Question

A research Van de Graaff generator has a 2.10 m diameter metal sphere with a charge of 40.0 C on it. (a) What is the potential (in kV) near its surface? 171.42 x kV (b) At what distance from its center (in m) is the potential 105 kV? m (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its kinetic energy in keV at the distance found in part (b)? keV

Answer 1

Answer:

distance is 3.428571 × $10^{-6}$ m

potential is 342.857143 × $10^{6}$ kV

kinetic energy is  713.571429 × $10^{6}$ KeV

Explanation:

Given data

diameter  = 2.10 m

charge = 40.0 C

to find out

potential and distance from its center  and kinetic energy

solution

we know diameter  = 2.10 m so radius will be = 2.10 / 2 = 1.05 m and charge = 40 C

so potential = KQ/R

here K is constant with value = 9× $10^{9}$ Nm²C²

so  potential =   9× $10^{9}$ (40) / 1.05

potential = 342.857143 × $10^{9}$ V

so  potential is 342.857143 × $10^{6}$ kV

and

in 2nd part we find radius

potential = KQ / R

put the value

105 × $10^{3}$  = 9× $10^{9}$ ( 40 )   / R

R = 3.428571 × $10^{-6}$ m

so distance is 3.428571 × $10^{-6}$ m

and

in 3rd part

we find KE

so change in KE  = change in V

so KE 2 - KE 1  = q (V2 - V1)

KE2 - 0 = 3 ( 342.857143 × $10^{6}$ - 105 )

so kinetic energy is  713.571429 × $10^{6}$ KeV