# Voltage required for penetration of an unknown element

• grantaere
In summary, the question asks for the amount of voltage needed to accelerate a proton to penetrate an unknown element with a positively charged nucleus. The potential difference is calculated by using the equation V = kQ/r, where Q is the charge of the nucleus and r is its radius. The force between two charges is then used to find the acceleration of the proton, and the potential difference is calculated by only considering the charge of the nucleus acting on the proton.

## 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.

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!

You are looking for the amount of energy (work) to bring two charges into contact. What is the force between two charges?

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?

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?

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!