Effects of Dielectric on capacitors

[Donnyboy]

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

 

[Helmholtz]

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.

 

[Donnyboy]

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

 

[berkeman]

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.

 

[Helmholtz]

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.

 

[Donnyboy]

err helmholtz, they are in parallel lol

 

[Helmholtz]

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.

 

[Donnyboy]

i'll be baffled, you are right O.O no wonder i went wrong in my ways

 

[Helmholtz]

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.

 

[Donnyboy]

hmmm the circuit is kinda correct... here is what i gathered

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