• Support PF! Buy your school textbooks, materials and every day products Here!

Voltage required for penetration of an unknown element

  • #1
15
0

Homework Statement


An unknown element has a nucleus with charge 14.00 e and a radius of about 3.60×10-15 m. How much voltage must be used to accelerate a proton (radius 1.20×10-15 m) so that is has sufficient energy to just penetrate the unknown element?
Assume that the potential is that for point charges.

Homework Equations


V = kQ/r

The Attempt at a Solution


I tried calculating the two potential differences caused by the incoming proton and the positively charged atom, then adding them separately, but it seems that this isn't the right answer/ right approach.
V = (8.99e9*(14*(1.6e-19))/3.6e-15) + (8.99e9(1.6e-19)/1.2e-15) = 6.79e6 V
Any pointers would be very much appreciated!
 

Answers and Replies

  • #2
Bystander
Science Advisor
Homework Helper
Gold Member
5,173
1,173
You are looking for the amount of energy (work) to bring two charges into contact. What is the force between two charges?
 
  • #3
15
0
So then I should use F=fqQ/r^2 to find the force between the two charges when they meet? I did that (and got 6.71E-13N) -- so then I can find the acceleration, but I'm still not sure how value this would be used to calculate voltage?
 
  • #4
Bystander
Science Advisor
Homework Helper
Gold Member
5,173
1,173
Just wanted to know if you were familiar with the general Coulomb repulsion/attraction. The problem statement has apparently done the integration for the work, but omitted one of the charges, perhaps to give you a potential (rather than work) in terms of volts per unit charge. Hint enough?
 
  • #5
15
0
Oh, I got the answer now. I forgot that the calculation for potential difference only involves the charge acting on the proton, and doesn't involve the charge of the proton itself. Thanks for your help!
 

Related Threads on Voltage required for penetration of an unknown element

Replies
2
Views
4K
  • Last Post
Replies
9
Views
13K
Replies
2
Views
4K
Replies
3
Views
5K
Replies
2
Views
3K
  • Last Post
Replies
5
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
0
Views
2K
Replies
3
Views
1K
Replies
5
Views
2K
Top