# Charging a parallel plate capacitor.

tntenigma
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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Homework Helper
welcome to pf!

hi tntenigma! welcome to pf! for a capacitor of area A and separation d …

C = Q/V

V = -Ed

E = D/ε

D = -Q/A

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

C = εA/d

Where
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