Anonymous
Anonymous asked in Science & MathematicsEngineering · 8 years ago

# What is the capacitance in pF?

Two 3.0cm x 3.0cm square aluminum electrodes, spaced 0.70mm apart are connected to a 400V battery.

I am using the C=eA/D but its not working.

Relevance

sure it is working, you just need to keep track of units - in metric system, unit of length is meter, not cm, mm, km etc.

so

A=0.03x0.03=0.0009 m^2

D=0.0007 m

ε0 ≈ 8.854187817620... × 10−12 F/m

or

ε0 ≈ 0.000000000008854187817620 F/m

you get

C=1.1384x10^-11F

• The capacitor

c = εo(A)/d where εo = 8.854 x 10^-12 Farad/meter

the plate area is (3cm)(3 cm) = 9 cm^2 which in square meters is

A = 9 x 10^-4 m^2

The distance between the plates is 0.7 mm which in meters is

d = 7 x 10^-4 m

c = (8.854 x 10^-12F/m)(9 x 10^-4 m^2)/ (7 x 10^-4 m) = 1.14 x 10^-11F ≈ 11 pF where pF is pico Farad which is 1 x 10^ -12 Farads

Source(s): The permittivity εo from Wikipedia
• Capacitance C = (0.0885 KA) divided by d, or C = (0.224 KA) divided by d.

In the first formula "A" is the area of the overlap and is in square centimetres and "d" the distance apart is in centimetres.

In the second formula "A" is in square inches and "d" is in inches. "K" is the dielectric constant.

The K of a vacumm is 1. Air is slightly more.

• You have to use the correct units. Distance in meters, area in meter squared and eo in F/meter. It will work then. BTW, the correct equation is C=eo er A/D. er is 1 for air.