Question

1.A 15 pF capacitor is charged to 1 kV and
then removed from the battery and connected
in parallel to an uncharged 65 pF capacitor.
What is the new charge on the second ca-
pacitor?
Answer in units of nC

2.What is the new charge on

3.Find the change in energy.
Answer in units of ?J

Answer 1

a similar question is solved below, but with different values. hope this helps you.

A 35 pF capacitor is chargedto 2 kV and then removed from the battery and connected in parallelto an uncharged 50 pF capacitor.

1. What is the new charge onthe second capacitor? Answer in units of nC.

2. What is the new charge on the first capacitor? Answer in unitsof nC.

3. Find the change in energy. Answerin units of ? J.

Charge on the 35 pF capacitor when it is connected 2kV batteryis

Q = CV = 35 x 10^-12 x 2 x 10³ C =70nC

now the battery is disconnected from the 35 pF capacitor andit is connected to the 50 pF capacitor.

Therefore, they redistribute the charges and attains commonpotential

common potential V =(C₁V₁+C₂V₂)/(C₁+C₂)

C₁ = 35pF ; C₂ = 50 pF ;V₁ = 2000 Volt ; V₂=0 Volt

plug the values and get the value of V

1.

now the charge on the 50pF capacitor is Q₂ =C₂V = 50 x 10^-12 x V

plug the value of common potential V and get the value ofcharge Q₂

2.

new charge on 35pF capacitor is

Q₁ = Q - Q₂

get the value of Q₂ from the (1) andplug it in the above and get the value of Q₁

3.

E_initial=1/2(C₁V_initial²) ( V_initial= 2000volt, C₁ = 35pF )

E_final = 1/2{C₁C₂ /(C₁+C₂) }V²)

C₁ = 35pF ; C₂ = 50 pF getthe value of common potential V from the above and plug it theabove equation and find the values of E_initial andE_final

and to find the change in energy ?E =E_initial- E_final