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

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In a parallel-plate capacitor with a plate area of 5.52 cm² and a separation of 5.52 mm, the left half is filled with a dielectric constant of 7.00 and the right half with 15.0. The capacitance can be calculated using the formula C = K(epsilon) A / d. There is confusion regarding the conversion of area and distance to meters, which may affect the final capacitance value. Participants are encouraged to verify their unit conversions for accuracy in calculating the capacitance. Accurate conversions are crucial for obtaining the correct capacitance value in this scenario.
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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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