Electron nucleus problem

Dec 2019
12
0
hello world
prow.PNG

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
661
333
Glasgow
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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