# Capacitor question

#### nov1ceElectrician

Joined Sep 7, 2011
1
Hi, Im a little bit stumped, i new to experimenting and was trying to build a recreation of Tesla's Utilization of Radiant Energy patent.

I am new to the use of capacitors so Im not sure if my readings are accurate.

With one lead of the capacitor connected to a suspended plate and the other to ground, I went to take a voltage reading. When i connected my DMM to the capacitor the voltage almost instantly went to zero. However with just one lead on a terminal it maintained a relatively constant voltage slowly incrementing similar to the patent description.

So is it when I connect both leads to the capacitor, is that the capacitor discharging? In this case the capacitor being a voltage source discharging to a resistor in series?

How are the effects explained for the one terminal connected to the capacitor and the other in free air?

Hope somebody can help!

#### #12

Joined Nov 30, 2010
18,210
The first thing to keep in mind is that a voltage is always a voltage compared to some other place. I can not fathom the voltage of a capacitor referenced to air.

I can say that when you measure the voltage of the capacitor plate compared to ground, some current is going through your meter, and that is discharging the plate. A good, modern volt meter usually has 10 million ohms of resistance to ground. Figure out your source impedance and see if 10 megohms is allowing enough current to defeat your voltage generator.