# Electron nucleus problem

#### nakatheking

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?

#### benit13

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

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

topsquark