two capacitors , one charged and the other uncharged are connected in parallel.Prove that when equilibrium is reached, each capacitor carries a fraction of the initial charge equal to the ratio of its capacitance of the two sum of capacitances . Show that the final energy is equal to initial energy, and derive a formula in terms of charge and capacitances of the two capacitor

A) The final energy has been shown to be equal to initial energy as;

(Q₁i)²/2C₁ = (Q₁f*C₁)²/2C₁ + (Q₂f* C₂)²/2 C₂

B) The formula in terms of charge and capacitances of the two capacitor is;

E_f = E_i(C₁/(C₁ + C₂))

A) We are told that the two capacitors are connected in parallel. One is charged and the other uncharged initially. Thus;

Let the charged one be Q₁i

While Q₂i = 0

From conservation of charges, we have;

Q₁i + Q₂i = Q₁f + Q₂f

Since  Q₂i = 0;

Q₁i = Q₁f + Q₂f

Since the two capacitors are connected in parallel, it means they will have the same voltage across them. Thus;

Vf = Q₁f/C₁ = Q₂f/C₂

Fraction of charge left on the first capacitor is; Q₁f/Q₁i

Putting Q₁f + Q₂f for Q₁i  gives;

Q₁f/(Q₁f + Q₂f)

Since Q₁f/C₁ = Q₂f/C₂, then;

Q₂f = (C₂/C₁)(Q₁f)

Thus;

Fraction of charge left on the first capacitor = Q₁f/(Q₁f + (C₂/C₁)(Q₁f))

This now gives;

(Q₁f*C₁)/((Q₁f*C₁) + (C₂*Q₁f))

This reduces to; C₁/(C₁ + C₂)

Thus, the second capacitor will have the remaining charge;

1 - (C₁/(C₁ + C₂))

⇒ C₂/(C₁ + C₂)

This means that the charge was conserved. Thus, applying conservation of energy with the formula for energy on a charged capacitor as  Q²/2C, we have;

Initial energy(E_i) = Final energy(E_f)

Thus;

(Q₁i)²/2C₁ = (Q₁f*C₁)²/2C₁ + (Q₂f* C₂)²/2 C₂

B) The final energy will be;

E_f = (Q₁f*C₁)²/2C₁ + (Q₂f* C₂)²/2C₂

⇒  [(Qi*C₁) /(C₁ + C₂)]²(1/2C₁) +  [(Qi*C₂) /(C₁ + C₂)]²(1/2C₂)

⇒ ((Q₁i)²/2C₁)(C₁/(C₁ + C₂))

From answer A above, we saw that; (Q₁i)²/2C₁ = E_i. Thus;

E_f = E_i(C₁/(C₁ + C₂))

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