In the figure the capacitances are C1 14 muF and C2 29 muF

In the figure the capacitances are C_1 = 1.4 muF and C_2 = 2.9 muF, and both capacitors are charged to a potential difference of V = 140 V but with opposite polarity as shown. Switches S_1 and S_2 are now closed, (a) What is now the potential difference between points a and b? What now is the charge on capacitor (b) 1 and (c) 2?

Solution

C1 and C2 are in series.So, resultant capacitance C = (C1xC2)/(C1+C2)

                                                                                  =     0.9441 uF

Charge on C1 is Q1 = C1V1

= 1.4uF x 140 volt

   = 196uC

Charge on C2 is Q2 = C2V2

= 2.9uF x 140 volt

   = 406uC

Required potential difference V \' = (Q1+Q2)/C

                                                     =(196uC+406uC)/0.9441uF

                                                      = 637.6 volt

New charge on C1 is q1 = C1 x V \' = 1.4 uF x 637.6 volt

                                             = 892.6 uC

New charge on C2 is q2 = C2 x V \' = 2.9 uF x 637.6 volt

                                             = 1849 uC