Effects of Dielectric on capacitors

In summary: When the battery is disconnected, the capacitors are in series and each gets half of the charge. This creates an electric field between them.
  • #1
Donnyboy
6
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
err helmholtz, they are in parallel lol
 
  • #7
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.
 
  • #8
i'll be baffled, you are right O.O no wonder i went wrong in my ways
 
  • #9
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..
..|...|...|...
..|...|...|...
..|_____|____|...
......
 

FAQ: Effects of Dielectric on capacitors

What is a dielectric and how does it affect capacitors?

A dielectric is a material that can store electrical energy. When placed between the plates of a capacitor, it increases the capacitance of the capacitor by reducing the electric field between the plates. This allows for more charge to be stored in the capacitor.

How does the type of dielectric used in a capacitor affect its performance?

The type of dielectric used in a capacitor can affect its performance in several ways. Different dielectric materials have different permittivity values, which determine how much the electric field is reduced. They also have different breakdown strengths, which determine the maximum voltage the capacitor can withstand. Additionally, some dielectrics may introduce losses, known as dielectric losses, which reduce the efficiency of the capacitor.

Can a dielectric material be used to increase the capacitance of a capacitor indefinitely?

No, there is a limit to how much a dielectric material can increase the capacitance of a capacitor. This limit is known as the dielectric constant or relative permittivity of the material. Once this limit is reached, further increases in capacitance are not possible.

Are there any disadvantages to using a dielectric in a capacitor?

While dielectrics can greatly increase the capacitance of a capacitor, they also have some drawbacks. They can introduce losses, as mentioned before, which can reduce the efficiency of the capacitor. They also add to the size and weight of the capacitor, which may be a disadvantage in certain applications.

How does temperature affect the performance of a capacitor with a dielectric?

The performance of a capacitor with a dielectric can be greatly affected by temperature. Some dielectric materials have a temperature coefficient, which means their permittivity value changes with temperature. This can lead to changes in capacitance and other performance characteristics of the capacitor. Additionally, temperature can affect the breakdown strength and stability of the dielectric material, which can also impact the capacitor's performance.

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