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

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

then rearrange for the unknown quantity and substitute everything else.


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:

12m(v22v12)=kq1q2(1r21r1)\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 v2\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

12mv12=kq1q2r2\displaystyle \frac{1}{2} m v_{1}^{2} = \frac{k q_{1} q_{2}}{r_{2}}.

Solving for final radius, we have

r2=2kq1q2mv12\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.