Energy Conservation for Moving Charges

The electric force is conservative, like gravity, so a charge moving among fixed charges keeps its total energy: whatever potential energy it loses becomes kinetic energy. With potentials in hand, this is often the quickest way to find a speed: no forces, no accelerations, no time.

Key idea

$$\Delta K + q\,\Delta V = 0 \qquad\text{or}\qquad \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = -q(V_f - V_i).$$ A positive charge speeds up moving to lower potential; a negative charge speeds up moving to higher potential. One electron-volt, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$, is the energy an electron gains crossing $1\ \text{V}$.

Why it works

The work the field does on the charge equals the drop in its potential energy, $-q\Delta V$, and by the work-energy theorem that work is $\Delta K$. Because the electrostatic force is conservative, the path does not matter, only the start and end potentials.

The electron-volt is handy because $q\,\Delta V$ is then just arithmetic: an electron crossing $500\ \text{V}$ gains $500\ \text{eV}$, and an alpha particle (charge $2e$) crossing the same $500\ \text{V}$ gains $1000\ \text{eV}$, whatever their masses. Convert to joules only when you need a speed. And remember that $\tfrac{1}{2}mv^2$ is the slow-speed formula: when a speed comes out above about a tenth of the speed of light, relativity gives a noticeably different answer.

An electron entering the gap between charged plates. The field does work on it; its kinetic energy grows by $e\,\Delta V$ as it moves toward the positive plate.

Worked examples

Worked example: An electron gun

An electron starts from rest and is accelerated through $500\ \text{V}$. How fast is it going?

It gains $K = e\,\Delta V = 500\ \text{eV} = (1.60 \times 10^{-19})(500) = 8.00 \times 10^{-17}\ \text{J}$.
$$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(8.00 \times 10^{-17})}{9.11 \times 10^{-31}}} = 1.33 \times 10^{7}\ \text{m/s}$$
About 4% of the speed of light, so the classical formula is still fine.

Answer: $v \approx 1.33 \times 10^{7}\ \text{m/s}$

Worked example: Pushed away by a fixed charge

A proton is released from rest $2.0\ \text{mm}$ from a fixed $+5.0\ \text{nC}$ charge. How fast is it moving when it is far away?

Far away the potential is zero, so the proton loses all of $U = \dfrac{kQe}{r}$: $$K = \frac{(8.99 \times 10^{9})(5.0 \times 10^{-9})(1.60 \times 10^{-19})}{0.0020} = 3.60 \times 10^{-15}\ \text{J}$$
$v = \sqrt{2K/m_p} = \sqrt{\dfrac{2(3.60 \times 10^{-15})}{1.67 \times 10^{-27}}} = 2.08 \times 10^{6}\ \text{m/s}$.

Answer: $v \approx 2.08 \times 10^{6}\ \text{m/s}$

Worked example: Rutherford's closest approach

An alpha particle ($+2e$) with $5.0\ \text{MeV}$ of kinetic energy heads straight for a gold nucleus ($+79e$). How close does it get?

At the turning point it is momentarily at rest, so all of its $K$ has become $U = \dfrac{k(2e)(79e)}{r}$.
$K = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-13}\ \text{J}$.
$$r = \frac{(8.99 \times 10^{9})(2)(79)(1.60 \times 10^{-19})^2}{8.0 \times 10^{-13}} = 4.55 \times 10^{-14}\ \text{m}$$ about $45.5\ \text{fm}$. That is thousands of times smaller than a gold atom (radius about $1.4 \times 10^{-10}\ \text{m}$), so the atom's positive charge must be packed into a tiny nucleus.

Answer: $r \approx 4.55 \times 10^{-14}\ \text{m}$

Common mistake

"Charges always speed up moving toward lower potential." That is true for positive charges only. For an electron, $\Delta K = -q\Delta V = +e\Delta V$: it gains energy moving toward higher potential, the way a bubble rises in water. Check with forces: the field points from high to low $V$, and a negative charge is pushed opposite the field, toward high $V$.

Energy Conservation for Moving Charges

Worked examples

Practice

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