# 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

$\displaystyle \Delta{} E_{\text{k}} = - \Delta{} E_{\text{el}}$,

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 × 10^{6} 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:

$\displaystyle \frac{1}{2} m \left ( v_{2}^{2} - v_{1}^{2} \right ) = - k q_{1} q_{2} \left ( \frac{1}{r_{2}} - \frac{1}{r_{1}} \right )$.

The particle will stop moving when it gets to the closest point, so we make $\displaystyle v_{2}$ 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

$\displaystyle \frac{1}{2} m v_{1}^{2} = \frac{k q_{1} q_{2}}{r_{2}}$.

Solving for final radius, we have

$\displaystyle r_{2} = \frac{2 k q_{1} q_{2}}{m v_{1}^{2}}$,

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