Dielectric Help: Solving Parallel Plate Capacitor

AI Thread Summary
The discussion revolves around calculating the capacitance of a parallel plate capacitor using different dielectric materials. The user initially miscalculated the area of the aluminum plates, using 0.180 m² instead of the correct value of 0.018 m². Upon correcting this error, the user successfully solved for the largest and smallest capacitance values using the provided dielectric constants and thicknesses. The largest capacitance was found using glass, while the smallest was determined using paraffin. The thread concludes with the user resolving their confusion and arriving at the correct calculations.
Bob Loblaw
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[SOLVED]Dielectric help

Homework Statement



To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 180 cm2), a sheet of paper (thickness = 0.10 mm, k= 3.5), a sheet of glass (thickness = 2.0 mm, . k= 7.0), and a slab of paraffin (thickness = 10.0 mm, k= 2.0).
(a) What is the largest capacitance possible using one of these dielectrics?
_________nF
(b) What is the smallest?
________pF

Homework Equations



C=k\epsilon_0 \frac{A}{d}

The Attempt at a Solution



I used the above equation to evaluate the capacitance of each in this manner:
1) 3.5*8.85e-12*.180m^2/.0001 = 5.5755e-8
2) 7.0*8.85e-12*.180m^2/.002 = 5.5755e-9
3) 2.0*8.85e-12*.180m^2/.010 = 3.186e-10

I recorded the largest as .55.755 and the smallest as 318.6 (in their appropriate units) both of which were wrong. Any ideas where I messed up?
 
Last edited:
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self-solved it. 180cm2 is .018m2, not .180m2 as I have recorded.
 
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