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

my prof said this wasn't a derivation? How isn't it? What is a PROPER derivation?

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?

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

my prof said this wasn't a derivation? How isn't it? What is a PROPER derivation?

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

my prof said this wasn't a derivation? How isn't it? What is a PROPER derivation?

So I had to derive the formula for capacitors 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 $V=IR \rightarrow \sum \frac{Q}{C_n} = V=\frac{Q}{C_{eq}}$ Since they are in series, we can apply kirchoffs voltage law, we get $ \frac{Q}{C_1}+\frac{Q}{C_2}+...\frac{Q}{C_n}= \frac{Q}{C_{eq}} \rightarrow Q \sum_{n=0}^{m} C_n^{-1} = \frac{Q}{C_{eq}}$ Thus we can conclude that $ \sum_{n=0}^{m} C_n^{-1} = \frac{1}{C_{eq}}$

my prof said this wasn't a derivation? How isn't it? What is a PROPER derivation?

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

my prof said this wasn't a derivation? How isn't it? What is a PROPER derivation?