A research Van de Graaff generator has a 360 m diameter meta
Solution
a) the electric potential at the surface of the van de graff generator is
V = kQ/ R ; Q and R are the charge and radius of the generator
= 8.89 * 10^9 * 1.08 * 10^-3 C / 1.8 m
= 5.388 *10^ 6 volt
b) if the potential is V = 3.0MV = kQ / R
R = k * Q / 3.0 * 10^6 V
= 8.89 * 10^9 * 1.08 * 10^-3 / 3.0 * 10^6
= 3.232 m
c) the charge on the oxygen atoms with 3 missing electrons is q = + 3 *e ; e is the charge on electron
= 3 * 1.602 * 10^-19 C
= 4.806 * 10 -19 C
the kinetic energy KE of the Oxygen atom must be equal to the potential energy of the potential energy of the charge placed on the surface of the generator
KE = electric potential energy = k * Q * q / R ; Q and q are the charges of generator and oxygen atom
= 8.89 * 10^9 * 1.08 * 10^-3 C* 4.806*10^-19 / 1.8 m
= 4.315*10-9 Joules
