- #1
jayjay112
- 36
- 0
Find the capacitance of a parrell plate capacitor in pF with A=0.04 m^2
distance of separation between plates = 2mm relative permittivity of dielectric = 6
and permittivity of free space = 8.854x10^-12
C = (6)(8.854x10^-12)(0.04) / 0.002
= 1.06248x10^-9 F
in pF = 1062.48 pF
Thanks
distance of separation between plates = 2mm relative permittivity of dielectric = 6
and permittivity of free space = 8.854x10^-12
C = (6)(8.854x10^-12)(0.04) / 0.002
= 1.06248x10^-9 F
in pF = 1062.48 pF
Thanks