Two capacitors C 1 = 5.5 F, C 2 = 13.4 F are charged individually to V 1 = 19.6

Question

Two capacitors C1 = 5.5 F, C2 = 13.4 F are charged individually to V1 = 19.6 V, V2 = 4.0 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.
a) Calculate the final potential difference across the plates of the capacitors once they are connected.

b) Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together

c) By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

a)

Charge on C1, Q1 = 5.5*10^-6*19.6 = 1.08*10^-4 C = 10.8*10^-5 C

Charge on C2 , Q2 = 13.4*10^-6*4 = 5.36*10^-5 C

Now, when they are connected , let the final voltage across each after connection be V

Now, let the charge on C1 be Q1' and on C2 be Q2'

As voltage across them are same : Q1'/C1 = Q2'/C2

Also, the total charge Q1' + Q2' = Q1 + Q2 = (10.8 + 5.36)*10^-5 C

So, (Q2'/C2)*C1 + Q2' = 1.616*10^-4

So, Q2'*(5.5/13.4) + Q2' = 1.616*10^-4

So, Q2' = 1.15*10^-4 C

and Q1' = 1.616*10^-4 - 1.15*10^-4 = 4.66*10^-5 C

So, final potential difference between them = Q1'/C1 = 4.66*10^-5/(5.5*10^-6) = 8.5 V <----answer

b)

Charge flowing = Q1 - Q1' = ( 10.8 - 4.66 )*10^-5 = 6.14*10^-5 C

c)

Total energy reduced = ( 0.5*C1*V1^2 + 0.5*C2*V2^2 ) - 0.5*(C1 + C2)*V^2

= ( 0.5*5.5*10^-6*19.6^2 + 0.5*13.4*10^-6*4^2 ) - 0.5*(5.5+13.4)*10^-6*8.5^2

= 4.81*10^-4 J

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