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 2.61. You have five 10.0 F capacitors. Show all the possible connections for the five capacitors toproduce a total capacitance of 50.0 F.Suppose the space between the plates of the capacitor in SampleProblem 2.6 item number 2 is filled with equal thicknesses of the samedielectrics but arranged as shown. What is its capacitance?2.++++++++glassmica3. The capacitors in Sample Problem 2.6 item number 3 are connected inparallel instead of being in series. The combination is connected to a100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potentialdifference across each capacitor. Sample Problems 2.61.Given that C,=10.0 F, C,=5.0 F, and C;=4.0 F, find the total capacitance for each connectionshown.(a)(Ь)(c)Given:C,=10.0 F C,=5.0 F C;=4.0 FSolution:C,, C2, and C, are in series. Therefore,а.1111CrotalC,C,1111Cotal4.0 F10.0 F5.0 F0.100.200.25FF10.55FCotalCal =1.8 Fb.C,, C2, and C3 are in parallel. Thus,Croual = G, +C, +C,totalCtal =10.0 F+5.0 F+4.0 F=19 FC, and C, are in series. Therefore, their combined capacitance C+2 isC.11110.100.200.30C210.0 F5.0 FFFFC+2 = 3.3 FThis series combination of C, and C, is parallel to C3. Therefore, the total capacitance isCroral = C+2 + C; = 3.3 F + 4.0 F= 7.3 FA 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.++++++++glassmicaEglass = 7x10-" C?/N•m?Emica = 4.8×10-1" C²/N•m²Given: A = 8.0×10-4 m²d= 5.00 mm = 0.005 mSolution:The setup is considered as a parallel combination of two capacitors, each witharea = 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 FCglass Eglass0.005 m4.0×10n'm²= 3.84×10¬1² F 4×10¯1² FACmica =E,= (4.8×10"C’/N -n2 /N•m³)Fmica 70.005 mC=C_+Cia = 5.6×10¬1² F + 3.84×10-1² F=9.44×10¬1² F=9.4×10-1² F"glassTwo capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in series andsubjected to a total potential difference of 100 V. Find the (a) total capacitance, (b) chargestored in each capacitor, and (c) potential difference across each capacitor.3.Given: C, = 2.0 FC2 = 3.0 FSolution:111110.50.330.83a.CcotalC ' C,2.0 F3.0 FFFFCcotal =1.2 F"A"I rotal = Cotal Vcoral = (1.2 F)(100 V)=120 CSince the capacitors are in series, qotal = 120 C=q, = q2.b.120 CV across 2.0 F capacitor == 60 Vс.2.0 F120 CV across 3.0 F capacitor =-C,= 40 V3.0 F

Question

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 2.61. You have five 10.0 F capacitors. Show all the possible connections for the five capacitors toproduce a total capacitance of 50.0 F.Suppose the space between the plates of the capacitor in SampleProblem 2.6 item number 2 is filled with equal thicknesses of the samedielectrics but arranged as shown. What is its capacitance?2.++++++++glassmica3. The capacitors in Sample Problem 2.6 item number 3 are connected inparallel instead of being in series. The combination is connected to a100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potentialdifference across each capacitor. Sample Problems 2.61.Given that C,=10.0 F, C,=5.0 F, and C;=4.0 F, find the total capacitance for each connectionshown.(a)(Ь)(c)Given:C,=10.0 F C,=5.0 F C;=4.0 FSolution:C,, C2, and C, are in series. Therefore,а.1111CrotalC,C,1111Cotal4.0 F10.0 F5.0 F0.100.200.25FF10.55FCotalCal =1.8 Fb.C,, C2, and C3 are in parallel. Thus,Croual = G, +C, +C,totalCtal =10.0 F+5.0 F+4.0 F=19 FC, and C, are in series. Therefore, their combined capacitance C+2 isC.11110.100.200.30C210.0 F5.0 FFFFC+2 = 3.3 FThis series combination of C, and C, is parallel to C3. Therefore, the total capacitance isCroral = C+2 + C; = 3.3 F + 4.0 F= 7.3 FA 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.++++++++glassmicaEglass = 7x10-&#34; C?/N•m?Emica = 4.8×10-1&#34; C²/N•m²Given: A = 8.0×10-4 m²d= 5.00 mm = 0.005 mSolution:The setup is considered as a parallel combination of two capacitors, each witharea = 4.0×10-“ m², one with mica as dielectric, and the other with glass.Using Eq. (2.7),A(7×10 &#34;C’/N ·m’)4.0x10 m?= 5.6×10 1² F 26×10 ² FCglass Eglass0.005 m4.0×10n'm²= 3.84×10¬1² F 4×10¯1² FACmica =E,= (4.8×10&#34;C’/N -n2 /N•m³)Fmica 70.005 mC=C_+Cia = 5.6×10¬1² F + 3.84×10-1² F=9.44×10¬1² F=9.4×10-1² F&#34;glassTwo capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in series andsubjected to a total potential difference of 100 V. Find the (a) total capacitance, (b) chargestored in each capacitor, and (c) potential difference across each capacitor.3.Given: C, = 2.0 FC2 = 3.0 FSolution:111110.50.330.83a.CcotalC ' C,2.0 F3.0 FFFFCcotal =1.2 F&#34;A&#34;I rotal = Cotal Vcoral = (1.2 F)(100 V)=120 CSince the capacitors are in series, qotal = 120 C=q, = q2.b.120 CV across 2.0 F capacitor == 60 Vс.2.0 F120 CV across 3.0 F capacitor =-C,= 40 V3.0 F

Accepted Answer

Given values:-
Capacitor,(C1)=2 F ----(1)
Capacitor,(C2)=3 F ----(2)
The potential…