Parallel plate capacitor with two dielectric materials?

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

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

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

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