# Finding the bound charge in a dielectric ATTEMPT 2

• xophergrunge
In summary, the space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant ϵr varies linearly from 1 at the bottom plate (x=0) to 2 at the top 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.
xophergrunge

## Homework Statement

The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant ϵr varies linearly from 1 at the bottom plate (x=0) to 2 at the top 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. 2. Homework Equations

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

$C=\frac{Aε_{0}}{d}$

$D=ϵE$

$D=ε_{0}E+P$

$\int D\bullet da = Q_{f}$

$σ_{b}=P⋅\widehat{n}$

$ρ_{b}=−∇⋅P$

## The Attempt at a Solution

First I found $ε_{r}$ as a function of x, $ε_{r}=\frac{x}{d}+1$.
Assuming that the area of each plate is A, I said that the bound charge must be $Q_b=A(\sigma_b+\int\rho_{b}dx)$.
Next, using $D=ε_{0}E+P$ and $D=εE$ I get that $P=D(\frac{x}{x+d})$.
Using $\int D\bullet da = Q_{f}$ I get that $D=\frac{Q_{f}}{A}$.
Nowing using $σ_{b}=P⋅\widehat{n}$ and $ρ_{b}=−∇⋅P$ I get that $\sigma_b=0$ at x=0, $\sigma_b=-\frac{Q_f}{2A}$ at x=d and $\rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}$.
I integrate $\rho_b$ from 0 to d and I get $\frac{Q_f}{A}(\ln 2 -\frac{1}{2})$.
So, now plugging those values into $Q_b=A(\sigma_b+\int\rho_{b}dx)$ I get $Q_f(\ln 2 -1)$, which only equals zero when there is no free charge on the plates of the capacitor, so I know I am doing something wrong. I am pretty sure I am going about this completely wrong, but I don't see any other way to do it. Any help would be really appreciated. Thank you.

Also, using $C=\frac{Q}{V}$, $C=\frac{Aε_{0}}{d}$, and $D=ϵE$ I can get the capacitance $C=\frac{A\epsilon_0}{d\ln 2}$ and $Q_f=\frac{A\epsilon_{0}V}{d\ln 2}$ but I don't see how that will help me find the bound charge.

Last edited:
xophergrunge said:
I get that $P=D(\frac{x}{x+d})$.
$D=\frac{Q_{f}}{A}$.

I think that's correct.

I get that $\sigma_b=0$ at x=0, $\sigma_b=-\frac{Q_f}{2A}$ at x=d
Should ##\sigma_b## have a negative sign at x = d? I guess it depends on your sign conventions and which plate is positively charged.

and $\rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}$.

I don't agree with the numerator ##x## in this expression.

1 person
I get negative because $\hat{n}=-\hat{i}$ since x is increasing as you move from the first plate (x=0) to the second (x=d).

I just re-did my calculation for $\rho_b$ and I got d in the numerator now, and the whole thing has a negative sign. This will change my $\int \rho_b dx$ and hopefully give me an answer that makes sense. Is that what you got?

Thanks.

Yes, I got ##d## in the numerator instead of ##x##. I got positive for ##\sigma_b## at ##x = d## and I got ##\rho_b## to be negative. I assumed the lower plate (x = 0) is the positively charged plate so that D, E, and P all point in the positive x direction.

1 person
Great, I am getting that $Q_b=0$ now, with the bound volume charge equal to negative one half and the bound surface charge to be negative one half. Thank you so much.

OK, but how do two negative quantities add to zero?

Sorry, that was a typo. Only the volume bound charge was meant to be negative.

And I meant one half $Q_f$ for both cases.

And I realized my mistake with the negative sign on $\sigma_b$, I was thinking of the surface of the plate rather than the surface of the dielectric.

Great. Good work.

Thanks again.

## 1. What is bound charge in a dielectric?

Bound charge in a dielectric refers to the electric charges that are bound to the atoms or molecules in a dielectric material. These charges do not move freely like in a conductor, but rather they are displaced slightly from their original positions.

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

It is important to find the bound charge in a dielectric because it helps in understanding the behavior of the material in an electric field. This information is useful in designing and optimizing electronic devices and circuits.

## 3. How is bound charge different from free charge?

Bound charge is different from free charge as it is not able to move freely in a material, while free charge can move within a material. Bound charge is also affected by the polarization of the material, while free charge is not.

## 4. What factors affect the magnitude of bound charge in a dielectric?

The magnitude of bound charge in a dielectric is affected by the polarization of the material, the strength of the applied electric field, and the dielectric constant of the material. The type of dielectric material and its temperature can also influence the magnitude of bound charge.

## 5. How do scientists calculate the bound charge in a dielectric?

Scientists use mathematical equations, such as Gauss's Law, to calculate the bound charge in a dielectric. These equations take into account the properties of the dielectric material, the electric field strength, and the geometry of the system in order to determine the bound charge.

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