Parallel Capacitors: Homework Solution for 9 μF Charge

[ZxcvbnM2000]

Homework Statement

A capacitor of capacitance 9 μF is fully charged from a 200 V dc supply. The
capacitor is now disconnected from the supply and connected in parallel with an
uncharged 3 μF capacitor. Once the 3 μF capacitor is fully charged, it is removed
and discharged. a) What charge remains on the 9 μF capacitor? b) How many
times would this process have to be repeated in order to reduce the charge on the 9
μF capacitor to below 50% of its initial charge? c) What would be the pd between
the plates of the 9 μF capacitor now be?

Homework Equations

Q=CV

The Attempt at a Solution

a) I said that the 3μF capacitor will be fully charged when the voltage across both capacitors is the same,let's say V_x.

Since energy has to be conserved then E_i=E_f therefore

C1V^2 = C1Vx^2 + C2Vx^2 <=> Q1' = 1.5588*10^-3 C . Am i doing this correct ?

 

[gneill]

Unfortunately, much as in elastic collisions of masses, connection of capacitors in parallel is not an operation that conserves energy; some energy is lost when the charges get shuffled into their new configuration.

What is conserved through the connection operation, though, is charge.

 

[ZxcvbnM2000]

Thank you very much !