Does charge get distributed according to ratio of capacitance in a parallel system?

Question

I was doing a question in which I was able to find the charges on each capacitor in parallel by assuming that the total charge would get distributed amongst each capacitor according to their capacitance. I want to know if this assumption is true, or if I just happened to get the right answer using the wrong method. The 'wrong method' that I am referring to is as follows:

The total charge in the system would be Q t= C equivalent x Voltage of battery.

Charge on the first capacitor in parallel would be Q1= Q t x C1/(C1+C2)

Similarly for capacitor 2, charge would be = Qt x C2/(C1+C2)

In this way the total charge Q1 +Q2 = Qt but I'm not sure if this logic is applicable to capacitor systems.

Answer 1

You happened to get the correct answer because that is what the definition of equivalent capacitance happens to be. Your thought process would be useless in the face of more complicated scenarios. It is easier to just learn it properly.

For the case of capacitors in parallel, each capacitor have one side connected together, and the other side also connected together, just different from the first junction. This way, it is necessary, for the uniqueness of voltages, to get the same voltage difference. Since each capacitor gets the same voltage, then you can find the individual charges by $Q_i=C_i V$, sum the charges, and thus derive what is the equivalent capacitance is.

If the capacitors are connected in series, then shifting any charge off one capacitor, will be forced onto the next capacitor. Then it is $Q = C_i V_i$, and then you can figure out that total voltage difference. That way, you can figure out the equivalent capacitance.

Here, you guessed the correct physical result (same voltage difference) and also guessed to use the correct equivalent capacitance, and that necessarily gives the correct result, even if just for the wrong reasoning.