Parallel plate capacitor with two dielectric layers

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Punchlinegirl
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A capacitor is constructed from 2 square metal plates. A dielectric k= 4.93 fills the upper half of the capacitor and a dielectric k=10 fills the lower half of the capacitor. Neglect edge effects. The plate are separated by a distance of 0.37 mm and the length of the plates is 27 cm. Calculate the capacitance C of the device. Answer in units of pF.
I used the equation for C of a parallel plate capacitor.
[tex]kE_o A/d + kE_o A/d[/tex]
Where the area= 3.7 x 10^-4 * .135 = 4.995 x 10^-5
So (4.93)(8.85 x 10^-12)(4.995 x 10^-5)/ 3.7 x 10^-4 + (10)(8.85 x 10^-12)(4.995 x 10^-5)/ (3.7 x 10^-4)
= 1.78 x 10^-11 F.
Converting to pF gave me 1.78 x 10^-23 which isn't right... help please?
 

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I think d has to be different for each one, but I'm not sure how to do that.
 
consider two capacitors in series each having distance d = half of 0.37 mm.The first has dielectric k=4.93,second has k=10.

Or you may use the following equation-

C=EA/[d/k + d'/k'] here d=d'=half of 0.37 mm
 
So would I do,
(8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?
 
Punchlinegirl said:
So would I do,
(8.85 x 10^-12)(.27)/ (3.7 x 10^-4/4.93) + ((1.85 x 10^-4)/10) = 2.55 x 10^-8 ?

area is (.27 m)^2;also d=1.85 x 10^-4 m.
 
Ok so it would be:
(8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
Am I doing this right now?
 
Punchlinegirl said:
Ok so it would be:
(8.85 x 10^-12)(.0729) / ((1.85 x 10^-4/4.93) + (1.85 x 10^-4/10)) =1.15 x 10^-8
Am I doing this right now?
yes,this should be the answer in farad.

If you are not allowed to use the formula directly,then it would be better to consider two capacitors in series rather than to deduce the formula.