A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it.

(a) What is the potential near its surface?

(b) At what distance from its center is the potential 1.00 MV?

(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

Answer:a)[tex]V=49.5MV[/tex]b)[tex]r=49.5m[/tex]c)[tex]\Delta U=3*(49.5-1)=145.5 MeV[/tex]Explanation:a)The potential equation is given by:[tex]V=k\frac{Q}{r}[/tex]

k is the electrostatic constant ([tex]k=9.9*10^{9}Nm^{2}/C^{2}[/tex])

Q is the charge Q = 5mC

r is the radius of the sphere r = 1 m

[tex]V=9.9*10^{9}\frac{5*10^{-3}}{1}=49.5MV[/tex]

b)We solve it using the same equation.Here we need to find r:

[tex]r=k\frac{Q}{V}[/tex]

[tex]r=9.9*10^{9}\frac{5*10^{-3}}{1*10^{6}}[/tex]

[tex]r=49.5m[/tex]

c)The relation between difference potential and electrical energy is:[tex]\Delta U=\Delta Vq[/tex]

here q is 3e becuase oxygen atom has three missing electrons

Therefore:

[tex]\Delta U=3*(49.5-1)=145.5 MeV[/tex]

I hope it heps you!