More conservation of energy

Now that we know about electric potential energy, we can revisit our old friend, the law of conservation of energy. It’s basically the same as before: start out with

$\mathrm{\Delta }{E}_{\text{k}}=-\mathrm{\Delta }{E}_{\text{el}}\text{,}$

then rearrange for the unknown quantity and substitute everything else.

Example

An alpha particle with a charge of 3.2 × 10⁻¹⁹ C moving at 1.0 × 10⁶ m/s from infinity approaches a gold nucleus whose charge is 1.3 × 10⁻¹⁷ C. How close does the alpha particle get?

We by substituting the energy formulae into the equation given above:

$\frac{1}{2}m(v_2^2-v_1^2)=-k{q}_1{q}_2(\frac{1}{r_2}-\frac{1}{r_1})\text{.}$

The particle will stop moving when it gets to the closest point, so we make $v_2$ zero. It starts at infinity, and one over infinity, for our purposes, is zero, so that gets rid of another term. We are left with

$\frac{1}{2}m{v}_1^2=\frac{k{q}_1{q}_2}{r_2}\text{.}$

Solving for final radius, we have

$r_2=\frac{2k{q}_1{q}_2}{m{v}_1^2}\text{,}$

and working that out gives us the answer, 1.1 × 10⁻¹¹ m.