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.

Answer 1

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=...