# About voltage and charge of van de graff generator

I have read that in case of Van de graff generator $V=kQ/r$ where $r$ is radius of the sphere. If that's the case, does the same voltage results in bigger charges in bigger radii?

-

If the charges (about some coulombs) numerically equalize or compensate (like doubling both the values simultaneously) the radius (in meters) of the sphere, then the potential indeed remains the same. $$V=\frac{kQ}r$$ The equation simply gives the relationship in $C/m$. Now, doubling the value of coulombs and meters would cancel each other out and so - you'd get the same value of potential...
@dodo: Hi dodo. Perhaps that's what I have been saying. When you double the radius (keeping $V$ as a constant), you're actually doubling the charge. That's all. So, Yes it does. Because the electric field intensity is high for greater coulombs. As far as your part is concerned, you're increasing $r$ to 10 times. Hence, $q$ must be 10'd so that the voltage remains the same :-) – Waffle's Crazy Peanut Dec 10 '12 at 17:33