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Capacitor with dielectric

  • #1
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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?
 
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Answers and Replies

  • #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.
 
  • #3
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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.
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?
 
  • #4
learningphysics
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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.
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...


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?
yes.
 
  • #5
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charge increases when the dielectric is inserted so shouldnt the field increse too?

to: [tex]\frac{kCV}{A\epsilon}[/tex]
 
  • #6
learningphysics
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charge increases when the dielectric is inserted so shouldnt the field increse too?

to: [tex]\frac{kCV}{A\epsilon}[/tex]
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.
 
  • #7
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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?
 
  • #8
learningphysics
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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?
Yes, that is correct.
 
  • #9
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but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?
 
  • #10
learningphysics
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but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?
No, just because the charge is the same doesn't mean the field is the same.
 
  • #11
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what is the field in the dielectric?
 
  • #12
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would it be:

[tex]E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}[/tex]
 
  • #13
learningphysics
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would it be:

[tex]E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}[/tex]
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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