Question

A 1.2mF capacitor and a 1.0mF capacitor are connected in series to a 12V battery. a) Calculate the potential difference across each capacitor and the charge on each. b) Find the electric energy stored in each capacitor. c) Place a paraffin dielectric (K=2.2) into the 1.2mF capacitor and redo a) and b).

Answer 1

C1 = 1.2*10^-3 F and C2 =1*10^-3 F

these capacitor are in series so charge will be same on both  

because Q = C*V

V1/C

V1 = ( 1) / ( 1.2+1) *12 =5.45454545 V
V2 = 1.2 /( 1.2+1) *12 =6.54545455 V

Q1 = C1*V1 = 1.2*10^-3 *5.45454545 =0.00654545454

Q2 = 00654545454  

energy U1 = 1/2*c1*v1^2 =0.5*1.2*10^-3 *5.45454545^2 =0.0178512396 J

U2 = 0.5*1*10^-3*6.54545455 ^2 =0.0214214876

after dielectric

C1 = 2.2*1.2*10^-3 =2.64*10^-3

C2 = 1 *10^-3

now same procedure

V1 = 1/(1+2.64)*12 =3.2967033 V , Q1 = 2.64*10^-3*3.2967033 =0.00870329671

U1 = 1/2*2.64*10^-3*3.2967033 ^2 =0.0143460935

V2 = 12 -3.2967033 =8.7032967

Q2 = 1*8.7032967 = 8.7*10^-3

U2 = 0.5*1*10^-3*8.7032967^2 = 0.0378736867

let me know in a comment if there is any problem or doubts