Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel plate capacitor question

  1. Nov 17, 2016 #1

    I’ve been reading for a while and decided to ask , suppose we have a parallel plate capacitor with a distance x between the plates , in between the plates are another two plates with a dielectric so in total 4 plates. The outer two are attached to a dc supply of 12volts for example.

    Now I understand that charge on the plates is given by their separation distance , surface area and the dielectric between them , what I don’t know for sure is what happens in the middle two plates in my example that are not attached to a power source ,

    They will get charge on them as well I suppose and the dielectric in the middle will get polarized two , as far as I understand inserting such two plates in the middle is effectively putting another capacitor in series.

    I want to know what happens on the two plates in the middle of the outer two which are attached to DC. I suppose the inner plates mimic the outer ones in terms of charge separation and + and – sides just that the charge on them is weaker because they are at a distance away?

    What would happen if I removed the inner plates with their dielectric , would they hold their charge just like any other parallel plate cap that has been charged , I suppose they should.

    My final question for now is this , what happens when I have my 4 plates , outer ones attached to DC , inner ones floating , now I charge the cap , then I connect a load like a light bulb between the inner ones , would the lightbulb flash as the charge flows through the load from plate to plate , and what would happen afterwards when my outer plates are still connected to DC but the inner ones have been attached to a load?

    Attached Files:

    • pic.png
      File size:
      54.6 KB
  2. jcsd
  3. Nov 17, 2016 #2


    User Avatar
    2016 Award

    Staff: Mentor

    This is an equivalent diagram:


    If you freshly connect the battery to the circuit, you will have some voltage in the middle capacitor, which then discharges via the resistor/lamp/other load, and after a while the middle capacitor is uncharged.
  4. Nov 17, 2016 #3
    Yes I believe that is true. and note that the net capacitance remains the same with the two inner sheets removed. I.e. three capacitors of separation x/3 in series has the same capacitance and one capacitor of separation x of same area A.

    The inner plates will be polarized but have net zero charge. The inner plates produce an equipotential surface at one third and two thirds the distance between the outer plates and the outer plates have a net charge being that of the 12 volts across a net capacitance of ε0A/x as noted above.

    If the battery was removed yes they would hold their charge . But if the battery remained the charge would be reduced because the capacitance would be decreased since you have now introduce an air gap of width x/3.

    The light bulb would flash since the two inner plates are at different potentials. But the load would eventually equalize the potential across the inner plates thus producing a different capacitance one of two capacitors in series with capacitance defined by separations of x/3 or ε02A/3x resulting in a new charge on the outer plates.
  5. Nov 18, 2016 #4
    Ok, thanks for the answers so far , I appreciate.

    I wonder could someone explain me in a step by step way what happens in this scenario that I will write down here , I sort of know myself but some things confuse me. Here goes.

    The same 4 plate parallel capacitor just that in this example the outer plates are stationary and they have a dielectric on each inner side to increase capacitance , but the inner plates that also have a dielectric between them are on a rotating axis and are capable of being either put fully in between the outer plates forming a 4 plate cap or they can be rotated out from the outer plates leaving just the two outer plates and an air gap.

    Now I wonder what happens here , the inner plates that are fixed on a rotating axis have a load attached to them , say I first connect the outer plates to a certain DC voltage they charge up, then I introduce the inner plates by rotating them in between, what happens now ? The inner plates and the dielectric get polarized and they charge up , does it require extra force for me to rotate the plates in between the outer ones or is no force exerted on the rotational movement as the plates slide in just except for some pure mechanical losses in bearings ?

    As I rotate the inner plates in , they charge up but if the load is constantly attached to them does the load experience some current through it while the plates are trying to charge up ? For example the lightbulb that is attached to those inner plates , does it flash?

    And my final one for now , if the inner plate do get some charge even with a load attached to them , the charge gets used up through the load , then as I move the plates out rotate them around and move them in once again will the same cycle repeat , will they get charged again ?

    I ask this because I understand how a capacitor can transfer energy when AC is attached to it’s plates , since the voltage/current is changing , but I wonder is there any case where DC can transfer some energy and how does it happen , that is why I made up this thought experiment with the inner two plates floating and not attached to anything and how they charge up and what happens to that charge and if indeed that charge can be used for a load then how did the energy transferred from the outer plates which had a static DC load on them to the inner ones which are inserted in between and polarized.
  6. Nov 22, 2016 #5
    BUMP please, i really need help guys, thank You in advance.
  7. Nov 23, 2016 #6
    @mfb I would welcome any criticism or comments on the following.

    First assume there is no load connected across the iinner plates in this case no force is required because a capacitor by virtue of the fringe field exerts a force on a dipole thus tending to pull the dipoles in the dielectric into it. This force results from the field gradient in the fringe field.. See an intermediate EM text.

    Now as the dielectric is pulled into the space between the two outer plates the capacitance begins to decrease thus lowering the energy stored in the capacitor. you can see this by calculating the capacitance of a capacitor with 1/3 of the dielectric removed and comparing that to the capacitance of effectively three identical capacitors connected in series each with the dielectric but only 1/3 the thickness.

    Attaching a load should have an affect in general . However if the inner plates are introduced slowly so as the difference in potential of the plates can be equalized quickly then nothing should happen the capacitance of the system will not change and there should not be any fringe field between these plates which could cause a pulling force. I guess you could say that the plates shield the dielectric between the inner plates from the outer plates field and cannot polarize the dielectric.

    If the inner plates are introduced much faster than the time constant of the load and inner plate configuration then the plates will have a potential difference for a time and charge must flow from the high to the low potential plate until the difference is zero. Now there is no net charge on these inner plates at any time but the potential difference will move charge from one plate to the other until the potential difference is zero. When the plates are removed the plates will have unequal charges on them so a current will flow again to equalize the potential difference outside of the outer plates. These inner plate sort of act like a battery in that there is no net charge but a potential difference.

    I think this cycle repeats if the rotation is slow enough to allow quick discharge. If it is fast then the inner plates will maintain some net charge and potential difference. dependent on the rotational speed.
  8. Dec 2, 2016 #7
    ok I have a few follow up questions , as you said and I understand that the inner plates sort of shield the very dielectric between them , then how do the inner rotating plates get charged at all if the dielectric between them can't get polarized and there is no field between them just outside them ?
    also I assume the capacitor had a higher capacitance when there were no plate in between the outer ones and the capacitance got lower when the two extra plates were rotated in between the outer ones because now the field has less distance to work through and it’s easier so less charge is required correct?
    And one more thing , I don’t quite get , the inner plates are moved in between the charged outer ones this makes the inner plates accumulate a charge themselves , now this moving in requires no extra force on the shaft rotation than simple bearing losses , now imagine I have a load attached to the inner plates and I move them in like before do now I have a force to push against or do they move in without any force like before ?
    It would be weird for them to move in without force because then theoretically I have a perpetual motion scenario if maybe not for the fact that the charge accumulate don the inner plates comes through the field from the outer ones and whatever source gives the charge to them , on the other hand moving them in with force doesn’t seem right because moving conductors parallel to an e field doesn’t seem to require any force.
  9. Dec 2, 2016 #8
    You can show that the capacitance of two parallel plates of area A separated by a distance x in which there is a dielectric of thickness t and dielectric coefficient K is given by

    C = ε0KA/[( x-t)K +t]

    along with CV =Q , 1/Ctot =1/C1 +1/C2 +1/C3 and the time constant of a capacitor shunted by a resistor R is RC and keeping in mind that the charge is constant with the battery removed and the voltage is constant when it isn't. You should be able to answer these questions

    Just run through the calculations for the different configurations and compare results. Keep in mind that you have two situation one in which the battery remain connected during the process and the other where it is removed.

    When you start thinking perpetual motion just stop and ask yourself "what am I missing" since you are starting to run up a blind alley.

    If there is a force that pulls it in then you got to think that same force will try to keep it from leaving too.

    And yes there is a force on the dipole moments of the polar molecules in the dielectric. The fringe field at the edge of the plates has a gradient that acts on the dipoles to pull them into the space between the plates.

    There is a force with and without a battery attached however the forces are small of the order of micro-Newtons to milli-Newtons depending on the capacitor and the force is constant when the battery remains attached but depends approximately inversely to the distance it is withdrawn if the battery is removed.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted