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Parallel plate capacitor find the volts

  1. Apr 23, 2008 #1
    A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

    A.) A voltmeter reads 50.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V .

    What is the dielectric constant of this material? If found the answer to the first part which was K= 4.35.

    B.) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

    This is the part i can't get, i know C=Q/V and also that C1 + C2 = Ceq, how to do go about finding the this part B ?
     
  2. jcsd
  3. Apr 23, 2008 #2

    G01

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    HINT: Think of this as two capacitors in parallel hooked up to the same battery. Now, remember Kirchoff's Loop Law?
     
  4. Apr 23, 2008 #3
    How come C = C1 + C2. They are in series.
     
  5. Apr 23, 2008 #4

    alphysicist

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    Hi atavistic,

    I don't think that's right. In this case, with the dielectric pulled out so that it only fills one third of the space, the part of the capacitor with dielectric and the part without dielectric can be thought of a two capacitors in parallel.

    (The difference between using series and parallel is whether the air/dielectric boundary is perpendicular to the plates, as it is here; or if the boundary is parallel to the plates, in which case you can think of two capacitors in series.)
     
  6. Apr 23, 2008 #5
    so i use C= q/V to and solve for V?
     
  7. Apr 23, 2008 #6

    G01

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    You shouldn't even need to do that. We know that the capacitor with the 1/3 dielectric can be treated as two capacitors in parallel connected to a battery. Draw the circuit diagram for this situation. What MUST the voltage across each capacitor be?
     
  8. Apr 24, 2008 #7
    shouldn't it be 7.7 because when the dielectric is completely fills the space it reads 11.5 so then if we pull 1/3 of it out then 2/3 is still fills the space so 11.5/3=3.83, 3.83 *2= 7.7. I know its wrong but is my assumption wrong as well?
     
  9. Apr 24, 2008 #8

    G01

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    Yes, your assumption is wrong.

    Your not following my advice. Are you thinking about the capacitor as now being two capacitors in parallel? Have you drawn the circuit diagram for two capacitors in parallel? What is the voltage across those two capacitors in the circuit (and thus your real capacitor) if they are hooked up to that battery of yours? HINT: Think Kirchoff's loop law.
     
  10. Apr 25, 2008 #9
    when there hooked up together the volt reads 50, when dielectric is put in volts reads 11.5 so a difference of 38.5 Volts. Does this have anything to do to get the answer?
     
  11. May 5, 2008 #10
    Finally Got it! Thanks
     
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