Two identical capacitors Ca and Cb of capacitance 1.50 F are connected in series

Question

Two identical capacitors Ca and Cb of capacitance 1.50 F are connected in series across a total potential of 220 V. A dielectric slab of dielectric constant 6.55 can fill Ca and is slowly inserted into that capacitor. (Note, account for any energy change by a corresponding energy change in some other part of the system.)

What is the change in charge, ?Qa, on Ca when the dielectric is added to Ca?

What is the change in potential drop, ?Va, across Ca when the dielectric is added to Ca?

What is the change in total electric energy, ?U, of the two capacitors when the dielectric is added to Ca?

Explanation / Answer

Given   Ca = Cb   = 1.5 F potential across the circuit     V = 220 V the capacitors in seriec combination              C = Ca Cb / Ca + Cb                    = 0.75 F Charge   Q   = CV                       = 0.75 *220 V                       = 165 C Total energy through the circuit                  U = Q2 / 2C                      = ( 165*165 ) / 2*0.75 = 18150 J Voltage across   Ca  Va = 165 / 1.5 = 110 V                when a dielectric slab of dielectric constant k = 6.55 can fill in the Ca       Capacitance of Ca is   Ca ' = k Ca                                                   = 6.55 * 1.5                                                   = 9.825 F the capacitors are in series combination    the equivalent capacitance is            1/ C = 1/ Ca' + 1/ Cb                     = 1/ 9.825 + 1/ 1.5                   C'   = 1.3 F when the capacitors are in series combination charge on each capacitor is same               Q' = C' V                     = 1.3 *220 V                      = 286.29 C    total energy U ' = Q'2 / 2C'                               = ( 286.29 ) 2 / 2*1.3                               = 31523 J Charge on Ca is   286.29 C    Voltage across Ca' Va'   =   Q' / Ca '                                            = 286.29 / 9.825                                             = 29.13 V change in potential drop across Ca                  Va = Va - Va '                            = 110 - 29.13                             = 80.87 V change in total electric energy                  U = U - U '                         = 31523-18150                           = 13373 J

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