# Capacitance of a parallel-plate Capacitor with non uniform dielectric

• approx12
The maxwell equation says: $$\vec{\nabla}\cdot \vec{D} = 4\pi\rho_f$$ But since we are in a dielectric with the only free charge being bound to the surface, ##\rho_f=0## inside the dielectric (atleast that's what I think). And that implies that ##\vec{D}=const.## in the dielectric. Obviously I'm wrong thinking about it this way, but I can't see the solution for it. Does the unevenly distributed charge on the surface kind of "induce" free charges into the dielectric?The surface charge density is

#### approx12

Homework Statement
I have to find the capacitance of a paralell plate capacitor which is filled with a dielectric ##\epsilon(r)## and has circular plates of Radius ##R_0##. The lower plate carries a charge ##-Q## and is situated at ##z=0## and the upper plate carries a charge ##Q## situated at ##z=d##. Boundary effects can be neglected and ##d<<R_0##. This cylindrical system is paralell to the ##z##-axis.
Relevant Equations
$$\epsilon(r)=\epsilon_0+\Delta\epsilon\frac{r}{R_0}$$ where $$r=\sqrt{x^2+y^2}$$
Hey guys! I'm having trouble with the solution that I arrived at.

Through boundary conditions I'm able to determine ##\vec{D}## as $$\vec{D}=-\frac{4Q}{R_0^2}\hat{e_z}$$ (In CGS units)

Trough that I'm able to get the electric field as $$\vec{E}=-\frac{1}{\epsilon(r)}\frac{4Q}{R_0^2}\hat{e_z}$$

Now I can integrate to get the potential difference between ##z=0## and ##z=d##: $$U=\int_0^d-\frac{1}{\epsilon_0+\Delta\epsilon\frac{\sqrt{x^2+y^2}}{R_0}}\frac{4Q}{R_0^2} dz$$

Finally I get the capacitance as: $$C=\frac{(\epsilon_0+\Delta\epsilon\frac{\sqrt{x^2+y^2}}{R_0})R_0^2}{4d}$$

I wondered if that solution can be correct as I always thought about the capacitance ##C## as a constant. Here it dependents on the radius of cylinder. Would I need to integrate the capacitance over ##\phi## from ##0\to2\pi## and ##r## from ##0\to R## (so in cylindrical coordinates) to get the right solution or is it fine as it stands?

I would appreciate any thoughts and comments on this problem! Thank you!

Twigg and Delta2
I think you need to go back and think about the scenario more.
If the dielectric varies radially but the voltage drop across it is constant, what does that say about the charge distributions on the plates?

Twigg, approx12 and Delta2
haruspex said:
I think you need to go back and think about the scenario more.
If the dielectric varies radially but the voltage drop across it is constant, what does that say about the charge distributions on the plates?
Thanks for taking the time to comment on my problem! So looks like I was on the wrong path.

If voltage drop is constant, then my Electric field ##\vec{E}## should not depend on the radius ##r## right? I dont't know what that would say about my free surface charge though. Shouldn't it per definiton be $$\sigma_f=\frac{Q}{A}=\frac{Q}{\pi R_0^2}$$? I can't seem to think of a reason why the dielectric would have an effect on my free surface charge...

Delta2
approx12 said:
Thanks for taking the time to comment on my problem! So looks like I was on the wrong path.

If voltage drop is constant, then my Electric field ##\vec{E}## should not depend on the radius ##r## right? I dont't know what that would say about my free surface charge though. Shouldn't it per definiton be $$\sigma_f=\frac{Q}{A}=\frac{Q}{\pi R_0^2}$$? I can't seem to think of a reason why the dielectric would have an effect on my free surface charge...
The voltage drop must be constant because each plate is a conductor. Therefore the charges are not uniformly distributed.
Think of each annular element as one of a set of parallel capacitors.

Steve4Physics, approx12 and Delta2
Interesting problem. At first glance I would have thought that the surface charge density of the plates is uniform, and proceed pretty much as the OP did.
That is that D is independent of ##\epsilon (r)## and E is dependent of it. But then I would 've come to a dead end as it would turn out that the potential difference depends on ##\epsilon (r)## which can't be the case since the plates are conductors and have everywhere the same potential.
So it turns out after some thought that D (and the surface charge density) is dependent of ##\epsilon (r)## and E is independent of it.

I think your first task would be to determine the surface charge density ##\sigma (r)##. It's dependence from r must be such that the Electric field E ends up independent of r.

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approx12
Delta2 said:
Interesting problem. At first glance I would have thought that the surface charge density of the plates is uniform, and proceed pretty much as the OP did.
That is that D is independent of ##\epsilon (r)## and E is dependent of it. But then I would 've come to a dead end as it would turn out that the potential difference depends on ##\epsilon (r)## which can't be the case since the plates are conductors and have everywhere the same potential.
So it turns out after some thought that D (and the surface charge density) is dependent of ##\epsilon (r)## and E is independent of it.
Thanks too for taking the time to comment!

I'm trying to think about this (that ##D## is dependent on ##\epsilon(r)##) but I can't seem to get to the point to see it.

The maxwell equation says: $$\vec{\nabla}\cdot \vec{D} = 4\pi\rho_f$$ But since we are in a dielectric with the only free charge being bound to the surface, ##\rho_f=0## inside the dielectric (atleast that's what I think). And that implies that ##\vec{D}=const.## in the dielectric.

Obviously I'm wrong thinking about it this way, but I can't see the solution for it. Does the unevenly distributed charge on the surface kind of "induce" free charges into the dielectric?

Also for @haspurex comment: Could you maybe specify the term "annular element" a little bit more? I'm also not a native english speaker and don't really have a picture for this word in my head. Thank you!

No free charges induced into the dielectric. As long as we accept that D will have only z-component is divergence will be equal to ##\frac{\partial D_z}{\partial z}## and in order for this to be zero, we are free to vary ##D_z## with respect to x and y in any possible way we want, we just have to keep ##D_z## independent of z

approx12
Delta2 said:
No free charges induced into the dielectric. As long as we accept that D will have only z-component is divergence will be equal to ##\frac{\partial D_z}{\partial z}## and in order for this to be zero, we are free to vary ##D_z## with respect to x and y in any possible way we want, we just have to keep ##D_z## independent of z
Thank you, I didn't thought about it that way, that makes good sense! Already learned something, thanks.

Would it possible if you could also give me a small hint on a starting point on how to get the ##\vec{E}## or ##\vec{D}##-Field mathmatically? What "Ansatz" should I try? I can't seem to get far with Gauss Law or boundary conditions since I've never seen such a problem before where ##\vec{D}## has some ##\epsilon## dependence...

Delta2
One possible ansatz i see is $$\vec{D}=C_0\epsilon(r)\hat z$$ where ##C_0## a proper constant which you can determine by using Gauss's law for D(in integral form and taking as surface a cylinder with radius equal to the radius of the plates, its one face inside the dielectric and the other face outside capacitor) .

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approx12
By ‘annular element’ I think @haruspex means this…
https://i.stack.imgur.com/cestm.png
with conductors on the annulus’s flat faces; the dielectric is in the gap between the flat faces. So each annular element is itself a capacitor.

Say an annular element has internal radius r, external radius r+dr and thickness d. The capacitor can be thought of as the 'sum' of the annular elements.

If you want to avoid using ##\vec D## and ##\vec E## there is a simple solution. You can find the capacitance of an annular element. The annular elements are electrically connected in parallel, so you can now find the total capacitance.

A couple of other points:

Your final expression for C must depend on only ##\epsilon_0, \Delta \epsilon, R_0## and ##d##. Having ‘x’ and ‘y’ (or ‘r’) in the expression doesn’t make sense.

You could keep the intermediate working steps a little neater by letting (say) ##\alpha = \frac{\Delta \epsilon}{R_0}## so that you can write ##\epsilon (r) = \epsilon_0 + \alpha r##.

approx12 and Delta2
approx12 said:
The maxwell equation says: $$\vec{\nabla}\cdot \vec{D} = 4\pi\rho_f$$ But since we are in a dielectric with the only free charge being bound to the surface, ##\rho_f=0## inside the dielectric (atleast that's what I think). And that implies that ##\vec{D}=const.## in the dielectric.
It's tempting to equate a zero divergence with a constant vector field, but as has been pointed out, you can't do that. Here's another example: ##\nabla \cdot \vec E=0## for the field of a point charge except at ##r=0##. The field in this case is not constant in magnitude or direction, yet its divergence still vanishes.

approx12 and Twigg
Steve4Physics said:
By ‘annular element’ I think @haruspex means this…
https://i.stack.imgur.com/cestm.png
with conductors on the annulus’s flat faces; the dielectric is in the gap between the flat faces. So each annular element is itself a capacitor.

Say an annular element has internal radius r, external radius r+dr and thickness d. The capacitor can be thought of as the 'sum' of the annular elements.

If you want to avoid using ##\vec D## and ##\vec E## there is a simple solution. You can find the capacitance of an annular element. The annular elements are electrically connected in parallel, so you can now find the total capacitance.

A couple of other points:

Your final expression for C must depend on only ##\epsilon_0, \Delta \epsilon, R_0## and ##d##. Having ‘x’ and ‘y’ (or ‘r’) in the expression doesn’t make sense.

You could keep the intermediate working steps a little neater by letting (say) ##\alpha = \frac{\Delta \epsilon}{R_0}## so that you can write ##\epsilon (r) = \epsilon_0 + \alpha r##.
Yes, that’s what I'm suggesting. Thanks for the diagram.

approx12
I did it with the ##\vec{D},\vec{E}## method and using the ansatz I suggest at post #9.
I got $$C_0=\frac{Q}{\int_0^R 2\pi r \epsilon(r)dr}$$, and
for the capacitance ##C## $$C=\frac{Q}{V}=\frac{Q}{C_0d}=\frac{\int_0^R 2\pi r \epsilon(r)dr}{d}$$
@haruspex ,@Steve4Physics what do you get using the "annular method"?

approx12
Delta2 said:
what do you get using the "annular method"?
The same.

approx12 and Delta2
Delta2 said:
I did it with the ##\vec{D},\vec{E}## method and using the ansatz I suggest at post #9.
I got $$C_0=\frac{Q}{\int_0^R 2\pi r \epsilon(r)dr}$$, and
for the capacitance ##C## $$C=\frac{Q}{V}=\frac{Q}{C_0d}=\frac{\int_0^R 2\pi r \epsilon(r)dr}{d}$$
@haruspex ,@Steve4Physics what do you get using the "annular method"?
Thanks to everyone for the discussion and the help with this problem, I learned quite a bit by that. Gonna be extra careful on the next problem.

I dont't want to be too annoying but I have two last questions:

1) In your first equation here, did you pull the constant ##4\pi## from ##div(\vec{D})=4\pi Q## into ##C_0##? Or am I missing something here.

2) The ansatz you posted was $$\vec{D}=C_0\epsilon(r)\hat{e_z}$$ Now, in every problem that I've encountered so far in physics I was able to derive pretty much everything from some kind of physical/mathmatical law (one exemption I could think of would be differential equations where you basically also guess the answer, but even there you have some kind of "recipe" on how to get to the answer).
So my question would be: Is there a way to derive this ansatz mathmatically (eg. by Gauss law, boundary conditions, multipole expansion,...) ? Or do I really have to kind of guess the answer by applying the thought process which is described in this thread?

approx12 said:
do I really have to kind of guess the answer by applying the thought process which is described in this thread?
As I outlined in post #4, my approach was to consider the annuli of the plates between radius r and r+dr. Each such pair of annuli forms a capacitor. They are all at the same voltage difference, V, but different local charge densities, q(r).
Write down the expression for the capacitance of this elemental capacitor in terms of r, dr and ##\epsilon(r)##. Hence get an expression for the charge on it in terms of those variables and V.
Write that Q is the integral of this charge across the positive plate,

Steve4Physics and approx12
approx12 said:
Thanks to everyone for the discussion and the help with this problem, I learned quite a bit by that. Gonna be extra careful on the next problem.

I dont't want to be too annoying but I have two last questions:

1) In your first equation here, did you pull the constant ##4\pi## from ##div(\vec{D})=4\pi Q## into ##C_0##? Or am I missing something here.

2) The ansatz you posted was $$\vec{D}=C_0\epsilon(r)\hat{e_z}$$ Now, in every problem that I've encountered so far in physics I was able to derive pretty much everything from some kind of physical/mathmatical law (one exemption I could think of would be differential equations where you basically also guess the answer, but even there you have some kind of "recipe" on how to get to the answer).
So my question would be: Is there a way to derive this ansatz mathmatically (eg. by Gauss law, boundary conditions, multipole expansion,...) ? Or do I really have to kind of guess the answer by applying the thought process which is described in this thread?
1) The constant ##4\pi## is used in Gaussian units. In SI units there isn't any such constant
2) This ansatz is derived by the requirement that $$V=\int \vec{E}\cdot dl=Ed=\frac{D}{\epsilon(r)}d$$ is independent of r (and that's because the capacitor's plate are conductors, so the potential difference between any two points that are on each plate should be the same, independent of r). So we set $$\frac{D}{\epsilon(r)}=C_0$$ ##C_0## a constant independent of r (and independent of z too, we need this in order for the divergence to be zero) from which the ansatz follows directly.

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approx12
haruspex said:
As I outlined in post #4, my approach was to consider the annuli of the plates between radius r and r+dr. Each such pair of annuli forms a capacitor. They are all at the same voltage difference, V, but different local charge densities, q(r).
Write down the expression for the capacitance of this elemental capacitor in terms of r, dr and ##\epsilon(r)##. Hence get an expression for the charge on it in terms of those variables and V.
Write that Q is the integral of this charge across the positive plate,
It’s probably worth noting that there is no need to find the charge on each elementary annular capacitor.

An elementary annular capacitor has capacitance ##dC = \frac {\epsilon(r) 2 \pi r dr}{d}## and the total capacitance is simply ##C = \int dC = \int_{0}^{R_0} \frac {\epsilon(r) 2 \pi r dr}{d}##.

It is not clear (to me anyway) why the original question states that the charges on the plates are +Q and -Q, since this is true of any charged capacitor.

Delta2