How Does a Dielectric Material Affect Electric Field and Surface Charge Density?

[unscientific]

Homework Statement

Find the electric field inside dielectric and surface charge density on plates.

Homework Equations

The Attempt at a Solution

$$\rho_b = -\vec {\nabla} . \vec{P} = -\vec {\nabla} . \epsilon_0 \chi \vec E = -\vec {\nabla} . (\epsilon_r - 1)\epsilon_0 \vec E$$

$$\int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec {\nabla} . \vec E dV$$

$$\int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec E . d\vec S$$

$$\rho_0 \int (d^2 - x^2) A dx = (1-\epsilon_r)\epsilon_0 EA$$

$$E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0} \int_{x'}^{d} d^2-x'^2 dx'$$

$$E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0}\left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right]$$

To find surface charge density,

$$\epsilon_0 \nabla . E = \rho_f + \rho_b$$
$$\epsilon_0 \int \nabla . E dV = \int \rho_f dV + \int \rho_b dV$$
$$\epsilon_0 \int E.dS = \sigma A + \int \rho_b A dx$$
$$\epsilon_0 EA = \sigma + \int \rho_b dx$$

Rearranging,

$$\sigma = \epsilon_0E - \epsilon_0(1-\epsilon_r)E = \epsilon_0 \epsilon_r E$$
$$\sigma = \frac{\rho_0 \epsilon_r}{1-\epsilon_r} \left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right]$$

 

[TSny]

The embedded charge density given in the problems is "free charge" density $\rho_f$ of the system, not bound charge density $\rho_b$.

In your integrals, you are integrating from $x'$ to $d$. But the upper limit can be tricky to deal with since there is bound surface charge density on the surface of the dielectric at $x = d$ as well as free surface charge density on the surface of the conductor at $x = d$.

There is quite a bit of symmetry in the problem. Can you use the symmetry to tell you anything about D and E at x = 0? If so, you might want to set up your integrals from $x = 0$ to $x = x'$ and avoid $x = d$.

 

[unscientific]

The embedded charge density given in the problems is "free charge" density $\rho_f$ of the system, not bound charge density $\rho_b$.

In your integrals, you are integrating from $x'$ to $d$. But the upper limit can be tricky to deal with since there is bound surface charge density on the surface of the dielectric at $x = d$ as well as free surface charge density on the surface of the conductor at $x = d$.

There is quite a bit of symmetry in the problem. Can you use the symmetry to tell you anything about D and E at x = 0? If so, you might want to set up your integrals from $x = 0$ to $x = x'$ and avoid $x = d$.

Ah, I see the problem now.

Gauss's law in Dielectric reads:
$$\nabla . D = \epsilon_0 \epsilon_r \nabla . E = \rho_f$$

$$\epsilon_0 \epsilon_r EA = \rho_0 \int_0^{x'} (d^2 - x'^2) A dx'$$

$$E = \frac{\rho_0}{\epsilon_0 \epsilon_r}\left[d^2x - \frac{1}{3}x^3\right]$$

For the surface charge, we must compute the bound surface charges on the dielectric first.

Gauss's Law reads:

$$\epsilon_0 EA = Q - Q_b$$
$$\epsilon_0 EA = \sigma A - A\vec P . \hat n$$
$$\epsilon_0 E = \sigma - \epsilon_0 (\epsilon_r -1)E$$
$$\sigma = \epsilon_0 \epsilon_r E$$

Or in other words, $E = \frac{\sigma}{\epsilon_0\epsilon_r}$, the electric field is reduced by a factor of $\epsilon_r$ due to the presence of the dielectric's polarized field.

Electric field at d is $\frac{2}{3}\frac{\rho_0}{\epsilon_0\epsilon_r} d^3$.
Thus, $\sigma = \frac{2}{3}\rho_0 d^3$.

 

[TSny]

I get essentially the same result except for some signs. I get a positive amount of bound surface charge at the surface of the dielectric (due to the polarization of the medium). I get a negative surface charge density on the surface of the metal plate.

If I pick a Gaussian pill box as you did, I would write Gauss' law as

$-\epsilon_0 E_x A = Q_{tot} = Q + Q_b = \sigma A + P_x A = \sigma A +\epsilon_0(\epsilon_r - 1)E_x$

 

[paranormal]

I would like to clarify the context of this problem. It appears to be a calculation for the electric field and surface charge density inside a dielectric material, specifically between two parallel plates. The equations used are based on the relationship between the electric displacement field (D), the electric field (E), and the polarization of the dielectric material (P). The first equation shows the relationship between the bound charge density (ρb) and the polarization vector (P), which is then related to the electric field through the dielectric constant (εr). The second equation is a general expression for calculating the electric field inside a dielectric, and the third equation is used to find the surface charge density on the plates. The final equation calculates the surface charge density using the values from the previous equations. Overall, this is a valid approach to solving for the electric field and surface charge density in a dielectric material.