Asked by nammilee_2012
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.
Answers
Answered by
Elena
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=...
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=...
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.