the increase in angular momentum on the downstroke will be less and the decrease in angular momentum on the upstroke will be more. If you take net electrical power out over one cycle you will lose net KE.

    Please see this link for a general proof that energy is always conserved in electromagnetism:
    http://farside.ph.utexas.edu/teaching/em/lectures/node89.html
    If you ever believe that you have a situation where energy is not conserved then you immediately know that you are not obeying Maxwell's equations.

    In any case, I am starting to agree with Vanadium 50. The physical principles have been explained to you quite clearly. This is starting to sound less like someone trying to learn and more like someone trying to prove that their pet perpetual motion machine could work.
     
  26. Dec 29, 2009 #25
    I apologize for the misunderstanding, A already had a charge.

    Why will I have to move it? Suppose A has been given a positive charge, then that will automatically develop a positive charge on the back side of B...which if earthed will make a current flow which will do work.

    I think I'm getting close to understanding this, thanks...with such effort, I will hopefully, understand.

    No no, again sorry for the misunderstanding, A was previously charged.

    Yes, I agree with that. But the charge stored in A does have a limit of it's energy...I mean there is some limited amount of energy in it that can be harvested...that is x...but we get back 3x doing the procedures...it should have been x.

    I know this is an argument, that's why I also posted the machine...the red plate there is A, and it has to be charged only one.

    If the infinite conductor is not grounded, then work will not be done on doing so...this is the actual case (art. Actual case)
     
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