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How electric fields store energy in a capacitor

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data

    The Question is here:

    wIlUZMM.jpg

    imgur link: http://i.imgur.com/wIlUZMM.jpg

    2. Relevant equations

    [tex]U_c = \frac{\kappa \epsilon_0 A d E^2}{2}[/tex]

    [tex]U_c = C \Delta V_c[/tex]

    [itex]\kappa[/itex]: dielectric constant

    [itex]\epsilon_0[/itex]: permittivity of free space

    [itex]C[/itex]: capacitance

    3. The attempt at a solution

    I understand why the answer to (a) is no, conservation of mass, and the charge can't flow anywhere since the system has been isolated.

    But I got completely the opposite answers to (b) and (c) from a solution I found posted online.

    My answer:

    (b) The electric field strength decreases by a factor of [itex]\frac{1}{\sqrt{2}}[/itex]. This is required by the conservation of energy and the equation [itex]U_c = \frac{\kappa \epsilon_0 A d E^2}{2}[/itex]. Since the [itex]U_c[/itex] cannot change, nor can the [itex]\frac{\kappa \epsilon_0 A}{2}[/itex] all that [itex]U_c[/itex] is stored in the electric field. After all, a certain amount of work had to be done on the capacitor by a source to charge it to a particular voltage, and that work/energy can't just disappear. Will increased separation there is now more space for the electric field to "change the shape of", so now any given set of equipotential surfaces are now more spread out, lower V per meter, but more distance, lower force per Coulomb, less force on a given charge, but over a greater distance to give an overall same [itex]\Delta K[/itex].

    (c) No, [itex]\Delta V_c[/itex] stays the same as it is proportional to the work done by the original energy source in separating the charges across the capacitor, so conservation of energy would prevent [itex]\Delta V_c[/itex] from changing.

    The answer in a solution I found online is:

    SMSdxiO.jpg

    imgur link: http://i.imgur.com/SMSdxiO.jpg

    The problem is, these solutions don't explain very much conceptually.

    Firstly, am I wrong? Secondly, why? Why is the answer the opposite of what I've answered?
     
  2. jcsd
  3. Oct 24, 2015 #2

    Simon Bridge

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    When you separate the plates, you do work... so you add energy to the system.
    Remember, opposite charges attract.
     
  4. Oct 25, 2015 #3
    b: If you mean the electric field between the plates- yes it changes with the distance. When u separate them further apart, the fistance between them is greater and the force of attraction weakens. You can see that fro coloumbs law where F=q1.q2/4p.E.r*2 where r is the distance between two charged particles. You can also see the meaning of distance from C=E. A/l(thats the capacitance of a capacitor formula). When the plates are separated further appart the attraction force between them gets weaker and weaker until eventualy they are so further apart that that force doesnt exist. AND STILL as u said the individual plates remain charged for there is nowhere that charge can go. About c: well you should realy have in mind that im not physisist but rather electro-man:-D so i might be wrong but i dont think so. So about c im not sure, but since the electric field is weakend, then when u connect the plates electricaly, the attraction force wont be that strong hence the voltage will fall down.
     
  5. Oct 25, 2015 #4
    Actualy about c, i think that the distance and attraction force takes part only in charging, and it bassicaly stores more or less charge depending on the distance between the plates. Already charged to some voltage, and since we said that the charge remains nearly the same, when u connect the plate together they still have that unequilibrium between them( the one plate is still positive and the other negative) they still habe that same potential difference between them and the answer is yes, the voltage will remain the same
     
  6. Oct 25, 2015 #5

    Simon Bridge

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  7. Oct 26, 2015 #6

    andrevdh

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    Maybe you should start out by considering what do capacitance measure?
     
  8. Oct 26, 2015 #7

    andrevdh

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    Ok the charge stays the same, but what happens to the voltage and why?
     
  9. Oct 26, 2015 #8
    "The charge stays the same" and since it is the same, that means that both plates still hold the same potential difference( one plate is still more negative than the other and the other still more positive, and when u connect them current will flow until both plates are at the same potential aka V=0. Its like in the batteries.
     
  10. Oct 26, 2015 #9

    andrevdh

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    Have another go at it after looking up what electric potential difference signifies.
     
  11. Oct 26, 2015 #10
    Uhh.. what?
     
  12. Oct 26, 2015 #11

    andrevdh

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    What does electric potential difference measure?
     
  13. Oct 26, 2015 #12

    andrevdh

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    As you separate the charges (or the plates), would the electric potential energy of the charges increase, decrease or stay the same?
     
  14. Oct 26, 2015 #13
    If you know what atom is, an atom has a core made of protons, hold together by the weak nuclear force.These protons have positive chage and that charge attracts electron(wich have negative charge) that began orbiting around the core. Every electron attracted makes the core less attractive, A core will attract as much electrons as it needs to cancel its positive charge. And then you have an atom(take in mind that becouse of that the atom is neutral). Now if you remove one electron from the atom, the positive charge of the core is now uncompensated wich means it is able to attract another free electron. Imagine two points( for an example A and B) connected with a wire. And lets say that A have more uncompensated atoms than B. These uncompensated atoms from A exsert a force of attraction on B and beggan taking electrons from it. BUT as B looses electrons its positive charge grows. And u can say the same abou A: as A accepts electrons, the amount of the uncompensated atoms slowly decreases aka. A moment will come when B have the same amount of positive charges as A.That means that both points have the same potential. Its almost that u cannot say where an electron will go now since both A and B exsert the same force of attraction on it. AT that point the potential difference between the points is 0v. An electric potential is the work done by the field to move an electron from B to A. But better open a new thread about that so we dont spam that guy's thread
     
  15. Oct 26, 2015 #14

    andrevdh

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  16. Oct 26, 2015 #15

    Simon Bridge

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    Although the charge is the same, the potential difference does not. This is easy to see by building a capacitor and charging it and measuring.

    It takes more wotk to move the same charge between more distant plates, therefore, the more distant the plates, the bigger the potential difference.
     
  17. Oct 27, 2015 #16

    andrevdh

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    For b make a drawing of the capacitor with a small + charge in the middle between the two plates.
    Would the force on the small charge increase, decrease or stay the same if distance between the plates is increased?
     
  18. Oct 29, 2015 #17
    Think about like this, the charges on those two plates are attracting to each other, so if those two isolated plates are reconnected into a circuit, the charges will flow back onto the plates to finally give V=0, but while this is happening, the force of attraction across the capacitor (between the plates) will also be resisting the force of motion to return the charges to an evenly distributed state. This force of attraction between the plates opposing the force moving the charges back through the circuit will reduce the amount of work a current flowing in that way would be able to do. So the further apart the plates are, the less force of attraction between the plates is giving the back current through the reconnected circuit less opposition, enabling it to do more work, thus the plates further apart must be holding a higher voltage.
     
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