Question

A 9.4 µF capacitor and a 23.4 µF capacitor are connected in series across the terminals of a 6.00-V battery

(a) What is the equivalence capacitance of this combination?
  µF

(b) Find the charge on each capacitor.
  µC across the 9.4 µF capacitor.
  µC across the 23.4 µF capacitor.

(c) Find the potential difference across each capacitor.
  V across the 9.4 µF capacitor.
  V across the 23.4 µF capacitor.

(d) Find the energy stored in each capacitor.
µJ stored in 9.4 µF capacitor
µJ stored in 23.4 µF capacitor

Answer 1

(a) Equivalent capacitance,

C=C1C2/(C1+C2)

C=9.4*23.4/(9.4+23.4)

C=6.706 micro farad

(b) total charge, Q=C*V

Q=6.706*10^-6*6

Q=40.24 micro coulamb

The charge across both capacitors is same. They are in series.

(c) the potential difference across C1,

V1=Q1/C1=40.24*10^-6/(9.4*10^-6)

V1=4.28 V

And across C2,

V2=40.24*10^-6/(23.4*10^-6)

V2=1.72 V

(d) the energy stored in C1,

E1=0.5*C1*(V1)²

E1=0.5*9.4*10^-6*(4.28)²

E1=8.61*10^-5 J

And in C2,

E2=0.5*23.4*(1.72)²

E2=3.46*10^-5 J