# Electrostatic induction in spheres

if we have a charged sphere with charge Q and radius r, the voltage on it's surface is calculated by gauss's law, what if we approach to it another sphere with radius R , R>r, earth it, remove earth, basically charging by induction, will the sphere have the same charge, which means less voltage on the surface or same voltage which means more charge? In other words my question is , which one is induced the voltage or the charge?

When you are approaching to the first sphere you have a potential $\phi_{from\ 1st\ sphere}$ on the second sphere. Grounding second sphere will give you opportunity to cancel this potential through negative charge from the Earth which has flown to 2nd sphere already. $\phi_{from\ 1st\ sphere}-\phi_{from\ charge\ from\ earth}=0$
• Not exactly, you mentioned that you can calculate $\phi_{from\ 1st\ sphere}=\frac {kQ}{r_{between\ spheres}}$. After grounding potential on the second sphere is $\frac {kQ}{r_{between\ spheres}}-\frac{kQ_{from\ earth}}{R}$ which should be zero. From this expression you can see that the new charge is with opposite sign, but not equal. Feb 8, 2015 at 19:47