Purely Capacitive AC - voltage and current relationship

Question

Studying electronics, capacitors in AC circuits currently. Can not understand two formula's.

I see it stated as fact: "In purely capacitive AC circuit Current leads Voltage by 90 degrees." I understand this relationship when viewing Waveform 90 degree phase difference Current leads Voltage.

I can't however find any example with circuit values to prove this is so.

I have two formula that prove this is so.

i = dq / dt

i = C * (dv / dt)

Can anybody show me how this works with actual circuit values? If I say (random example) AC 10V 200Hz with 0.0005F capacitor

Then i = dq / dt = ?? / ??

or

i = C * (dv / dt) = ? * ?? / ??

Example with number please for above circuit?

I am so confused I don't know if I just take "Current leads Voltage 90 degrees" at face value, but can it be proven with actual numbers? Maybe there is no example because there is no circuit without resistance or truly purely capacitive? Maybe it is just a theory not real world formula?

Answer 1

Never mind numbers, just go with hand-waving.

Consider an initially uncharged capacitor. Now slam in a large current, that causes its voltage to rise. Now let the current decrease, and gradually become zero, this means the rate of rise of voltage will slow down and eventually the voltage will stop rising.

What I have just described there is the first quadrant of a sine wave voltage starting at zero, and a cosine wave current starting at max and falling to zero.

What the current does, the voltage does a quarter cycle later. The current leads the voltage.

Answer 2

Let me try.
$$ I=\frac{U}{j\omega\;C}$$ $$ |I|=\frac{10}{2\pi*200*0.0005}$$