Parallel plate capacitor with two dielectric materials?

AI Thread Summary
The discussion focuses on using Gauss' law to derive the electric field in a parallel plate capacitor with two dielectric materials. The initial approach involves treating the capacitor as two series capacitors, calculating the electric fields for each dielectric separately. The proposed equations for the electric fields are E1 = Q/(ε0*εr1*A) and E2 = Q/(ε0*εr2*A), leading to a total electric field expression. An alternative method is suggested, averaging the permittivities of the two dielectrics based on their proportions in the gap, resulting in a combined permittivity formula. This alternative approach appears to be more promising for finding the electric field.
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Homework Statement


Use Gauss' law to find an expression for the electric field within a charged parallel plate capacitor with two layers of dielectric materials (?r1 and ?r2).
(See attached image, this is for question 2a).

The Attempt at a Solution


My approach would be to split the capacitor into two capacitors, each with r1 and r2 dielectrics - connected in series. Then the electric field across them both would be the sum of the electric fields across each one?

So:
For The first dielectric:
The integral of: E1.dA = Q/(e0*er1)
E1 = Q/(e0*er1*A)
E2 = Q/(e0*er2*A)

Therefore
Etotal = (Q/(?0*A)) * (1/?r1 + 1/?r2)

Am I taking the right approach?

Any help would be appreciated,

Thanks.
 

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Just had a brain wave: is it possible to treat both dielectrics as a single material by averaging their respective permittivities divided by their proportion of the gap?

So
\epsilon_{T}=\epsilon_{r1}*1/3 + \epsilon_{r2}*2/3 = (\epsilon_{r1} +2\epsilon_{r2})/3

Therefore:
E = 3Q/(A\epsilon_{0}(\epsilon_{r1}+2\epsilon_{r2}))

This seems more likely... I think...
 
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