Question

Two capacitors, one of 5 μF and the other of 8 μF, connected in parallel, are charged with a 12 volt battery. The battery is then disconnected, and while the capacitors remain connected in parallel, a dielectric with constant 2 is inserted between the plates of the 8 μF capacitor. What is the voltage across this capacitor after the dielectric is inserted?

Answer 1

Given

   capacitors c1 = 5 μF and c2 = 8 μF,

battery potential difference v = 12 V
battery disconnected and

dielectric constant k = 2, inserted between the plates of capacitor c2

we know that if the dielectric material of constant k placed between the plates of capacitor , the capacitance increases by k times

and potential will decreases by k times

   and Q = C*V

   V = Q/C

as the battery disconnected then the charge on each plate is constant (does not change)

   Q1 = c1*v = 5 μF*12 = 60 μF
  
   Q2 = c2*v = 8 μF*12 = 96 μF

after the battery disconnected

   v1 = Q1/c1 = 60/5 = 12 V

   v2 = Q2 /(kc2) = 96/(2*8) = 6 V

it would be redistributed (because they are in parallel) then the potential across each capacitor is 9 V