Asked by Anonymous
The plates of a parallel-plate capacitor are
separated by 0.303 mm.
If the material between the plates is air,
what plate area is required to provide a ca-
pacitance of 3.78 pF ? The permittivity of a
vacuum is 8.8542 × 10−12 F/m.
Answer in units of m2
separated by 0.303 mm.
If the material between the plates is air,
what plate area is required to provide a ca-
pacitance of 3.78 pF ? The permittivity of a
vacuum is 8.8542 × 10−12 F/m.
Answer in units of m2
Answers
Answered by
Elena
C=εε₀A/d
ε=1
ε₀=8.85 •10⁻¹² F/m
A=Cd/ ε₀=
=3.78•10⁻¹²•0.303•10⁻³/8.85•10⁻¹²=
=1.29•10⁻⁴ m²
ε=1
ε₀=8.85 •10⁻¹² F/m
A=Cd/ ε₀=
=3.78•10⁻¹²•0.303•10⁻³/8.85•10⁻¹²=
=1.29•10⁻⁴ m²
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