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十++ A parallel plate capacitor is made up of two plates, each having an area of 8.0×10-4 m? 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. glass mica Eglass = 7x10-1" C²/N•m? Emica = 4.8x10-1" C?/N•m? Given: A = 8.0×10-4 m? d = 5.00 mm = 0.005 m

十++ A parallel plate capacitor is made up of two plates, each having an area of 8.0×10-4 m? 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. glass mica Eglass = 7x10-1" C²/N•m? Emica = 4.8x10-1" C?/N•m? Given: A = 8.0×10-4 m? d = 5.00 mm = 0.005 m

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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
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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