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=...
The electric field between the plates of a paper-separated (K = 3.75) capacitor is 9.21 104 V/m. The plates are 2.05 mm apart and the charge on each plate is 0.775 µC. Determine the capacitance of this capacitor.
1 answer