Bob Loblaw
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[SOLVED]Dielectric help
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
[tex]C=k\epsilon_0 \frac{A}{d}[/tex]
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?
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
[tex]C=k\epsilon_0 \frac{A}{d}[/tex]
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?
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