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Find Electric Field between Two Plates with Varying Permittivity 
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#1
Jan2711, 09:58 PM

P: 7

We have a problem for our Electromagnetics class. It is the following
"A parallelplat capacitor of crosssectional area A an thickness d is filled with a dielectric material whose relative permittivity varies linearly from [tex]\epsilon_{r}=1[/tex] at one plate to [tex]\epsilon_{r}=10[/tex] at the other plate. a) Find the electric field in the capacitor b) What is the potential difference between the plates" For a we attempted to use Gauss's law [tex]\oint D \bullet ds = Q_{enclosed}[/tex] We took a cylindrical surface over one of the plates. Using the formula [tex]D = \epsilon E[/tex] we rewrote the surface integral as [tex]\int \epsilon_{0} ds + \int \epsilon_{r0} E ds[/tex] We did a second Gauss's law for the other plate of the capacitor. For the charge enclosed we assumed [tex]\pi r^2 \rho_s[/tex] and [tex] \pi r^2 \rho_s[/tex] respectively. This gave us two equations [tex]\rho_s = \epsilon_0 E_1 + \epsilon_{r0} E_{2}[/tex] [tex] \rho_s = \epsilon_0 E_3 + \epsilon_{rd} E_{4}[/tex] We assumed that [tex]E_1 = E_4[/tex] So we solved each equation for those E and then set them equal to each other to get [tex]\rho_s  \epsilon_{r0} E_{2} =  \rho_s  \epsilon_{rd} E_{rd}[/tex] We know from the problem that [tex]\epsilon_{r0}[/tex] equals [tex]\epsilon_0[/tex] and we set considered [tex]E_2[/tex] to be an initial value which we called [tex]E_0[/tex]. We also detemined an equation relation [tex]\epsilon_r[/tex] to x. [tex]\epsilon_r = 1 + \frac {9x} {d}[/tex] Plugging this in and solving for [tex]E_{rd}[/tex] gave us the following equation [tex]E_{rd} = \frac {\epsilon_0 E_0  2 \rho_s} {(1 + \frac {9x} {d}) \epsilon_0}[/tex] Our question is whether our approach is right. If it isn't we would like some direction to head in. For the second part of the problem we were going to use the relation [tex]E =  \frac {dV} {dx} [/tex] to find the potential. Any help is appreciated. 


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