Two parallel plate capacitors, C1 and C2, are connected in series with a 40.0 -

Question

Two parallel plate capacitors, C1 and C2, are connected in series with a 40.0 - V battery and a 290 - komega resistor, as shown in the figure. Both capacitors have plates with an area of 2.27 cm2 and a separation of 0.190 mm. Capacitor C1 has air between its plates, and capacitor C2 has the gap filled with porcelain (dielectric constant of 7 and dielectric strength of 5.70 kV/mm). The switch is closed, and a long time passes. a) What is the charge on capacitor C1? b) What is the charge on capacitor C2? c) What is the total energy stored in the two capacitors? d) What is the electric field inside capacitor C2?

Explanation / Answer

Given Area of plates of each capacitor is A = 2.27 cm2                                                         = ( 2.27 cm2 ) ( 10-4 m2 / 1 cm2 )                                                         = 2.27*10-4 m2 Seperation between the plates    d = 0.190 mm                                                     = ( 0.190 mm ) ( 10-3 m / 1 mm )                                                     = 0.190*10-3 m   Capacitor C1 has air between the plates Capacitance of capacitor                        C1 = o A / d                              = ( 8.85 *10-12 C2 / Nm2 )( 2.27*10-4 m2 ) / ( 0.190*10-3 m )                              = 1.0573*10-11 F Capacitance of the capacitor is filled with  porcelain has a dielectric constant k = 7                         C2 = o A k / d                              = ( 8.85 *10-12 C2 / Nm2 )( 2.27*10-4 m2 ) (7) / ( 0.190*10-3 m )                              =7.4011*10-11 F Equivalent capacitance is                        1 / C = 1 /C 1 + 1 / C2                            C = C1 C 2 / C 1 + C 2                                  = 9.251 *10-12 F After a long time voltage across capacitor in RC is equal to voltage across the battery                       Q / C   = V Equivalent capacitance is                        1 / C = 1 /C 1 + 1 / C2                            C = C1 C 2 / C 1 + C 2                                  = 9.251 *10-12 F After a long time voltage across capacitor in RC is equal to voltage across the battery                       Q / C   = V When capacitors are connected in series combination charge on each capacitor is same. Charge on capacitor C1 is                                 Q1     = V C                                           = ( 40.0 V ) (  9.251 *10-12 F )                                           = 0.37*10-9 C                                           = ( 0.37 *10-9 C ) ( 1 nC / 10-9 C )                                           = 0.37 nC ------------------------------------------------------------------------------------------- ------------------------------------------------------------------------------------------- Charge on capacitor C2 is                                Q2       = 0.37 nC ------------------------------------------------------------------------------------------- Total energy stored in the capacitors is                               U   = Q2 / 2 C                                     = ( 0.37 *10-9 C ) 2 / 2( 9.251 *10-12 F )                                     = 7.3*10-9 J ____________________________________________________________________ ____________________________________________________________________ Electric field in side the capacitor C2                            E    = V2 / d Voltage across C2 is                           V2 = Q / C2                                  = ( 0.37 *10-9 C ) / ( 7.4011*10-11 F )                                  =4.99 V                             E   = 4.99 V / 0.190*10-3 m                                  =   2.263*104 V/m                                                           =4.99 V                             E   = 4.99 V / 0.190*10-3 m                                  =   2.263*104 V/m                                                                    

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