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Changes in capacitor after dielectric inserted

  • #1
Krushnaraj Pandya
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Homework Statement


A parallel plate capacitor (capacitance C) is charged with a battery of emf V volts. A dielectric slab of dielectric constant K is placed between the plates to fully occupy the space. The battery remains connected.
What are the changes in-1)C 2)Q (charge on capacitor) 3)E(E. field) 4)V(potential difference) 5)U (energy of capacitor) 6)F(Force of attraction between the plates). Symbols have usual meanings

Homework Equations


All applicable to capacitors

The Attempt at a Solution


I know that when a dielectric is inserted C2=CK (by adding 2, I'm indicating the new value). Since battery is still connected, V remains V(?). Here comes my first confusion, since E2=E/K, shouldn't V change as well. But if we assume for now that V remains V, then Q must have become KQ to make capacitance KC, either that or V became V/K..which one is happening here?
also, I don't know how to approach the 6th question ,as to how the force will change because of dielectric. I'd appreciate some help.
 

Answers and Replies

  • #2
Delta2
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When we have a battery in a circuit the voltage difference between the two ends of the battery is always (no matter what happens to the rest of the circuit) ##V'=V-Ir## where ##V## the emf of the battery and ##r## the internal resistance of the battery , and ##I## the current that flows through the battery. In our case we consider steady states or equilibrium states where ##I=0##, so the voltage difference between the two ends of the battery is simply ##V'=V##.
 
  • #3
Krushnaraj Pandya
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When we have a battery in a circuit the voltage difference between the two ends of the battery is always (no matter what happens to the rest of the circuit) ##V'=V-Ir## where ##V## the emf of the battery and ##r## the internal resistance of the battery , and ##I## the current that flows through the battery. In our case we consider steady states or equilibrium states where ##I=0##, so the voltage difference between the two ends of the battery is simply ##V'=V##.
Alright, so voltage definitely remains constant...
I calculated that if charge changes to KQ, then Electric field also remains unchanged.
And potential energy becomes KU.
The Force of attraction is Q*E, E=Q/2Aepsilon. Since charge changes to KQ, Force changes to K^2(F).
Thank you very much for your help
 
  • #4
Delta2
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I think you did a mistake regarding the Force, the force changes to ##F'=Q'E'=Q'E=kQE=kF##
 
  • #5
Krushnaraj Pandya
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I think you did a mistake regarding the Force, the force changes to ##F'=Q'E'=Q'E=kQE=kF##
I see my mistake, I forgot to factor in K in the denominator for E, but my textbook says the answer is K^2(F) which seems puzzling now
 
  • #6
Delta2
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That's strange, does your textbook agrees that the electric field remains unchanged?
 
  • #7
Krushnaraj Pandya
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That's strange, does your textbook agrees that the electric field remains unchanged?
yes it does.
 
  • #8
Krushnaraj Pandya
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  • #9
Delta2
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That's strange.. Does it also agree that the new charge is Q'=kQ?
 
  • #10
Krushnaraj Pandya
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That's strange.. Does it also agree that the new charge is Q'=kQ?
It agrees to the following, new charge is KQ, new Capacitance is KC, V remains same, E remains same, U becomes kU. And finally, that F becomes (K^2)F
 
  • #11
Delta2
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It agrees to the following, new charge is KQ, new Capacitance is KC, V remains same, E remains same, U becomes kU. And finally, that F becomes (K^2)F
It must have an error regarding the force. Cant understand it otherwise.
 
  • #12
Krushnaraj Pandya
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It must have an error regarding the force. Cant understand it otherwise.
Alright. Thank you very much.
 
  • #14
Krushnaraj Pandya
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It must have an error regarding the force. Cant understand it otherwise.
its definitely not an error, but I don't see a concrete evaluation in the above link, just a correction in the comments that it is proportional to k^2
perhaps @CWatters knows
 
  • #15
Delta2
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I think where my post #4 goes wrong is that force is not ##Q'E'## when we insert the dielectric. Intuitively I can understand why it will be## k^2F##, because we have ##kQ## charge in the positive plate and ##-kQ ##charge in the negative so the coulomb force (if those charges where point charges) would be ##-k^2Q^2/r^2##
 
  • #16
Krushnaraj Pandya
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force is not Q′E′
I can follow your intuition, but its still not a rigorous way to see it.
I found the following statement-
The reason is that the field acting on the capacitor plate is entirely due to the other capacitor plate; the field due to dielectric is zero outside the dielectric

This statement implies that the K we factored in the denominator is just an expression for E inside the dielectric, but the F due to the plate on the other plate is still due to E, not E/K
 
  • #17
Delta2
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I can follow your intuition, but its still not a rigorous way to see it.
I found the following statement-
The reason is that the field acting on the capacitor plate is entirely due to the other capacitor plate; the field due to dielectric is zero outside the dielectric

This statement implies that the K we factored in the denominator is just an expression for E inside the dielectric, but the F due to the plate on the other plate is still due to E, not E/K
Yes I guess that's the catch, that the field on the surface of the plate is not equal to the field inside the dielectric.
 
  • #18
Krushnaraj Pandya
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Thanks for the help :D
 

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