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
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
To find the plate area required to provide a capacitance of 3.78 pF with a separation of 0.303 mm and air as the dielectric material, we can use the formula for the capacitance of a parallel-plate capacitor:
C = (ε₀ * A) / d
Where:
C is the capacitance (in Farads)
ε₀ is the permittivity of vacuum (8.8542 × 10−12 F/m)
A is the area of the plates (in square meters)
d is the separation between the plates (in meters)
We need to rearrange the formula to solve for A:
A = (C * d) / ε₀
Now we can substitute the given values into the formula to find the plate area:
C = 3.78 pF = 3.78 × 10^(-12) F (convert from pico to Farads)
d = 0.303 mm = 0.303 × 10^(-3) m (convert from millimeters to meters)
ε₀ = 8.8542 × 10−12 F/m (given)
A = (3.78 × 10^(-12) F * 0.303 × 10^(-3) m) / (8.8542 × 10−12 F/m)
Now, simplifying the expression:
A = 1.147 × 10^(-15) m^2
Therefore, the plate area required to provide a capacitance of 3.78 pF with a separation of 0.303 mm and air as the dielectric material is approximately 1.147 × 10^(-15) square meters.