Why isn't this a valid derivation of the formula for capacitors in series?

Question

I had to derive the formula for capacitors (I decided to use m capacitors in my derivation) in series, and this is what I did.

The formula for a capacitor is $$Q=CV,$$ which is the same as saying $$\int \dfrac{dI}{dt} dt= C_n V_n.$$ Since they are all in series, $I_1=I_2=...=I_n=I$, thus $Q_1=Q_2....=Q$.

We also know that $$V=IR \Longrightarrow \sum \frac{Q}{C_n} = V=\frac{Q}{C_{eq}}.$$ Since they are in series, we can apply kirchoffs voltage law, obtaining $$ \frac{Q}{C_1}+\frac{Q}{C_2}+...\frac{Q}{C_n}= \frac{Q}{C_{eq}} \Longrightarrow Q \sum_{n=0}^{m} C_n^{-1} = \frac{Q}{C_{eq}}.$$ Thus we conclude that $$ \sum_{n=0}^{m} C_n^{-1} = \frac{1}{C_{eq}}.$$

Is this a proper derivation of the result?

Answer 1

You bring up an $R$, for no known reason, and immediately after that write down something equivalent to what you want to show that also comes out of nowhere, so it's very very confusing as written.

In series, the voltage of the unit is the sum of the voltages of the parts.

In series, the charge of each one of the plates has to be the same because in the region between two plates, the charge from one side/plate comes from the other side (the one before/after in series). Thus, the $Q$ of the unit is the $Q$ of the parts.

These two facts give you what you want, write the voltage as the sum of voltages, express those voltages in terms of $Q$ and $C_i$, then divide by the $Q$ of the whole unit (which is the regular $Q$ of the parts), to get an expression for the total capacitance.