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I Having trouble conceptualizing capacitors

  1. Apr 1, 2016 #1
    So I basically understand the C = ε0(A/d) formula. Larger area means there is room for more electrons, smaller d means larger E field, which pulls electrons onto the opposite plate more strongly.

    What's tripping me up is when dielectrics come into the picture. When a dielectric is inserted, the E field goes down, but the capacitance goes up. I feel like I get the reasons for each of these effects individually, but why does dropping the E field between the plates not reduce capacitance? Is it just cancelled out by the positive effect of the dielectric?
     
  2. jcsd
  3. Apr 1, 2016 #2
    A dielectric contains molecules that become electrically polarised in an electric field. This acts like screening for want of a better term.

    Hopefully the pros will chime in.
     
  4. Apr 1, 2016 #3
    The insertion of a dielectric into the space between two charged parallel plates reduces the electric field in this space due to the induced polarization of the molecules of the dielectric. The polarized molecules produce a local electric field which opposes the field due to the charge on the plates thus reducing the electric field between the plates. This cause a reduction in the potential difference between the two plates since V = E⋅s where s is the separation. Since the charge on the plates has not changed and since the capacitance in the charge per volt on the plates as the voltage decreases the capacitance must increase.
     
  5. Apr 1, 2016 #4
    Thank you for your reply.

    So if you have a battery in the circuit, the voltage/electric field can't go down and therefore the charge goes up instead?

    I think I was confusing capacitance with actual energy stored in the capacitor.
     
  6. Apr 1, 2016 #5
    Yes with a battery in the circuit the charge goes up.
     
  7. Apr 1, 2016 #6
    When a given Physical quantity, in this case C< depends on more than one quantity, whether it increases or decreases with respect to any one of the quantity can be ascertained only when the other variables are held constant. So capacity increases with increase of A when d and kε0 are held constant. Similarly it decreases with increase of d when A and increase with increase of k, the dielectric constant, ε0, being the universal constant. This is just mathematical and not much Physics involved here. But when you say Increase of area means more electrons, the physical significance of capacitance comes into force, the number of free electrons available even with a small part of one of the plate is very large of the order of Avogadro number (10²⁴). So whether I double this area triple it not much order of magnitude change occurs in that number. But that number does not constitute Charge. Equal and opposite Charge on the plates is the after effect of applying potential difference between the plate. For a given potential difference if more charge is stored we say the capacitance is more, here again we need to keep d same Only then we can say that larger area of plate larger is the charge stored. If area is constant the smaller the distance larger will be the electric field between the plates hence more charge will be stored for the same potential difference when k is also fixed.But we must note that the applied voltage is also fixed. Now if A and d the mechanical parameters of capacitance are also fixed and I wish to increase the stored charge for the same applied potential difference we need to introduce the material with higher dielectric constant. This can be viewed in two alternative ways. If you fix the voltage difference higher dielectric constant will store higher cahrge and if you fix charge by first charging the capacitor and then disconnecting the supply, introduction of dielectric material will reduce the potential difference between the plates.
     
    Last edited: Apr 1, 2016
  8. Apr 1, 2016 #7

    sophiecentaur

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    Another way of thinking about it is : Using the dielectric is really just equivalent to bringing the metal plates closer together. A large + charge ends up on the side of the dielectric in contact with the - plate and a - charge ends up by the + plate. All the molecules of the dielectric are relatively easily polarised and that's where the displaced charges appear on each side. The only reason for using a dielectric is to avoid flashover and to allow the accuracy of spacing of the plates (foils) to be less critical. Put in twice the permittivity and it's the same as halving the separation.
    Dielectrics can be a nuisance because they introduce loss at high frequencies. Picking the right material for the required C value and the required operating frequency can be difficult / impossible / expensive.
     
  9. Apr 2, 2016 #8

    sophiecentaur

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    I know not everybody likes leaping into Maths but the formulae
    Q=CV
    and
    C = ε0(A/d)
    are really good shorthand descriptions of what goes on and it's easy to apply one or both of them when testing out a more arm waving, physical model that one can have in one's head. Easy enough to remember, they can be really good friends in a confusing situation.
     
  10. Apr 2, 2016 #9
    What i wished to emphasize is that the formula C = kε₀(A/d) is a mathematical formula with A and d the mechanical characteristics of what is C and k an electrical characteristic. Whereas the formula C = Q/V, defines the concept of capacitance in terms of the observable electrical parameters, V and Q. If V is fixed then the value of C decides what will be Q. And if Q is fixed then again C decides what will be the value of V. The first is easy to visualize because we consider Q as the effect of V. But it is possible to set the apparatus to make V the effect of Q. Charge a capacitor to some value of Q and then disconnect te nattery and insulate teh plates from leaking into each other. Now if you change C by changing A, d or k, you will find that he potential difference beween the plates will accordingly change as per teh formula V = Q/C.
     
  11. Apr 2, 2016 #10

    sophiecentaur

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    I wonder whether you can really make that distinction. V and Q are measurable and C is just the ratio between them. kε₀ , A and d are also measurable and the formula gives a value for C. The only thing about kε₀, afaiks, is that it's an intensive variable. Difficult to change in many cases but not impossible if, for instance, you make a dielectric foam with different sized bubbles. A formula is a mathematical description.
     
  12. Apr 2, 2016 #11
    To simplify matters and focus on concepts let us take k = 1, The we have two as you say mathematical descriptions for C. First C = (ε₀A/d) and another = V/C. The first give you the mechanical description of the gadget C as ε₀ is a universal constant. But V/C gives you the physical significance of C and tells you that it is an electrical quantity.
     
  13. Apr 2, 2016 #12

    sophiecentaur

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    I'm not sure what you're really saying. I'll wait to see if anyone else has a reaction.
     
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