College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter16: Electrical Energy And Capacitance
Section: Chapter Questions
Problem 56P: A parallel-plate capacitor has plates of area A = 7.00 102 m2 separated by distance d = 2.00 104...
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1. You have five 10.0 F capacitors. show all the possible connections for the five capacitors to produce a total capacitance of 50.0 F.
2. Suppose the space between the plates of the capacitor in sample problems 2.6 item number 2 is filled with equal thickness of the same dielectrics but arranged as shown. What is its capacitance?
![+++++ +++
A parallel plate capacitor is made up of two plates, each having an area
of 8.0×10-4 m² and separated from each other by 5.0 mm. Half of the
space between the plates is filled with glass and the other with mica.
Find the capacitance of this capacitor.
2.
glass
mica
Given: A = 8.0×10-4 m?
Eplass = 7×10-11 C²/N•m²
d = 5.00 mm = 0.005 m
Emica = 4.8×10-11 C?/N•m?
Solution:
The setup is considered as a parallel combination of two capacitors, each with
area = 4.0×10-4 m², one with mica as dielectric, and the other with glass.
Using Eq. (2.8),
A
4.0x10m
Colass =E,
glass d
= (7x10-1" C/N•m²)
= 5.6x10-12 F ~6x10-12 F
--
0.005 m
A
4.0×10m?
(4.8×10-" C²/N m²)
d
-11
Cmica =Emica
= 3.84x10 12 F 4x10-12 F
%3D
0.005 m
C=Cs +Cie =5.6x10-12 F+ 3.84x1012 F = 9.44×10-12 F9.4×10-12 F.
%3D
mica
%3D
'glass
ra Tith 20 Fond 3 0 P](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e9b7fd0-895e-4646-a4a6-e00c42974f5f%2F0be7d2eb-c1e6-482f-8b49-d71bc74362cb%2Ft5h097r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:+++++ +++
A parallel plate capacitor is made up of two plates, each having an area
of 8.0×10-4 m² and separated from each other by 5.0 mm. Half of the
space between the plates is filled with glass and the other with mica.
Find the capacitance of this capacitor.
2.
glass
mica
Given: A = 8.0×10-4 m?
Eplass = 7×10-11 C²/N•m²
d = 5.00 mm = 0.005 m
Emica = 4.8×10-11 C?/N•m?
Solution:
The setup is considered as a parallel combination of two capacitors, each with
area = 4.0×10-4 m², one with mica as dielectric, and the other with glass.
Using Eq. (2.8),
A
4.0x10m
Colass =E,
glass d
= (7x10-1" C/N•m²)
= 5.6x10-12 F ~6x10-12 F
--
0.005 m
A
4.0×10m?
(4.8×10-" C²/N m²)
d
-11
Cmica =Emica
= 3.84x10 12 F 4x10-12 F
%3D
0.005 m
C=Cs +Cie =5.6x10-12 F+ 3.84x1012 F = 9.44×10-12 F9.4×10-12 F.
%3D
mica
%3D
'glass
ra Tith 20 Fond 3 0 P
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