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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?

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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...

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No, i meant if i had 2 spheres, first 1 has a radius of 1 decimeter and voltage of 320 kv for example and second one radius of 1 meter and also 320 kv, will the charge differ? – dodo Dec 10 '12 at 16:59
@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 :-) – Ϛѓăʑɏ βµԂԃϔ Dec 10 '12 at 17:33
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Ok, thanks very much for help ;-) – dodo Dec 10 '12 at 20:39

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