Capacitance with changes in dielectric material

In summary: Essentially, you are solving for the potential ##V## and electric field ##\mathbf{E}## in both configurations and then taking the difference to find the change in capacitance.In summary, the capacitance of a capacitor with a dielectric material inserted between the two conductors is affected by the position of the dielectric material. This is because the electric field and potential will change with the position of the dielectric material, resulting in a change in capacitance. The amount of change in capacitance can be quantified by solving for the potential and electric field in both configurations and taking the difference.
  • #1
mnb96
715
5
Hello,

Let's consider a capacitor simply made of two conductors with arbitrary shape in the vacuum (http://www.kshitij-iitjee.com/Study/Physics/Part4/Chapter26/3.jpg).
Now, if I place a small piece of dielectric material (for example a tiny sphere of glass) between the two conductors, the capacitance changes.

However, it seems to me that the observed change in capacitance depends on the position of the piece of glass. If I hypothetically place the piece of glass very far away from both conductors, the change in capacitance will be basically negligible.

Is it so that the change in capacitance is somewhat proportional to the magnitude of the electric field? If so, how can I prove it mathematically?
 
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  • #2
mnb96 said:
Is it so that the change in capacitance is somewhat proportional to the magnitude of the electric field? If so, how can I prove it mathematically?
The capacitance of an arbitrarily shaped conductor can be written as
$$C=\frac{q}{V}=\int\frac{dq}{V}=\int\frac{\sigma da}{V}$$
where ##\sigma## is the surface charge density and ##V## is the electrical potential. ##V## is related to the electric field by
$$\mathbf{E}=-\nabla V$$
So yes, the capacitance is related to the electric field between the conductors however it is generally not necessary to know ##\mathbf{E}## since we are only concerned with the potential at the surface of the conductor.
$$V=\frac{1}{4\pi\epsilon}\int_{S}\frac{\sigma da}{r}$$
So ##V## is only dependent on the shape of the surface ##S##, the permittivity ##\epsilon##, ##\sigma##. When this is plugged into the above equation for ##C##, the dependence on the charge density usually cancels out so the capacitance tends to be only a function of ##\epsilon## and the shape of the conductor ##S##.
 
  • #3
Hi NFuller,

thanks for your reply.
I have the feeling that you answered a different question than the one I asked. You basically explained why capacitance in a "static" configuration does not in general depend on E.

The scenario I was describing in my OP was the following: I was considering two completely different configurations: 1) two conductors in the vacuum, 2) the same two conductors with a small piece of glass placed between them. We probably agree that the capacitance in the latter configuration will be slightly different than the capacitance in the first configuration. Hence, if we switch from configuration #1 to configuration #2, we would observe a change in capacitance.

At this point I was observing that, intuitively, this change in capacitance must be dependent on the position of the piece of glass: if I place the piece of glass right between the two conductors, then I would expect a reasonable change in capacitance; while if I put the piece of glass one kilometer away from both conductors I would not expect any significant change.

This position-dependence made me suspect that the change in capacitance is somehow related to the intensity of the electric field, but I am not sure of this. That's why I was asking for help.
 
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  • #4
mnb96 said:
You basically explained why capacitance in a "static" configuration does not in general depend on E.
I did not say that. I showed that the capacitance does depend on ##\mathbf{E}## but noted that ##\mathbf{E}## is not usually considered when solving these problems.

When you stick the piece of glass in between the conductors both ##\mathbf{E}## and ##V## will change so capacitance can also change.
 
  • #5
NFuller said:
When you stick the piece of glass in between the conductors both ##\mathbf{E}## and ##V## will change so capacitance can also change.

Ok. Any hint about my original question about the change in capacitance being dependent on the position of the piece of glass?
 
  • #6
mnb96 said:
Ok. Any hint about my original question about the change in capacitance being dependent on the position of the piece of glass?
Yes, the position is also important since this will affect ##\mathbf{E}## and ##V##.
 
  • #7
Thanks. We are slowly getting to the point of my question:

mnb96 said:
...if I place the piece of glass right between the two conductors, then I would expect a reasonable change in capacitance; while if I put the piece of glass one kilometer away from both conductors I would not expect any significant change.

Is there a way to quantify the amount of change in capacitance as a function of the position of the piece of glass?

What I want to obtain is basically a scalar field that would represent the "sensitivity" of the original capacitor to changes in the medium. My guess was that such "sensitivity" would be somewhat proportional to |E|, but I might be wrong.
 
  • #8
mnb96 said:
Is there a way to quantify the amount of change in capacitance as a function of the position of the piece of glass?
Yes, see the equations in post #2. You will need to know the shape and location of the conductors to do this.
 

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in farads (F) and is dependent on the physical characteristics of the system, such as the distance between the plates of a capacitor and the type of material between them.

2. How does capacitance change with changes in dielectric material?

Capacitance is directly proportional to the dielectric constant of the material between the plates of a capacitor. Therefore, as the dielectric constant increases, the capacitance also increases, and vice versa.

3. What is dielectric material?

Dielectric material is an insulating material that is placed between the plates of a capacitor to increase its capacitance. It can be solid, liquid, or gas and is chosen based on its dielectric constant and other physical properties.

4. Can capacitance change due to other factors besides dielectric material?

Yes, capacitance can also be affected by the distance between the plates, the surface area of the plates, and the type of material used for the plates. These factors also play a role in determining the capacitance of a system.

5. How can changes in capacitance affect a circuit?

Changes in capacitance can affect the overall behavior of a circuit, such as the amount of charge stored and the time it takes to charge or discharge. It can also impact the frequency response of a circuit, particularly in high-frequency applications.

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