# Current in a fluorescent tube that is not in a circuit

In Walter Lewin's 8.02 Electricity and Magnetism course, he places a fluorescent tube pointing radially outwards from a large Van de Graaff (VDG) generator. Due to the VDG's E-field, this causes a large potential difference between both ends of the tube. He then says that this high voltage causes a current in the tube, which creates light.

Subsequently, he then touches the end of the tube furthest the VDG, receives a shock, and by grounding this end, allows more current to pass, thus creating more light.

If the tube is not in a circuit, that must mean all the charges that were moved by the potential difference are accumulating in one end of the tube, causing the tube to act like a capacitor and eventually stopping current flow. Obviously this is not what is happening because the tube creates light continuously.

So what exactly is the path of the current, and why should the tube light up more if Dr. Lewin grounds one end through his body?

EDIT: He also does this experiment with a neon flash tube, and it creates light as well

• There's always a current flowing if there's a voltage potential difference. It may be very small (high impedance), but if there were no current loop, after a possible start-up pulse, both ends of the tube would have to be at the same potential. They aren't, and it does light up. Dec 31, 2014 at 0:30
• Dec 31, 2014 at 1:43

When the VDG sphere is 'charged' to say $300kV$ or greater, the moisture in the air around it is able to conduct a very small $<100\mu A$ electric current, which is enough to dimly light the fluorescent tube.