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Capacitor with dielectric

  1. Oct 11, 2007 #1
    there is a capacitor with a constant potential difference of V in air. When a dielectric is completely inserted between the plates of the capacitor completely filling it, additional charge flows onto the positive plate.

    The additional charge should be Q=kCV-CV right?

    Would the induced charge on either faces of the dielectric be Q=kCV?

    Since the dielectric doesnt change the electric field (V=ED), why doesnt the increase in charge on the plates increase the field?
     
    Last edited: Oct 11, 2007
  2. jcsd
  3. Oct 11, 2007 #2

    learningphysics

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    Yes, kCV is correct. extra charge is kCV - CV.

    V = Ed

    V is the same. d is the same. So E must be the same.
     
  4. Oct 11, 2007 #3
    since there is an increase in charge on the plates, why doesnt this give rise to an increase in the E field since the increase in the chrage on the plates should serve to increase the field between the plates? I know V is the same and so E must be the same, I'm looking more for a conceptual idea.

    also, there is kCV charge on the top on the dielectric right? does that mean that there is a -kCV charge on the bottom of the dielectric?
     
  5. Oct 11, 2007 #4

    learningphysics

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    Well, the field is [tex]\frac{\sigma}{\epsilon}[/tex] which is [tex]\frac{Q}{A\epsilon}[/tex]

    so how does this change when the dielectric is inserted...


    yes.
     
  6. Oct 11, 2007 #5
    charge increases when the dielectric is inserted so shouldnt the field increse too?

    to: [tex]\frac{kCV}{A\epsilon}[/tex]
     
  7. Oct 11, 2007 #6

    learningphysics

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    k is the dielectric constant. so [tex]\epsilon = k\epsilon_0[/tex]

    plug this in and the k's cancel in the numerator and denominator...

    so you're left with

    [tex]\frac{CV}{A\epsilon_0}[/tex], which is just like without the dielectric.
     
  8. Oct 11, 2007 #7
    okay, so i want to understand this in a non mathematical way.

    in the same situation as above, we insert a dielectric and then we get an increase in charge on the plates. the reason why we dont get an increase in the E-field is because the dielectric creates an opposing field that makes it so the E-field does not change. is the right?
     
  9. Oct 11, 2007 #8

    learningphysics

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    Yes, that is correct.
     
  10. Oct 11, 2007 #9
    but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?
     
  11. Oct 11, 2007 #10

    learningphysics

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    No, just because the charge is the same doesn't mean the field is the same.
     
  12. Oct 11, 2007 #11
    what is the field in the dielectric?
     
  13. Oct 11, 2007 #12
    would it be:

    [tex]E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}[/tex]
     
  14. Oct 11, 2007 #13

    learningphysics

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    Yeah... I would write it as: [tex]E=\frac{Q}{A\epsilon_o}-\frac{kQ}{A\epsilon_o}[/tex]

    so when you add the field due to just the charge... you get the net field...
     
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