Effects of Dielectric on capacitors

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Homework Help Overview

The discussion revolves around two parallel plate capacitors connected in parallel, with one capacitor having a dielectric inserted after the voltage source is disconnected. Participants are exploring the effects of the dielectric on charge and electric field calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between charge, capacitance, and voltage, particularly how inserting a dielectric affects these variables. Some express confusion about the state of the capacitors after disconnecting from the voltage source, questioning whether they remain in parallel or behave differently.

Discussion Status

There is an ongoing exploration of the implications of inserting a dielectric and the resulting changes in charge and electric field. Some participants are attempting to clarify the conditions under which the capacitors operate, while others are reflecting on their initial misunderstandings.

Contextual Notes

Participants are navigating the complexities of capacitor behavior when disconnected from a power source, with some misinterpretations regarding series and parallel configurations being discussed.

Donnyboy
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Homework Statement


2 parallel plate capacitors of distance separation d between the plates are connected in parallel of capacitance C_1 and C_2 respectively and charged with a voltage V

after that the Voltage is disconnected and a dielectric of κ is inserted into C_2

I) Express the charge on C_1 and C_2 in terms of V, d , C_1, C_2, κ where relevant
II) Express the electric field between the 2 capacitors in terms of V, d , C_1, C_2, κ where relevant

Homework Equations



Q=CV, C_new=C_0 * κ

V_0=E_0*d
 
Last edited:
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The charge on the capacitors do not change when you insert a dielectric. So as the capacitance goes up, the voltage goes down proportionally. So V_2 = V_0/k and C_2 new = C_2 * k

Q_1=C_1 * V
Q_2 = C_2 * V

E_1 = V/d
E_2 = V/(k*d)

I think these are right.
 
well the charge Q_1= C_1*V and Q_2=C_2*V applies to which the capacitors are still connected to the power supply. after it disconnects and one of them is inserted with a dielectric the capacitance changes

Furthermore the Electric Field is the same when it is connected (charging) in parallel

ie: E_1= E_2 = V/d

when the circuit only consists of the charged capacitors... that's where i got stuck :S

as of now my mind is in a jumbled state and not thinking properly so i apologized before hand :S
 
Helmholtz said:
The charge on the capacitors do not change when you insert a dielectric. So as the capacitance goes up, the voltage goes down proportionally. So V_2 = V_0/k and C_2 new = C_2 * k

Q_1=C_1 * V
Q_2 = C_2 * V

E_1 = V/d
E_2 = V/(k*d)

I think these are right.

Please do not do the OP's work for them. The Rules link at the top of the page is explicit about how to provide Homework Help. You may ask questions, give hints, find errors, etc. But you may NOT do the student's work for them.
 
I misread the question. I think it might be helpful to find the total charge placed on both capacitors. Then we know that this is the charge in the circuit after the battery is also disconnected.

Then with the fact that capacitors in series have the same charge as one another it would be reasonable to say that each get's half of the total charge. Then we can figure out the electric field with the charges placed on each, keeping in mind what effect the dielectric has.

Extra emphasis on them being in series meaning they have equal charge now, instead of equal voltage when in parallel.
 
err helmholtz, they are in parallel lol
 
Even when you disconnect them from the battery? I was thinking of a circuit like this:

......
...__________...
..|...|...
..V...__|___...
..|...|...|...
..|...C_1...C_2..
..|...|_____|...
..|...|...
..|__________|...
......

And then when you disconnect the battery it's in series in the square.
 
i'll be baffled, you are right O.O no wonder i went wrong in my ways
 
Does this make sense to you now then? I was trying to imagine a circuit in which they could still be in parallel after disconnecting, but it would only lead to a short circuit such as:

......
...____________...
..|...|...|...
..V...|...__|___...
..|...|...|...|...
..|...|...C_1...C_2..
..|...|...|_____|...
..|...|...|...
..|_____|______|...
......

But then when you disconnect the battery, there is nothing in the middle branch and it's just still a circuit of capacitors in series.
 
  • #10
hmmm the circuit is kinda correct... here is what i gathered

......
...__________...
..|...|...|...
..V...|...|...
..|...|...|...
..|...C_1...C_2..
..|...|...|...
..|...|...|...
..|_____|____|...
......
 

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