[Physics] About voltage and charge of van de graff generator

Question

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?

Answer 1

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...