Finding the bound charge in a dielectric

In summary, the problem involves a parallel plate capacitor filled with a dielectric material with a varying dielectric constant from 1 to 2. The capacitor is connected to a battery of voltage V. One approach is to slice the capacitor horizontally and use the fact that the normal component of the auxiliary field D is the same at every slice, and the potential difference of the slice capacitors is dU=Edx=(D/ε)dx. The total voltage across the capacitor can be found by integrating E.
  • #1
xophergrunge
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MODERATOR'S NOTE: problem statement is mistaken, please see https://www.physicsforums.com/showthread.php?t=723342

Homework Statement



The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant $\epsilon_r$ varies linearly from 1 at the bottom plate (x=0) to 2 at the top of the plate (x=d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero.

Homework Equations



$C=\frac{Q}{V}$
$C=\frac{A\espilon_0}{d}$
$D=\epsilon E$
$\sigma_b=P\cdot\hat{n}$
$\rho_b=-\nabla \cdot P$

The Attempt at a Solution



I am not sure if those are the relevant equations, I am really confused about what to do for this problem. What I have tried, with no success is to find the capacitance by assuming the plates are a distance L apart and have area A. First I find $\epsilon_r$ as a function of x, and it is $\epsilon_r = (\frac{x}{d}+1)$ and then saying $C=\int_0d (\frac{x}{d}+1)\frac{\epsilon_0A}{L}dx=\frac{3\epsilon_0 Ad}{2L}$ And since $C=\frac{Q}{V}$ then the free charge is $Q=\frac{3\epsilon_0 AdV}{2L}$. I don't really know where to go from here, or even if what I have done so far is how I want to approach it. I also thought about "slicing" the capacitor horizontally so that each little slice looked like a capacitor with a linear dielectric inside, so that I could use $P=\epsilon_0(\epsilon_r-1)E$. But, I again, don't know where I would go from there. Any help would be appreciated.
 
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  • #2
How do I delete a post? I thought latex worked here and I think it would be easier just to start this over?
 
  • #3
Hi there!

We have a few relevant equations here indeed. So first off we want to focus on the auxiliary field ##\vec{D}## because we can use it to treat this just like any other parallel plate capacitor problem meaning that ##\vec{\nabla} \cdot \vec{D} = \rho_f\Rightarrow \oint \vec{D}\cdot d\vec{A} = Q_f## combined with the obvious symmetry of the system (assuming we can ignore edge effects) will straightforwardly give us ##\vec{D}##. Furthermore we know that ##\vec{D} = \epsilon\vec{E}## and ## \vec{D} = \epsilon_{0}\vec{E} + \vec{P}## as well as ##\rho_{b} = -\vec{\nabla}\cdot \vec{P}## so you just have to piece everything together.
 
  • #4
xophergrunge said:

Homework Statement



The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant ##\epsilon_r## varies linearly from 1 at the bottom plate (x=0) to 2 at the top of the plate (x=d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero.

Homework Equations



##C=\frac{Q}{V}##

##C=\frac{A\epsilon_0}{d}##

##D=\epsilon E##

##\sigma_b=P\cdot\hat{n}##

##\rho_b=-\nabla \cdot P##

The Attempt at a Solution



I am not sure if those are the relevant equations, I am really confused about what to do for this problem. What I have tried, with no success is to find the capacitance by assuming the plates are a distance L apart and have area A. First I find ##\epsilon_r## as a function of x, and it is ##\epsilon_r = (\frac{x}{d}+1)## and then saying ##C=\int_0^d (\frac{x}{d}+1)\frac{\epsilon_0A}{L}dx=\frac{3\epsilon_0 Ad}{2L}## And since ##C=\frac{Q}{V}## then the free charge is ##Q=\frac{3\epsilon_0 AdV}{2L}##. I don't really know where to go from here, or even if what I have done so far is how I want to approach it. I also thought about "slicing" the capacitor horizontally so that each little slice looked like a capacitor with a linear dielectric inside, so that I could use ##P=\epsilon_0(\epsilon_r-1)E##. But, I again, don't know where I would go from there. Any help would be appreciated.
I made your post readable. The slicing is a good idea, but the slices are capacitors connected in series, and they do not add. Use that the normal component of D is the same at every x, and the potential difference of the slice capacitors is dU = Edx=(D/ε) dx. Get the total voltage across the capacitor by integrating E.

ehild
 
Last edited:
  • #5
ehild

I was actually thinking of slicing perpendicularly to the plates, so that the dielectric in each slice had the same ε[itex]_{r}[/itex], and that way the slice capacitors would be connected in parallel.
 
  • #6
The dielectric constant varies perpendicularly to the plates from 1 at x=0 (bottom plate) to 2 at x=d (top plate). At a given x, ε is constant on the plane parallel with the plates. You have to slice the capacitor parallel to its plates.

ehild
 
  • #7
No, you are misreading the question. The ε is constant in the plane perpendicular to the plates. The x-axis is running along the plate.
 
  • #8
I am sorry. I was misreading the question. I was thinking "bottom OF THE plate" and "top OF THE plate." Your interpretation of it is correct, it is the "bottom plate" and "top plate."

Thanks. This may have been where all my confusion came from. I'll have to start fresh on it.
 
  • #9
Yeah, I realized right after I posted, sorry. I feel like an idiot, I wasted a lot of time attempting the wrong problem.
 
  • #10
No worries. Start it again. It is simpler than the "vertical" slicing.

ehild
 
  • #11
ehild,

I was looking at this problem again, and your suggestion. I don't understand how finding the voltage the way you describe helps me find bound charge. Could you please elaborate?
 

What is a dielectric?

A dielectric is a material that can store electric charge without conducting it. Examples of dielectric materials include rubber, glass, and plastic. They are commonly used in capacitors to increase their capacitance.

What is bound charge?

Bound charge is the electric charge that is stored in a dielectric material when it is subjected to an electric field. It is the result of the polarization of the material's molecules, which causes them to align in the direction of the electric field.

Why is it important to find the bound charge in a dielectric?

Knowing the bound charge in a dielectric is important for understanding the behavior of capacitors and other electronic devices that use dielectric materials. It also helps in the design and optimization of these devices.

How can the bound charge in a dielectric be calculated?

The bound charge in a dielectric can be calculated using the electric displacement vector, which is related to the electric field and the permittivity of the material. It can also be found by subtracting the free charge from the total charge in the material.

What factors affect the bound charge in a dielectric?

The bound charge in a dielectric is affected by the strength of the electric field, the permittivity of the material, and the temperature of the material. It is also influenced by the type of dielectric and its physical properties, such as its thickness and surface area.

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