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Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 5 µF capacitor whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

The source impedance of your generator is two of these in series, and this puts a limit on how much power you can get out of it. If you match the source impedance with a load of 1060 Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 400 mA through the combination, for a total power delivered of about 170 W.

Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 1200 V, 5 µF capacitor¹ whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

The source impedance of your generator is two of these in series, and this puts a limit on how much power you can get out of it. If you match the source impedance with a load of 1060 Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 400 mA through the combination, for a total power delivered of about 170 W.

¹ Assuming air is the dielectric, you'll need a spacing on the order of 1 mm in order to withstand the voltage, which means that the area of each plate needs to be about 566 m², roughly the same order of magnitude as a baseball diamond.

Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 5 µF capacitor whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

This is the source impedance of your generator, and it puts a limit on how much power you can get out of it. If you match the source impedance with a load of 530Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 800 mA through the combination, for a total power delivered of about 680 W.

Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 5 µF capacitor whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

The source impedance of your generator is two of these in series, and this puts a limit on how much power you can get out of it. If you match the source impedance with a load of 1060 Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 400 mA through the combination, for a total power delivered of about 170 W.

Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 5 µF capacitor whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

This is the source impedance of your generator, and it puts a limit on how much power you can get out of it. If you match the source impedance with a load of 530Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 800 mA through the combination, for a total power delivered of about 680 W.

Your unit analysis is off.

5 µF × 1200 V = 6 mC (millicoulombs)

6 mC × 120 Hz = 720 mA

So yes, if you can build a 5 µF capacitor whose plates can be swapped 120 times per second, you will create a generator. Some of the mechanical work you put into moving the plates will be converted into electrical power. There will also be a significant amount of power drawn from the source that's keeping the fixed plates charged to 1200 V.

Note that the impedance of a 5 µF capacitor at 60 Hz is rather high:

$$\frac{1}{2\pi f C}=530 \Omega$$

This is the source impedance of your generator, and it puts a limit on how much power you can get out of it. If you match the source impedance with a load of 530Ω and your drive voltage is 1200 V peak (850 Vrms), you'll drive about 800 mA through the combination, for a total power delivered of about 680 W.