Charging a parallel plate capacitor.

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SUMMARY

The capacitance of a parallel-plate capacitor is determined by the formula C = εA/d, where C is capacitance in Farads, ε is the permittivity of the dielectric (8.854 * 10^-12 F/m), A is the area of the plates in square meters, and d is the separation distance in meters. The capacitance remains unaffected by the voltage applied to the capacitor, relying solely on the physical dimensions and dielectric properties. For air as the dielectric, the formula can be simplified to C (in pF) = 0.0885 * Area (in square cm) / separation distance (in cm), allowing for straightforward calculations based on plate area and distance.

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  • Understanding of capacitance and its units (Farads)
  • Familiarity with the formula C = εA/d
  • Knowledge of dielectric materials and their permittivity
  • Basic geometry for calculating area of plates
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  • Learn about the implications of varying plate area and separation distance on capacitor performance
  • Investigate practical applications of capacitance sensing circuits
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tntenigma
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I'm building a parallel-plate capacitor for a capacitance sensing circuit. If I let the capacitor fully charge, and then pulled the plates apart by 5 cm, how would that change the capacitance? Is there a formula to figure that out?
 
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welcome to pf!

hi tntenigma! welcome to pf! :smile:

for a capacitor of area A and separation d …

C = Q/V

V = -Ed

E = D/ε

D = -Q/A

so C = εA/d :wink:
 
You can calculate the capacitance as follows:

C = εA/d

Where
C= capacitance in Farads
ε = 8.854 * 10 ^-12 Farads per meter
A = area of plates in square meters
d = separation of plates in meters

The capacitance does not depend on the voltage on the capacitor, only on the dimensions and permittivity of the dielectric.

If the dielectric is air, this formula can be simplified as follows:
C (in pF) = 0.0885 * Area ( in square cm) / separation distance (in cm)

So, if the plates were 10 cm * 10 cm that is 100 sq cm
If these were 0.5 cm apart,
C = 0.0885 * 100 / 0.5 = 17.5 pF
 

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