# Finding final charge distribution on capacitors on grounding

I am trying to solve the following problem on capacitors:

Here's the circuit

I have to find the total energy loss when all three switches are closed, and subsequent equilibrium attained.

I had very little idea of how to proceed, but this is what I did.

Even when the switches are closed, the sum of charges on right plate of $C_1$ and the left plate of $C_2$ $(-30\ mC)$ will remain conserved. Let the final charge appearing on on the right plate of $C_1$ be $q\ mC$, then the charge on the left plate of $C_2$ will be $(-30-q)\ mC$.

Now, will the charge on the left plate of $C_1$ and the right plate of $C_2$ be $-q\ mC$ and $(30+q)\ mC$ ?

If this is so, then I think I can obtain the value of $q$ using the Voltage Law (and therefore the energy lost).

If I am incorrect (which I suspect I am), I hope someone can point me in the right direction.

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With all the switches closed you just have two capacitors in parallel with the same voltage drop across them. The ratio of the charges on the two capacitors will be the same as the ratio of their capacitances i.e. $Q_1/C_1 = Q_2/C_2$, and as you said $Q_1 + Q_2$ = -30mC. – John Rennie Mar 14 '13 at 19:18
OP's question is not about how to do this question, which John Rennie has done, but instead about why the charges on the two plates of a single capacitor must be equal in magnitude and opposite in sign. Indeed, why? It is a non-trivial question and the answer is straightforward but somewhat involved. See kwantlen.ca/science/physics/faculty/mcoombes/P1102_Solutions/… . The upshot of this is that at the end of the day, we find that a capacitor has equal and opposite charges on its two plates, so it is unambiguous to say a capacitor has charge $Q$. – nervxxx Mar 14 '13 at 19:35
@JohnRennie So, essentially, was I correct? Applying the voltage law will give me $q_1= -20\ mC$ and $q_2= -10\ mC$, and an energy loss of $150\ mJ$. – darkv0id Mar 14 '13 at 19:36
@darkv0id: the site policy is not to answer homework questions, so the best I can do is discuss general methods. Your working looks fine to me. – John Rennie Mar 14 '13 at 19:40
@JohnRennie Thanks. I am aware that StackExchange is not a homework help site. What prompted me to ask the question was the fact that the answer to this problem in my textbook is $625/3\ mJ$, and not $150\ mJ$, so I assumed there was a flaw in either my concepts or my methods. So, thanks for helping me out! – darkv0id Mar 14 '13 at 19:47