# Homework Help: Van de Graaff generator questions?

1. Dec 31, 2013

### AFSteph

Hello~ I've got a confusing question;

"If a Van de Graaff generator is charged to 50,000 volts, how much energy does it take to add an additional electron to the charge on the sphere?"
(The charge of an electron is given as -1.6 x 10$^{-19}$C)
My answer: (50,000 V)( -1.6 x 10$^{-19}$C) = 8.0 x 10$^{-15}$ J
Er, I actually don't understand this at all, but if it's correct then meh.

Second question:
"How would the voltage of the generator in Part (A) compare to the voltage of a larger Van de Graaff generator with the same amount of charge? To which generator could an electron be added with the least expenditure of energy?"
According to my textbook,
So then by that reasoning, the larger generator should have more voltage for the same amount of charge. And by using the formula of volts x charge of an electron to find the energy needed to add an electron then it should take more energy to add to the bigger, higher voltage, generator.

BBUUUUUUUUTT... this is a bit contradictory to me because here https://www.physicsforums.com/showthread.php?t=229046 Doc Al has a different (opposite!) answer that makes sense but doesn't fit.

Can someone help me in clear English here? Thanks for any assistance!

2. Jan 1, 2014

### rude man

Q = CV.
If Q stays constant and C increases, what does V do? Then what does qV do?

(Note: the capacitance of a sphere = 4 pi epsilon R, R = radius of sphere. Adding the gas also increases C).

Last edited: Jan 1, 2014
3. Jan 1, 2014

### AFSteph

Hmm maybe I should mention that the course I'm doing is a conceptual Physics one and we don't learn JACK about the mathematical side of things. "Q = CV" is nowhere in my textbook so I guess I've got some googling to do... Thanks for your input :D

4. Jan 1, 2014

### AFSteph

Q = CV
Where Q is charge; C is capacitance and V is voltage.
So I think I know what you're saying... I had capacitance and voltage a bit backwards here.

So as the radius increases, capacitance, not voltage, increases and the voltage decreases. And by that formula given earlier, (voltage x electron charge) with less voltage it is easier to add an extra electron. That makes a lot more sense now :D

Thank goodness for the internet because this textbook really is not that helpful. They don't give the formula for capacitance nor do they even mention the frickin' word capacitance in the chapter on Van de Graaff generators!

If I'm understanding this correctly, then the quote in my textbook is incorrect
The word "voltage" should be replaced with the word "capacitance", no?

Am I understanding this correctly?

5. Jan 1, 2014

### rude man

You are doing fine.

But your textbook could still be right: the larger generators can probably hold much more charge so even though the capacitance is higher, you can add even more Q with the result that the voltage is higher also than on a smaller unit. Nevertheless, I find the statement confusing at best.

6. Jan 1, 2014

### voko

I believe the statement in the book has to do with electrical breakdown or perhaps corona discharge. Voltages in van der Graaff generators are limited by that.

7. Jan 1, 2014

### dlgoff

That was my impression as well.

http://en.wikipedia.org/wiki/Robert_J._Van_de_Graaff

8. Jan 1, 2014

### AFSteph

Thank you so much for clarifying. The course I'm doing is the American School of Correspondence's Conceptual Physics course, which is meant for homeschooled students (as I am).

They develop the exams and "study guide" themselves, and they obviously have no influence on the course book -- they just built around it. That said, it seems bizarre that they would ask a question when the course book would lead you to a contradictory answer. When I mail off this exam (correspondence course -- things are sent though the mail) should I include a letter with a link to this thread suggesting they have a section in the study guide on capacitance clarifying this? Seems like legitimate constructive criticism / feedback.

Again, thank you for helping me through this. I appreciate the hinting approach :)

9. Jan 1, 2014

### rude man

10. Jan 1, 2014

### rude man

I would think that's a good idea.

11. Jan 1, 2014

Yes it's to do with electric field strength ( E=-V/r r= radius of dome). If this exceeds a certain value(I think its between 3MV to 4MV for dry air at normal pressure )the air breaks down and you get sparking. Of course the dome is not perfectly spherical and there are smaller radii bits surrounding the hollow where the support and the belt enter. The field is stronger at these more "pointy bits".

12. Jan 1, 2014

### voko

I assume you wanted to say MV/m, not just MV.

13. Jan 1, 2014

Yes, thanks for pointing that out.