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Energy stored in a capacitor

  1. Aug 31, 2012 #1
    This page

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

    talks about how the battery does work to move a charge from one plate to the other plate. Can charge jump across the gap between the plates? I was under the impression that the plates get charged due to electrons traveling towards/away from the plates via the wires which are connected to the battery and not due to electrons jumping between the plates.

    Thanks for your help!!
     
  2. jcsd
  3. Aug 31, 2012 #2

    Jano L.

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    Gold Member

    You are right, there is no current throught the space between the plates. The battery works against the electrostatic forces that tend to return the electrons to the positive plate through the wire.
     
  4. Aug 31, 2012 #3
    No where in your link does it say that the electrons 'jump' from one plate to the other.
     
  5. Sep 1, 2012 #4

    vanhees71

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    2016 Award

    Of course, when you charge a capacitor the electrons flow through the wires and not through free space. There's a "displacement current" however, which is the most important contribution of Maxwell's work to electromagnetism.

    Let's answer your question by a little calculation considering the charging of a capacitor in quasistationary approximation, which is well justified for this problem under most practical circumstances, i.e., here we neglect the displacement current and inductance effects.

    Take a battery plugged to an initially uncharged capacitor. Let [itex]R[/itex] be the total resistance of the wires. Then the current is given by the time derivative of the charge on the capacitor plate connected to the positive charged plate (charge conservation). Kirchhoff's Law thus tells us
    [tex]CQ+R \dot{Q}=U.[/tex]
    Here [itex]U[/itex] is the voltage of the battery (considered constant over time, which is an approximation too) and [itex]C[/itex] is the capcitance of the capacitor. This equation with the initial condition [itex]Q(0)=0[/itex] has the solution
    [tex]Q(t)=\frac{U}{C} \left [1-\exp \left(-\frac{R}{C} t \right ) \right ].[/tex]
    Now the current is given by
    [tex]i(t)=\dot{Q}(t)=\frac{U}{R} \exp \left(-\frac{R}{C} t \right ).[/tex]
    The total power put into the system is [itex]P_{\text{tot}}(t)=U i(t).[/itex] Thus the total energy delivered by the battery is
    [tex]E_{\text{tot}}=U \int_0^{\infty} \mathrm{d} t i(t)=\frac{U^2}{C}.[/tex]
    One part of this energy is stored in the capacitor, the other is converted to heat in the wires. The latter dissipative part is given by Ohm's law: [tex]P_R(t)=R i^2(t)[/tex] and the total energy converted to heat is
    [tex]E_{\text{heat}}=R \int_0^{\infty} \mathrm{d} t i^2(t)=\frac{U^2}{2C}.[/tex]
    Thus the rest of the total energy must be stored in the electric field between the capacitor's plates,
    [tex]E_{\text{el. field}}=E_{\text{tot}}-E_{\text{heat}}=\frac{U^2}{2C}.[/tex]
     
  6. Sep 1, 2012 #5
    If charge does jump, it's arcing like a spark plug, not the normal design plan for a capacitor.
    What DOES traverse the gap is the electromagnetic field...that is what enables charge accumulation on one plate to push around charge on the other plate....
     
  7. Sep 1, 2012 #6
    Not all of the energy that is not stored on the capacitor (1/2QV) is lost as heat due to resistance. The explanation must include a description of what happens when resistance is zero.
    There must be another means of energy dissipation. This is not shown in the equations here.
     
    Last edited: Sep 1, 2012
  8. Sep 1, 2012 #7

    Dale

    Staff: Mentor

    It doesn't matter how small the resistance is, half of the energy is lost. The loss is not a function of the resistance.
     
  9. Sep 1, 2012 #8
    I know that, it is in the maths....
    .where is the missing energy? I don't think it is all heat from resistance. Vanhees equation only has resistance in it as a means of energy loss.
    What if there s no loss of energy by heating because of resistance ?
     
  10. Sep 1, 2012 #9

    Dale

    Staff: Mentor

    If there is no heating then it radiates away. As the resistance becomes lower the time constant becomes smaller and the radiation becomes greater. In circuits, energy loss due to radiation is treated as an equivalent resistance.
     
  11. Sep 1, 2012 #10
    If the energy radiated away is 'not heat' what kind of energy is it?
    Is there any easy way to calculate the equivalent resistance that is resposible for the radiated energy?
    Does this mean that the lost 1/2QV is not equal to I^2R in the capacitor/resistor circuit?
     
  12. Sep 1, 2012 #11

    Dale

    Staff: Mentor

    Electromagnetic.

    Not that I know of. Calculating the resistance of an antenna structure is not trivial in general.

    No, it means that R can be complicated for very high frequency (short duration) discharges.
     
  13. Sep 1, 2012 #12

    vanhees71

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    Of course, the effect of "energy loss" through radiation is not included in my little exercise. This is what I explicitly neglected when I've thrown away the discplacement current. For a complete calculation, also the assumption of "compact" electric circuit elements like resistors capacitors, etc. doesn't make sense anymore since if there are effects that are so fast that they lead to noticeable radiation the corresponding wave lenghts are not large compared to the spatial extension of the circuit, and one must take into account the full geometry in terms of boundary conditions of the electromagnetic field.
     
  14. Sep 2, 2012 #13
    I think I now get it!!
    Is it fair to say that the energy not stored on the capacitor is lost as heat and radiation(electromagnetic)
    The smaller the resistance the less the heat and the greater the radiation ??
    And the greater the resistance the less the radiation and the greater the heat ??
    I can understand how to calculate the heat generated if the resistance is known but is it correct to say that calculating the radiation is more difficult?
     
  15. Sep 2, 2012 #14

    Dale

    Staff: Mentor

    Yes, this seems all correct to me.

    If the radiation resistance is known then calculating the energy lost to radiation is just as easy as calculating the energy lost to Ohmic dissipation. Calculating the radiation resistance is difficult since it depends on the geometry of the structure and requires Maxwells equations. Here is a brief description.

    http://en.wikipedia.org/wiki/Radiation_resistance
     
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