Electron nucleus problem

Dec 2019
hello world

So what I did is I had Fd = energy. So k*Qnucleus*qproton/(r+R)^2 * r= q(deltaV). Where r and R are the radii of the proton and the nucleus of the unknown, respectively. What is wrong with this?
Oct 2017
If the two charges are far apart initially (\(\displaystyle r >> r_1 + r_2\)), then we can say that the initial potential energy is zero.

\(\displaystyle U_0 = 0\)

If the proton is "just" penetrating the unknown element, it is situated in its electric field and must gain a potential energy equal to:

\(\displaystyle U_1 = \frac{q Q}{4 \pi \epsilon_0 (r + R)}\)

Therefore, the energy that needs to be bestowed to the proton is equal to

\(\displaystyle \Delta U = U_1 - U_0 = \frac{q Q}{4 \pi \epsilon_0 (r + R)}\)

The definition of the potential difference is the work done by an electric field per Coulomb of charge operated on. In other words,

\(\displaystyle V = \frac{W}{q}\)

We can set \(\displaystyle W = \Delta U\), so we get

\(\displaystyle V = \frac{\Delta U}{q} = \frac{Q}{4 \pi \epsilon_0 (r + R)}\)
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