# Capacitor, stored energy versus reactive power

When an AC voltage U is put onto a loss less capacitor C the capacitor will require a certain reactive power Q (current 90 degress out of phase with the voltage):

$$Q=UI = \omega C U^2$$ in VAr (Only reactive power, no real power)

I can also determine the reactive power strating from the stored energy in a capacitor:

$$E= (1/2) C U^2$$

With $$Q=dE/dt$$ using Fourier transformer to go to the frequency domain: $$Q=\omega E$$

Filling in the equations results into: $$Q=(1/2) \omega C U^2$$

You can now see my problem. Deriving the reactive power of a capacitor from voltage / current / capacitance results in factor 1/2 difference compared to deriving reactive power derived starting from energy stored in a capacitor.

Can you help me with what I'm missing or doing wrong.

When you use complex notation, you have to use a 1/2. The energy stored in a capacitor is given by: $$E = \frac{1}{2} Re(U I^{*})$$ with * the complex conjugate and $$Re$$ denote the real part.