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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