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QUESTION: thank you,
so the way i see it (including studying the schematics on the link you gave me of the voltage multiplier) is this:
yes the multiplier build up the voltage at the end terminal without the ground BUT;
to simplify it, the voltage multiplier ends up with the final capacitor charged to full potential, but as any capacitor, it needs both the terminals connected to draw a current, that is, if i have a charged capacitor (normal), and i connect a wire to only one terminal and earth it, it will not discharge the capacitor -as the oppositely charged terminals of the capacitor are attracted to each other and grounding them individually does nothing.
but with a van da graff the terminal is just one, and grounding it will discharge the charge to ground.
thank you,

ANSWER: Not quite, but close.  With the Van de Graaf generator, the other terminal IS the ground.  It's charged relative to the ground, just like the terminal at the end of the voltage multiplier is (see the ground in the diagrams).  If you ground either one of those, you zero out the voltage on the high voltage terminal.  A voltage multiplier may have steps which are made of two-terminal capacitors, but in the end it is more similar to the Van de Graaf generator...a high-voltage terminal which is at a high potential relative to ground.

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QUESTION: thank you very much for clearing that up, so on another different aspect of this same issue.
so now i need to know now, in the case of a Van Da Graaff (VDG) that has a sphere capable of sustaining say 600kV. now i conductively attach/contact a needle to the sphere. this will leak most of the charge into the air. but will the terminal sphere still be at 600kV even though it is leaking? (normally) it will not; correct?
but with a 100kV multiplier (because of the much higher amperage relative to a normal VDG) the 100kV will be sustained despite the leak from a needle; correct?
so then (if you could tell me) i am trying to figure out this:
for ~400USD i can either buy a 600kV VDG or a 100kV multiplier. the VDG is 600kV, ~60 ľA. the multiplier is 100kv, 200ľa. which one would most likely be better at ionising the air with a needle (same needle for each) attached to the negative terminal? will the 600kV VDG with the needle be reduced down to around 100kV, and therefore make the multiplier a better choice?

The voltage on the terminals is the same in both case, it depends on the size of the terminal and the charge on it.  Leaking?  Depends on the rate of replenishment, since the leakage will depend on the voltage. This is how we control voltage on a tandem accelerator, no matter the charging mechanism (both are used).  It sounds, in this case, like what you want depends on the needle you want to put on it.  If your goal is to maximize the air ionization, then go with the higher current.  If your goal is higher voltage, then go with that.  Simple.


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Dr. Stephen O. Nelson


I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.


I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.

Ph. D. from Duke University in physics, research in nuclear astrophysics reactions, gamma-ray astronomy technology, and advanced nuclear reactors.

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