How to show Q1/Q2 = k in a Dielectric

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The discussion focuses on demonstrating the relationship between charge ratios (Q1/Q2) and the dielectric constant (k) in capacitors under constant voltage conditions. The user seeks to establish that the capacitance remains constant regardless of whether the scenario involves constant charge or constant voltage. Key equations referenced include E1 = E2 * k and C = Q/V, with the user attempting to derive Q1/Q2 = k without relying on the known relationship of capacitance. The conversation emphasizes the importance of Maxwell's equations and the definitions of capacitance in understanding these relationships.

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Electric to be
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Hello. I've been doing some personal learning on the topic of dielectrics and I've come across a bit of a problem.

Say I charged up a cap. plate with a battery and then I disconnected in the battery. Then I inserted some dielectric material in the middle.

It is then given that E1 = E2 * k.

Then, since Q must remain constant, V1 / V2 = k. From this you can deduce that C1 and C2 and are a ratio of k. (Since C = Q/V)

Now I know that the capacitance will not change if there is a constant voltage applied so long as the same geometry and dielectric material applies, and therefore in the case of a constant voltage (such as putting in a dielectric material without disconnecting a battery) then k will then be a ratio of Q1 and Q2.

Now my problem:

Even though I understand this, I wanted to work backwards and prove C1 and C2 are ratios of k by showing that Q1 and Q2 are ratios of k in a constant voltage scenario. I'm having a bit of trouble doing this. I tried doing this by comparing one capacitor not connected to a battery with a dielectric plate and another capacitor connected to a battery with no dielectric plate such that both voltages (and therefore fields) are the same across each capacitor.

So then I have E1= E2, V1 = V2 and consequently (Q) = (Q1 - Qi) Well then I just get that Q = Q1 - Qi Where Q is the charge on the first capacitor plate, Q1 is the charge on the second capacitor plate, and Qi is the induced charge near the surface of the dielectric/metal plate interface. No k anywhere in sight.

How could i show that k is a ratio of Q1 and Q2 without invoking the fact that C is constant? (Like I did above to show that k is a ratio of V1 and V2)

Basically what I'm doing here is choosing to prove that C only depends on geometry and dielectric material mathematically, even though this is a known fact.

I'm probably missing something easy but would appreciate any help.

Thank you.
 
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The question is very unclear to me. Perhaps a diagram with values labeled?
 
Jeff Rosenbury said:
The question is very unclear to me. Perhaps a diagram with values labeled?

Hi I'll try to explain it a bit better.

If there is a capacitor with some charge Q disconnected from a battery it has some field strength E1. Then if you slide a dielectric between the capacitor plates, there will be a new field strength E2.

Then E1 = (E2 * some constant k).

Using this and the fact that charge on each plate doesn't change you can show that the voltage between the plates is V1/V2 = k.
I was wondering how you would show, in a constant potential situation (instead of constant charge like above) that Q1/Q2 = k. In addition I would like to show this without relying on the fact C1/C2 = k.

(So that I can prove to myself mathematically that Capacitance will always remain constant regardless of if the situation is constant Q or constant V)
 
From Maxwell's we get: E = σ/2εo. Converting from E to V, V = ∫ E dl, and σ to Q, Q = ∫ σ dA.

C ≡ Q/V. So for two plates:

C = kεoA/l for a || plate capacitor.

k = relative permittivity;
εo = permittivity of free space;
A = plate area;
l = plate separation.

I'm not sure how you can get around using k here. Maxwell's equations are from experiment and C is from definition. K is experimentally derived for the material in question.
 
Jeff Rosenbury said:
From Maxwell's we get: E = σ/2εo. Converting from E to V, V = ∫ E dl, and σ to Q, Q = ∫ σ dA.

C ≡ Q/V. So for two plates:

C = kεoA/l for a || plate capacitor.

k = relative permittivity;
εo = permittivity of free space;
A = plate area;
l = plate separation.

I'm not sure how you can get around using k here. Maxwell's equations are from experiment and C is from definition. K is experimentally derived for the material in question.

My question is how to show that k is a ratio of Q1 and Q2 in a constant voltage situation? Similar to how we showed that k is a ratio of V1 and V2 in a constant charge situation?

For constant charge:

Cap 1 is original Capacitor. Cap 2 is capacitor with dielectric.

E1 = E2 * k

So V1 = V2 * k

k = V1/V2How do we do this for constant voltage to show k = Q1/Q2?
 
C is defined as Q/V. Hold V constant.

Other things being equal, C ∝ k. So C = kK where K is held constant due to a constant geometry. C1 = k1K, etc.

So C1=Q1/ V, and C2=Q2/ V.

Q1/Q2 = VKk1/ VKk2 = k1/k2
 

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