Dielectric Help: Solving Parallel Plate Capacitor

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SUMMARY

The discussion focuses on calculating the capacitance of a parallel plate capacitor using various dielectrics: paper, glass, and paraffin. The formula used is C = kε₀(A/d), where k is the dielectric constant, ε₀ is the permittivity of free space, A is the area, and d is the thickness of the dielectric. The user initially miscalculated the area of the plates, using 0.180 m² instead of the correct value of 0.018 m². After correcting this error, the user successfully solved for the largest and smallest capacitance values.

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  • Understanding of capacitance and dielectric materials
  • Familiarity with the formula C = kε₀(A/d)
  • Knowledge of units for capacitance (farads, nanofarads, picofarads)
  • Basic skills in unit conversion (cm² to m²)
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Students in electrical engineering, physics enthusiasts, and anyone involved in capacitor design and analysis will benefit from this discussion.

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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