How to show Q1/Q2 = k in a Dielectric

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In summary: So k = Q1/Q2.In summary, in a constant voltage scenario, k can be shown as a ratio of Q1 and Q2, similar to how it is shown as a ratio of V1 and V2 in a constant charge scenario. This is done by holding the voltage constant and showing that the capacitance is directly proportional to k, which is a constant dependent on the geometry and dielectric material.
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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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  • #2
The question is very unclear to me. Perhaps a diagram with values labeled?
 
  • #3
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)
 
  • #4
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
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?
 
  • #6
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
 

What is Q1/Q2 in a dielectric?

Q1/Q2 refers to the ratio of the electric charges on the two plates of a capacitor when a dielectric material is inserted between them. It is also known as the dielectric constant (k) or relative permittivity.

What is the significance of Q1/Q2 = k in a dielectric?

Q1/Q2 = k is a mathematical representation of the capacitance of a capacitor with a dielectric material. It shows the relationship between the electric charges on the plates and the dielectric constant, which affects the capacitance of the capacitor.

How is Q1/Q2 = k calculated in a dielectric?

The dielectric constant (k) is calculated by dividing the capacitance of a capacitor with a dielectric by the capacitance of the same capacitor without the dielectric. This gives the ratio of the charges on the plates (Q1/Q2) when a dielectric is present.

Why does a dielectric material affect Q1/Q2?

A dielectric material is an insulating material that reduces the electric field between the plates of a capacitor. This reduces the force between the charges on the plates, thus decreasing the amount of charge that can be stored on the plates and affecting the Q1/Q2 ratio.

What factors can affect the value of Q1/Q2 = k in a dielectric?

The value of Q1/Q2 = k can be affected by the type of dielectric material used, its thickness, and the distance between the plates. Temperature and frequency of the electric field can also influence the dielectric constant and, therefore, the Q1/Q2 ratio.

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