The electric field between the plates of a paper-separated capacitor is 8.18×104 . The plates are 2.20 apart, and the charge on each plate is 0.795 .Determine the capacitance of this capacitor. Determine the area of each plate.
E= 8.18•10⁴ V/m
d=2.2 mm
q=0.775 μ C
ε (paper)= 2.3
(a) C=q/U
E=U/d
C=q/Ed = ...
(b)
ε₀=8.85•10⁻¹² F/m
C= ε₀εA/d =>
A=Cd/ε₀ε=...
To determine the capacitance of the capacitor, we can use the formula:
C = Q / V
Where C is the capacitance, Q is the charge on each plate, and V is the potential difference (voltage) between the plates.
In this case, Q = 0.795 C and V = 8.18×10^4 V.
Substituting these values into the formula:
C = 0.795 / (8.18×10^4)
C ≈ 9.71×10^(-6) F
Therefore, the capacitance of the capacitor is approximately 9.71×10^(-6) Farads.
To determine the area of each plate, we can rearrange the formula for capacitance:
C = ε₀ * A / d
Where C is the capacitance, ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
In this case, C ≈ 9.71×10^(-6) F and d = 2.20 m. The permittivity of free space, ε₀, is approximately 8.85×10^(-12) F/m.
Substituting these values into the formula and rearranging for A:
A = (C * d) / ε₀
A = (9.71×10^(-6) * 2.20) / (8.85×10^(-12))
A ≈ 2.42×10^(-5) square meters
Therefore, the area of each plate is approximately 2.42×10^(-5) square meters.