- 28 Dec '09 15:25I'm having a problem regarding an ideal capacitor.

Consider an ideal capacitor made up of two parallel plates. Between these plates we see a uniform electrical field.

Now, when we apply a sinusoidal voltage over the plates, the electrical field should vary in time.

Does this send out an electromagnet wave?

And if so, given the fact that an EM wave has an energy associated with it, shouldn't there be energy lost to this wave, thus making the ideal capacitor have losses? - 29 Dec '09 21:01

not if you uncouple the heisenberg compensators*Originally posted by Meneer Dries***I'm having a problem regarding an ideal capacitor.**

Consider an ideal capacitor made up of two parallel plates. Between these plates we see a uniform electrical field.

Now, when we apply a sinusoidal voltage over the plates, the electrical field should vary in time.

Does this send out an electromagnet wave?

And if so, given the fact that an EM w ...[text shortened]... it, shouldn't there be energy lost to this wave, thus making the ideal capacitor have losses? - 30 Dec '09 18:28

Some energy will be radiated, depends on how long the wires are that are connected to the plates and the frequency of the AC going to the plates. It will act like a resistor with a phase shift, the formula for the instantaneous current is I=C *(de/dt), e voltage, t time, C capacitance in Farads. The impedance of a cap is Xc=1/2Pi*F*C, F is frequency in hertz (cycles per second) and C capacitance also in Farads. Here is a link:*Originally posted by Meneer Dries***I'm having a problem regarding an ideal capacitor.**

Consider an ideal capacitor made up of two parallel plates. Between these plates we see a uniform electrical field.

Now, when we apply a sinusoidal voltage over the plates, the electrical field should vary in time.

Does this send out an electromagnet wave?

And if so, given the fact that an EM w ...[text shortened]... it, shouldn't there be energy lost to this wave, thus making the ideal capacitor have losses?

http://www.allaboutcircuits.com/vol_2/chpt_4/2.html

They lay it all out, an example given for a 100 microfarad cap, the impedance is 26 ohms at 60 Hertz, 23 Ohms at 120 hertz, and 0.6 ohms at 2500 hertz. Higher frequency =lower impedance. So if you have a 100 microfarad cap in series with a with a resistor, the resultant will be governed by a vector equation, here is a link for that: http://www.allaboutcircuits.com/vol_2/chpt_4/3.html - 31 Dec '09 10:33

Thanks for the info, but you misinterpreted my question.*Originally posted by sonhouse***Some energy will be radiated, depends on how long the wires are that are connected to the plates and the frequency of the AC going to the plates. It will act like a resistor with a phase shift, the formula for the instantaneous current is I=C *(de/dt), e voltage, t time, C capacitance in Farads. The impedance of a cap is Xc=1/2Pi*F*C, F is frequency in hert ...[text shortened]... a vector equation, here is a link for that: http://www.allaboutcircuits.com/vol_2/chpt_4/3.html**

Waht you are talking about is energy lost through Ohmic heating in the resistor and the energy going to the capacitor is reactive energy, which is not lost but regained when the capacitor is decharged. And it is indeed this reactive energy you see in the phase shift.

That's the stuff I know too

What I wondered about was the fact if a charging capacitor emits electromagnetic waves and thus loses energy (not reactive energy) to them?