Power factor correction

Hi
I'm doing my lv3 electricial course. I don't get the following question. But I have a feeling it's a time constant. If so I just earned myself my browney points and perhaps am not as thick as I think I am.

A pf correction capacitor is required to draw a current of 10 A from a 240 volt 50hz supply. Calculate the capacitance of the capacitor?

The impedance of a capacitor is -j/ωC
where ω is the angular frequency, or 2π times the frequency in Hz

I'll ignore the phase of the impedance, because that isn't asked for, so we can call it 1/ωC

so 240 = 10 * 1/ωC

C = 10 / (240 * 2 * π * 50) = 133 µF

thanks alot. I found something on google. I did it a different way.

So Q is the same as var because thats where the capacitor gets it's power from right? I got the same answer as you 0.132629119 x10(to power -3). Browney points for both of us I think

In this case, Q is used to represent the power as a vector quantity, and the capacitor Q is measured in Volt-Amps (Reactive), abbreviated to VAR