Calculate Capacitance of Flat Plate Capacitor

[phys-lexic]

the problem is:

To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 190 cm2), a sheet of paper (thickness = 0.10 mm, κ = 3.5), a sheet of glass (thickness = 2.0 mm, κ. = 7.0), and a slab of paraffin (thickness = 10.0 mm, κ = 2.0). Find each's capacitance.

my answers are:

5.885E-7 farads for paper
5.885E-8 farads for glass
3.363E-9 farads for paraffin

...now the question asks to enter the largest capacitance possible, of the three, in nF and the smallest in pF...

-smallest= 3363 pF (paraffin)
-largest= 588.5 nF (paper)

am i right with my values and just converting wrong or am i totally off... i converted all the values to meters before plugging them into the equation below:


*****here is the equation i am using*****

(capacitance) = ((epsilon not)x(A)x(k)) / (d=thickness?)

 

[AvroArrow]

where epsilon not is 8.85E-12, A is area of plates, k is the dielectric constant, and d is the distance between the plates (thickness).

 

[nlightner]

Yes, your values and conversions are correct. The equation you are using is the correct formula for calculating capacitance for a parallel plate capacitor. The only thing I would suggest is to use the correct units for capacitance, which is Farads (F), instead of the scientific notation you used (E-7 and E-8). Other than that, your calculations are correct.