Capacitance with changes in dielectric material

AI Thread Summary
The discussion centers on how the placement of a dielectric material, such as a piece of glass, affects the capacitance between two conductors. It is established that the change in capacitance is dependent on the position of the dielectric; placing it closer to the conductors results in a more significant change compared to positioning it far away. The relationship between capacitance, electric field, and potential is explored, indicating that both the electric field and potential change when the dielectric is introduced. The participants seek to quantify this change in capacitance as a function of the dielectric's position, suggesting that this "sensitivity" could be related to the magnitude of the electric field. Ultimately, understanding this relationship requires knowledge of the conductors' shape and location.
mnb96
Messages
711
Reaction score
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?
 
Physics news on Phys.org
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##.
 
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.
 
Last edited:
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.
 
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?
 
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##.
 
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.
 
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.
 
Back
Top