How Do Different Dielectrics Affect Capacitance in a Parallel-Plate Capacitor?

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SUMMARY

The discussion focuses on calculating the capacitance of a parallel-plate capacitor with two different dielectrics. The capacitor has a plate area of 5.52 cm² and a plate separation of 5.52 mm, with dielectric constants κ1 = 7.00 and κ2 = 15.0. The capacitance formula used is C = K * ε * A / d. Participants noted discrepancies in their calculations, particularly in the exponent of the resulting capacitance value, suggesting potential issues with unit conversions.

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Homework Statement


There is a parallel-plate capacitor with a plate area A = 5.52 cm2 and a plate separation d = 5.52 mm. The left half of the gap is filled with material of dielectric constant κ1 = 7.00; the right half is filled with material of dielectric constant κ2 = 15.0. What is the capacitance?


Homework Equations


C= K(epsil) A/ d


The Attempt at a Solution


I tried to add the two values of K, I also converted both A and d to m^2 and m, respectively. I got 9.74 x 10^-14 F
 
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I'm getting a similar answer, except it disagrees in the 10^-14 part. I suspect the conversions to m^2 and m may be off; what did you get for A and d?
 

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