Capacitance (farads) am I doing this right?

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Homework Help Overview

The discussion revolves around calculating the capacitance of a parallel plate capacitor using various dielectric materials, specifically paper, glass, and paraffin. Participants are exploring the application of the capacitance formula and the conversion of units.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the capacitance formula and the necessary parameters, questioning the correct interpretation of the thickness of the dielectric materials as the distance in the formula. There are attempts to verify calculations and unit conversions, particularly regarding area and its impact on capacitance results.

Discussion Status

Some participants have provided specific calculations and expressed discrepancies in their results, prompting requests for clarification on the calculation process. There is acknowledgment of potential errors in unit conversions, particularly concerning area, which has led to differing capacitance values.

Contextual Notes

Participants are working under the constraints of homework guidelines that require accurate unit conversions and application of the capacitance formula. There is an emphasis on ensuring all values are correctly converted to SI units before calculations.

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1. 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; then enter the largest capacitance possible, of the three, in nF and the smallest in pF.



2. (capacitance) = ((epsilon not)x(Area)x(k)) / (d)

**d, or distance between the capacitors, would be the thickness of the paper/glass/paraffin right?**



3. my answers are:

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

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

when i enter the values on my webassign it says I am wrong... are the values correct and just converted wrong or am i totally off [i did converted all the given values to meters before plugging them into the equation]
 
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phys-lexic said:
3. my answers are:

5.885E-7 farads for paper
5.885E-8 farads for glass
3.363E-9 farads for paraffin
This is not what I get. I get different exponents. Can you show one example of how you calculated the capacitance?
 
in example, for paper i did

(Eo) (A) (k) (d)
(8.85E-12)x(1.9)x(3.5)... then divided by .0001 (thickness)
 
the equation i though to use was C = ( (Eo)x(A)x(k) ) / (d)
 
The error lies in the conversion of the area; 190 cm[tex]^{2}\neq[/tex] 1.9 m[tex]^{2}[/tex].
 
oh wow... i was in a rush and forgot it was area... hha thanks a lot
 

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