Derivation: how is angular frequency ralated to electrical resistance and capacitance

simplify notation

$ Q=I \cdot t \qquad [Q]=C,\quad [I]=A,\quad [t]=s $

$ I=\frac{E}{R} \qquad [E]=V,\quad [I]=A,\quad [R]=\Omega $

$ C=\frac{Q}{E} \qquad [E]=V,\quad [Q]=C,\quad [C]=F $

$ I=\frac{Q}{t} \qquad E=\frac{Q}{C} $

$ I=\frac{E}{R} \mid E=\frac{Q}{C} \quad \rightarrow \quad I=\frac{Q}{R \cdot C} $

$ I=\frac{Q}{R \cdot C} \mid I=\frac{Q}{t} \quad \rightarrow \quad \frac{Q}{t}=\frac{Q}{R \cdot C} $

$ \frac{Q}{t}=\frac{Q}{R \cdot C} \mid :Q \quad \rightarrow \quad \frac{1}{t}=\frac{1}{R \cdot C} $

$ \frac{1}{t}=f \qquad [t]=s,\quad [f]=\frac{1}{s} $

$ \omega = 2 \Pi f \qquad [\omega]=\frac{1}{s},\quad [\Pi]=\varnothing ,\quad [f]=\frac{1}{s} $

$ \frac{1}{t}=\frac{1}{R \cdot C} = \omega $