Two identical capacitors Ca and Cb of capacitance 5.10 F are connected in series

Question

Two identical capacitors Ca and Cb of capacitance 5.10 F are connected in series across a total potential of 690 V. A dielectric slab of dielectric constant 4.35 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.)

 

hint:

Initially we have two capacitors C1 and C2 and with total voltage V.

Finally we have two capacitors kC1 and C2 and with total voltage V.

You can treat them as two separate problems and figure out all the charges and voltages.

 

 

 

 

I have answered these:

 

 

 

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

 

 

 

Answer: 1.10×10-3 C

 

 

 

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

 

 

 

Answer: 1.10×10-3 C

 

 

 

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

 

 

 

3.80×10-1 J

 

 

 

I need Answers for these two parts

 

 

 

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

 

 

 

b. What is the change in potential drop, Vb, across Cb when the dielectric is added to Ca? 

 

 

 

 

 

 

 

 

 

 

 

Please supply the correct answer and show work. If correct, I will rate lifesaver right away.

Initially we have two capacitors C1 and C2 and with total voltage V.

Finally we have two capacitors kC1 and C2 and with total voltage V.

You can treat them as two separate problems and figure out all the charges and voltages.

Explanation / Answer

Given C_a = C_b = 5.1F total potential   V = 690 V The capacitors   C_a   and C_b are in series combination      equivalent Capacitance                   C _eq   = C_a C_b / C_a + C_b                                 = ( 5.1 ) ( 5.1 ) / ( 5.1 + 5.1 )                               = 2.55F      the capacitors are in series combination so charge on each capaciotr is same                Q =    C_eq V                     = ( 2.55F ) ( 690 V )                     = 1.759 mC Voltage across C_a    V_a =   Q / C_a                                             = 1.759*10-3 C / ( 5.1*10-6 F )                                             = 345 V   Voltage across C_b = 345 V   _______________________________________________________________ when the dielectric is added to C_a                Cpacitance of C_a'   = kC_a                                               = 4.35*5.10 = 22.185F     equivalent capacitance                     C'_eq=   C_a' C_b / ( C'_a +C_b )                            =   4.1520F           Charge on each capaciotr is                  Q' = C'_eq V                       = (4.1520*10-6 ) ( 690)                       = 2.864 mC      Voltage across C'_a                V'_a =   Q' / C'_a                      =   2.864 mC / 22.185 F                        = 129 V          Voltage across C'_b              V'_b = Q' / C_b                       = 2.864mC / 5.1F                        = 561.5 V      change in potential drop, Va, across Ca when the dielectric is added to Ca                      V_a = V_a - V'_a                                   = 345 - 129 = 216 V    change in potential drop, Vb, across Cb when the dielectric is added to Ca                                                          V_b = 561.5 - 345 =    216.5 V          

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