Calculating Proton Speed in an Accelerator and Repulsive Force at 8 x 10^-14 m

[cathoderay]

In an accelerator, protons are accelerated through a potential difference of 1.0 x 10^6 volts.
a)what is the speed of the proton when they leave the accelerator?

b)When the protons enter a zinc target, the closest distance they come to a zinc nucleus is 8 x 10^-14 m. show that the force of repulsion between the proton and nucleus at this distance is about 1N. The charge on the Zinc nucleus has value 30 times the charge on the electron.
(Picture 1-attached)
c)On the diagram below, complete the path of three protons.
-> Aimed directly at the nucleus of Zn
-> Aimed towards the edge of a nucleus of Zn
-> Aimed in the area between two nuclei of Zn.
(Picture 2-attached)



for a) i use E=vq and then v= sqr(2Ek/m) to get the answer




3. for B&C I am not sure wht i have to do?? could some one splaind me please thanks..

 

[Irid]

For part B, try to use Coulomb's law.

 

[cathoderay]

part A & B solve thanks...
part c i would like some help figuring out what I am supous to do..
thanks ..