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e capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capa

e capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capa

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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The capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor.

Practice Exercises 2.6 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. Suppose the space between the plates of the capacitor in Sample Problem 2.6 item number 2 is filled with equal thicknesses of the same dielectrics but arranged as shown. What is its capacitance? 2. ++++++++ glass mica 3. The capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor.
Transcribed Image Text:Practice Exercises 2.6 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. Suppose the space between the plates of the capacitor in Sample Problem 2.6 item number 2 is filled with equal thicknesses of the same dielectrics but arranged as shown. What is its capacitance? 2. ++++++++ glass mica 3. The capacitors in Sample Problem 2.6 item number 3 are connected in parallel instead of being in series. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor.
Sample Problems 2.6 1. Given that C,=10.0 F, C,=5.0 F, and C;=4.0 F, find the total capacitance for each connection shown. (a) (Ь) (c) Given: C,=10.0 F C,=5.0 F C;=4.0 F Solution: C,, C2, and C, are in series. Therefore, а. 1 1 1 1 Crotal C, C, 1 1 1 1 Cotal 4.0 F 10.0 F 5.0 F 0.10 0.20 0.25 F F 1 0.55 F Cotal Cal =1.8 F b. C,, C2, and C3 are in parallel. Thus, Croual = G, +C, +C, total Ctal =10.0 F+5.0 F+4.0 F =19 F C, and C, are in series. Therefore, their combined capacitance C+2 is C. 1 1 1 1 0.10 0.20 0.30 C2 10.0 F 5.0 F F F F C+2 = 3.3 F This series combination of C, and C, is parallel to C3. Therefore, the total capacitance is Croral = C+2 + C; = 3.3 F + 4.0 F = 7.3 F 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 Eglass = 7x10-" C?/N•m? Emica = 4.8×10-1" C²/N•m² Given: A = 8.0×10-4 m² d= 5.00 mm = 0.005 m Solution: The setup is considered as a parallel combination of two capacitors, each with area = 4.0×10-“ m², one with mica as dielectric, and the other with glass. Using Eq. (2.7), A (7×10 "C’/N ·m’) 4.0x10 m? = 5.6×10 1² F 26×10 ² F Cglass Eglass 0.005 m 4.0×10n 'm² = 3.84×10¬1² F 4×10¯1² F A Cmica =E, = (4.8×10"C’/N -n 2 /N•m³) Fmica 7 0.005 m C=C_ +Cia = 5.6×10¬1² F + 3.84×10-1² F=9.44×10¬1² F=9.4×10-1² F "glass Two capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in series and subjected to a total potential difference of 100 V. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor. 3. Given: C, = 2.0 F C2 = 3.0 F Solution: 1 1 1 1 1 0.5 0.33 0.83 a. Ccotal C ' C, 2.0 F 3.0 F F F F Ccotal =1.2 F "A" I rotal = Cotal Vcoral = (1.2 F)(100 V)=120 C Since the capacitors are in series, qotal = 120 C=q, = q2. b. 120 C V across 2.0 F capacitor = = 60 V с. 2.0 F 120 C V across 3.0 F capacitor =- C, = 40 V 3.0 F
Transcribed Image Text:Sample Problems 2.6 1. Given that C,=10.0 F, C,=5.0 F, and C;=4.0 F, find the total capacitance for each connection shown. (a) (Ь) (c) Given: C,=10.0 F C,=5.0 F C;=4.0 F Solution: C,, C2, and C, are in series. Therefore, а. 1 1 1 1 Crotal C, C, 1 1 1 1 Cotal 4.0 F 10.0 F 5.0 F 0.10 0.20 0.25 F F 1 0.55 F Cotal Cal =1.8 F b. C,, C2, and C3 are in parallel. Thus, Croual = G, +C, +C, total Ctal =10.0 F+5.0 F+4.0 F =19 F C, and C, are in series. Therefore, their combined capacitance C+2 is C. 1 1 1 1 0.10 0.20 0.30 C2 10.0 F 5.0 F F F F C+2 = 3.3 F This series combination of C, and C, is parallel to C3. Therefore, the total capacitance is Croral = C+2 + C; = 3.3 F + 4.0 F = 7.3 F 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 Eglass = 7x10-" C?/N•m? Emica = 4.8×10-1" C²/N•m² Given: A = 8.0×10-4 m² d= 5.00 mm = 0.005 m Solution: The setup is considered as a parallel combination of two capacitors, each with area = 4.0×10-“ m², one with mica as dielectric, and the other with glass. Using Eq. (2.7), A (7×10 "C’/N ·m’) 4.0x10 m? = 5.6×10 1² F 26×10 ² F Cglass Eglass 0.005 m 4.0×10n 'm² = 3.84×10¬1² F 4×10¯1² F A Cmica =E, = (4.8×10"C’/N -n 2 /N•m³) Fmica 7 0.005 m C=C_ +Cia = 5.6×10¬1² F + 3.84×10-1² F=9.44×10¬1² F=9.4×10-1² F "glass Two capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in series and subjected to a total potential difference of 100 V. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor. 3. Given: C, = 2.0 F C2 = 3.0 F Solution: 1 1 1 1 1 0.5 0.33 0.83 a. Ccotal C ' C, 2.0 F 3.0 F F F F Ccotal =1.2 F "A" I rotal = Cotal Vcoral = (1.2 F)(100 V)=120 C Since the capacitors are in series, qotal = 120 C=q, = q2. b. 120 C V across 2.0 F capacitor = = 60 V с. 2.0 F 120 C V across 3.0 F capacitor =- C, = 40 V 3.0 F
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