Current through capacitor

We already know that after charging a capacitor, the charge q and voltage $v_C$ across capacitor's plates is formulated with the following equation: $$ \bbox[10px,border:1px solid grey]{q = Cv} \tag{1}$$ By taking derivative w.r.t t of the above equation, we will get $$ {dq \over dt} = C{dv \over dt} \tag{2}$$ where $$ {dq \over dt} = i = i_C$$ The current $i_C$ associated with a capacitance C is related to the voltage across the capacitor. Therefore eq. (2) becomes, $$ \bbox[10px,border:1px solid grey]{i_C = C{dv_C \over dt}} \tag{3}$$ where $C{dv \over dt}$ is a measure of the change in $v_C$ in a vanishingly small period of time. The function ${dv \over dt}$ is called the derivative of the voltage $v_C$ with respect to time t. If the voltage fails to change at a particular instant, then $$dv_C = 0$$ and $$ i_C = C{dv \over dt} = 0 A$$ In other words, if the voltage across a capacitor fails to change with time, the current $i_C$ associated with the capacitor is zero. To take this a step further, the equation also states that the more rapid the change in voltage across the capacitor, the greater the resulting current. In an effort to develop a clearer understanding of Eq. (3), let us calculate the average current associated with a capacitor for various voltages impressed across the capacitor. The average current is defined by the equation $$ \bbox[10px,border:1px solid grey]{i_{Cav} = C{\Delta v_C \over \Delta t}} \tag{4}$$ where $\Delta$ indicates a finite (measurable) change in charge, voltage, or time. The instantaneous current can be derived from Eq. (4) by letting $\Delta t$ become vanishingly small; that is, $$ i_{Cins} = \lim_{t \to 0} C{\Delta v_C \over \Delta t} = C {dv \over dt}$$