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 areaof 8.0×10-4 m² and separated from each other by 5.0 mm. Half of thespace between the plates is filled with glass and the other with mica.Find the capacitance of this capacitor.2.glassmicaGiven: A = 8.0×10-4 m?Eplass = 7×10-11 C²/N•m²d = 5.00 mm = 0.005 mEmica = 4.8×10-11 C?/N•m?Solution:The setup is considered as a parallel combination of two capacitors, each witharea = 4.0×10-4 m², one with mica as dielectric, and the other with glass.Using Eq. (2.8),A4.0x10mColass =E,glass d= (7x10-1" C/N•m²)= 5.6x10-12 F ~6x10-12 F--0.005 mA4.0×10m?(4.8×10-" C²/N m²)d-11Cmica =Emica= 3.84x10 12 F 4x10-12 F%3D0.005 mC=Cs +Cie =5.6x10-12 F+ 3.84x1012 F = 9.44×10-12 F9.4×10-12 F.%3Dmica%3D'glassra Tith 20 Fond 3 0 P

Given: The capacitance of each capacitor is 10 F. The number of the capacitors is 5. The equivalent…

Answered: You have five 10.0 F capacitors. show…

College Physics

11th Edition

ISBN:9781305952300

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

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