# How to show Q1/Q2 = k in a Dielectric

1. Feb 7, 2016

### Electric to be

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.

2. Feb 7, 2016

### Jeff Rosenbury

The question is very unclear to me. Perhaps a diagram with values labeled?

3. Feb 7, 2016

### Electric to be

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)

4. Feb 7, 2016

### Jeff Rosenbury

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.

5. Feb 7, 2016

### Electric to be

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/V2

How do we do this for constant voltage to show k = Q1/Q2?

6. Feb 8, 2016

### Jeff Rosenbury

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