Electrostatics - Capacitance Of A Parallel Plate Capacitor

Discussion in 'Physics' started by Nirvana, Aug 22, 2006.

  1. Nirvana

    Thread Starter Well-Known Member

    Jan 18, 2005
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    Looking back over the Electrostatics described earlier, we have enough tools to analyze the behaviour of a Parallel Plate Capacitor, and in particular it's Capacitance (C).

    Consider the two plates in a Parallel Plate Capacitor , we will assume that these plates are an exact copy of each other. Lets say that the top plate is connected to the positive terminal of a battery whilst the bottom plate is connected to the negative terminal of a battery.
    To calculate the total charge stored on the top and bottom plate we would take the amount of charge and divide it by the area of the plate - this would tell us how much charge is spread over the plate.
    The top plate in this case would have Q/A amount of charge on its surface. The bottom plate would have
    - Q/A charge on its surface, the negative sign shows negative charge from negative terminal of the battery.

    Keeping the above in mind I'd like to point out a useful concept that is the Flux density (D) = ε E , Therefore
    E = D/ε, Now what might not have been discussed earlier it that the electric flux density (D) is how dense the charge is or how much charge there is in a given area so D = Q/A.

    With that in mind E becomes; E = Q/εA.

    Ok now back to Capacitance, well as stated in another thread Capacitance is the ratio of Charge (Q) to Voltage (V) or
    C = Q/V. Now also from previous discussions is was found that V = E. d, and E (From above) = Q/εA therefore putting it all together we get: C = Q/V = Q / (Qd/εA) flippling side Qd/εA to give a product as you would do in fractions to give:
    C = Q * (εA/Qd) = QεA/Qd, now the Q's cancel out giving the result, C = εA/d.
    Here we see that the Capacitance then is not dependant by the amount of charge there is or voltage, but by the Geometry (Shape) of the component itself.

    Nirvana.
     
  2. aliashar86

    Active Member

    Nov 23, 2006
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    can u plz explain me two things:

    i) why and how in series combination the charge remain same?

    ii)
    i) why and how in parallel combination the voltage remain same?
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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    i) For capacitors in series if you charge the first plate on the first capacitor, this causes an equal and opposite charge on the opposite plate of the first capacitor. The first plate of the second capacitor (which is in series) will respond by ensuring that equal charge is deposited on this plate as the first plate of the first capacitor (consider the charge redistribution from the opposite plate of the first capacitor to the first plate of the second capacitor, which again causes an equal and opposite charge on the opposite plate of the second capacitor. And so on for each subsequent capacitor in series. You should be able to see that whatever charge is deposited on the first plate of the first capacitor, the charge deposited on the first plate of subsequent capacitors in series is equal.

    ii) For capacitors in parallel the voltage is the same because they share common potentials on each plate, i.e. the voltage across each of the capacitors in parallel is the same.

    Dave
     
  4. aliashar86

    Active Member

    Nov 23, 2006
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    can u plz explain

    "whn we charge a capacitor from a battery of (say 5V) so whn capacitor get charged so both capacitor and battery have equal amount of charge"
    Now it is accepted tht after charging capacitor from battery both have equal amount of charges. now my question is that wt if the capacitor is small i.e., it can hold only 2 V so in this both wouldn't have equal amount of charges.
    plz explain.
     
  5. Papabravo

    Expert

    Feb 24, 2006
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    Can you explain what you mean by a small capacitor only being able to hold 2 Volts? If you put 5V on it it will eventually get very close to 5V.
     
  6. aliashar86

    Active Member

    Nov 23, 2006
    71
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    i mean if capacitor can hold only 2v and battery contains 5 v then in the tranferring of voltage from battery to capacitor where the remaining 3 v will be left
     
  7. beenthere

    Retired Moderator

    Apr 20, 2004
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    Hi,

    The voltage a capacitor can withstand is determined by the diaelectric and/or spacing between the plates. Possibly you mean the amount of charge on the capacitor (capacity), which is determined by its rating in Farads. Capacity and voltage rating are not related.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
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    And the area of the plates.

    Dave
     
  9. Papabravo

    Expert

    Feb 24, 2006
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    A capacitor will hold any amount of voltage which is applied until the dielectric material between the plates breaks down. When this happens the voltage across the capacitor is reduced by the current flow between the two plates. Under these conditions the capacitor may or may not discharge completely. Its future performance is compromized by the breakdown event so it's not something you want to happen very often if at all..

    In 5V logic circuits we often use monolithic ceramic capacitors with a voltage rating of 50 VDC or 100 VDC. This rating is far in excess of anything expected in a 5V logic circuit.
     
  10. kanan

    New Member

    May 4, 2007
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  11. recca02

    Senior Member

    Apr 2, 2007
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    probably since when a voltage is applied across a capacitor the plates get charged according to polarity of source since there is a gap between two surface of capacitor (like parallel plates with air gap in between) so the charges
    are unable to flow further and are stored on the plates of capacitor.
     
  12. Papabravo

    Expert

    Feb 24, 2006
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    A capacitor will store a charge because no DC current can flow. The charges are added to one plate and removed from the other. AC current is an entirely different matter.
     
  13. kanan

    New Member

    May 4, 2007
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    What mechanism makes a capacitor store charges ? Is it electric field spread over parallel plates ? Is it distance of seperation of plates ? Is it type of dielectric ?
    2) What factors are responsible for charge density ?
     
  14. pebe

    AAC Fanatic!

    Oct 11, 2004
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    How does the area of the plates affect the voltage proof?
     
  15. Papabravo

    Expert

    Feb 24, 2006
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    I already explained that. Charge is stored on a conductor because it is placed there by a potential (voltage), and because of that potential it cannot leave. What is it about this explanation that you do not understand?

    BTW - what is stored in the Electric Field between the two plates is energy. Was that the question you were trying to ask?
     
  16. recca02

    Senior Member

    Apr 2, 2007
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    like mr papabravo said the charges are stored on the plate due to applied voltage to put it in layman's term the plates are charged to the potential
    applied across them since the charges stop at the plates.

    electric field in the air gap is caused by these charges.
     
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