# Homework Help: Electric Field in dielectric

1. May 5, 2014

### unscientific

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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,

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

2. May 6, 2014

### 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 $x = d$.

3. May 6, 2014

### unscientific

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, $\sigma = \frac{2}{3}\rho_0 d^3$.

Last edited: May 6, 2014
4. May 6, 2014

### 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

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

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