Conservation of Electric Energy

An X-ray tube accelerates electrons from rest across a gap held at $30\ \text{kV}$, and they hit the target at about a third of the speed of light. You can find that speed without knowing the shape of the field in the gap, how long the gap is, or the path the electrons take. All you need is the potential difference and conservation of energy, because the electric force is conservative and the work it does depends only on the endpoints.

The energy change across a potential difference

When a charge $q$ moves from a point at potential $V_i$ to a point at potential $V_f$, the potential energy of the charge-field system changes by the amount below. As in topic 9.1, that energy belongs to the moving charge together with the charges that produce the field, not to the moving charge alone:

This is topic 9.2's definition of potential, $\Delta V = \frac{\Delta U_E}{q}$, solved for the energy. The four sign combinations are worth knowing without computing:

Left alone, a positive charge accelerates toward lower potential, the way a ball rolls downhill. A negative charge accelerates toward higher potential. In both cases the system's potential energy falls.

Because $q$ for a particle is a small multiple of $e$, atomic and nuclear energies are often given in electron volts. One electron volt is the energy change of a charge $e$ moving through a potential difference of $1\ \text{V}$:

A proton moving through $-2000\ \text{V}$ loses $2000\ \text{eV}$ of potential energy; an alpha particle, with charge $2e$, moving through the same difference loses $4000\ \text{eV}$. Convert to joules before you put an energy into $\frac{1}{2}mv^2$.

Energy conservation for a moving charge

If the electric force is the only force doing work, the total of kinetic and electric potential energy stays fixed:

Applied to the X-ray tube, $\frac{1}{2}mv^2 = (1.60\times10^{-19}\ \text{C})(3.0\times10^{4}\ \text{V})$ gives $v = 1.0\times10^{8}\ \text{m/s}$, about a third of $c$.

The formula $v = \sqrt{\frac{2|q\,\Delta V|}{m}}$ carries the factor-of-change answers:

• the speed grows as $\sqrt{\Delta V}$, so four times the potential difference gives twice the speed;
• the kinetic energy is proportional to the charge, so an alpha particle ($2e$) gains twice the energy of a proton through the same $\Delta V$;
• the speed goes as $\sqrt{\frac{q}{m}}$, so the alpha particle, with twice the charge but about four times the mass, ends up slower by a factor of $\frac{1}{\sqrt{2}}$.

For particles, gravity is almost always negligible. A proton between plates $1.0\ \text{cm}$ apart with $100\ \text{V}$ across them feels an electric force of $qE = (1.60\times10^{-19})(1.0\times10^{4}) = 1.6\times10^{-15}\ \text{N}$ and a weight of $1.6\times10^{-26}\ \text{N}$, a hundred billion times smaller.

Energy conservation also finds turning points. An alpha particle fired straight at a gold nucleus ($+79e$, heavy enough to treat as fixed) with $5.0\ \text{MeV} = 8.0\times10^{-13}\ \text{J}$ of kinetic energy slows as it climbs the potential hill around the nucleus and stops, for an instant, where all its kinetic energy has become potential energy:

When both charges are free to move, neither one sits in a fixed potential. Then use topic 9.1's energy of the whole system, $\frac{q_1q_2}{4\pi\varepsilon_0 r}$, together with conservation of momentum, as Example 2 does.

Showing the energy with bar charts

An energy bar chart draws one bar for each form of energy at two moments, with height proportional to the energy. Potential energy can be negative, so its bar can hang below the zero line. The total (the sum of the bars, with signs) is the same at both moments when no outside work is done. Drawing one before calculating catches sign mistakes: if the potential energy bar grows, the kinetic energy bar must shrink by the same amount.

Worked examples

Worked exampleExample 3: measuring charge-to-mass ratio (experimental design style)

A lab has an ion source that releases ions nearly at rest, a pair of grids across which a known, adjustable potential difference $\Delta V$ can be applied, and a field-free tube $0.40\ \text{m}$ long after the grids, with detectors at each end that record how long each ion takes to cross it.

(a) Describe a procedure to determine the charge-to-mass ratio $\frac{q}{m}$ of the ions, including what to graph.

(b) A student collects the data below. Use it to determine $\frac{q}{m}$.

$\Delta V$ (V)	Time to cross the tube ($\mu\text{s}$)
100	4.14
200	2.86
300	2.37
400	2.03
500	1.83

(c) The ion source turns out to release ions with a small, fixed kinetic energy rather than at rest. How does that affect the graph and the value from (b)?

Show solution

(a) Set $\Delta V$ with the power supply and measure it with a voltmeter across the grids. For each setting, record the ions' crossing time $t$ for the $0.40\ \text{m}$ tube (averaging several ions), and compute their speed $v = \frac{0.40\ \text{m}}{t}$; the tube is field-free, so the speed is constant there. Repeat for at least five values of $\Delta V$. Energy conservation for an ion that starts from rest gives $\frac{1}{2}mv^2 = q\,\Delta V$, so

$$v^2 = \left(\frac{2q}{m}\right)\Delta V.$$

Graph $v^2$ on the vertical axis against $\Delta V$ on the horizontal axis. The graph should be a straight line through the origin with slope $\frac{2q}{m}$. Using a best-fit line through all the points, rather than one measurement, averages out random timing errors.

(b) Compute $v = \frac{0.40}{t}$ and $v^2$ for each row:

$\Delta V$ (V)	$v$ ($10^{5}$ m/s)	$v^2$ ($10^{10}$ m$^2$/s$^2$)
100	0.966	0.934
200	1.40	1.96
300	1.69	2.85
400	1.97	3.88
500	2.19	4.78

The best-fit line has slope $9.6\times10^{7}\ \text{m}^2/(\text{s}^2\cdot\text{V})$ and passes very close to the origin, as the model predicts. Its units work out to C/kg, since $1\ \text{V} = 1\ \text{J/C}$ and $1\ \text{J} = 1\ \text{kg}\cdot\text{m}^2/\text{s}^2$. So

$$\frac{q}{m} = \frac{\text{slope}}{2} = 4.8\times10^{7}\ \text{C/kg}.$$

That is half the proton's value, $\frac{e}{m_p} = 9.58\times10^{7}\ \text{C/kg}$: the ions carry one elementary charge on about twice the proton's mass, as a deuteron (one proton and one neutron) does.

(c) With initial kinetic energy $K_0$, energy conservation becomes $\frac{1}{2}mv^2 = K_0 + q\,\Delta V$, so $v^2 = \frac{2q}{m}\Delta V + \frac{2K_0}{m}$. The points shift up by the same amount: the line gets a positive vertical intercept, but its slope, and so the value of $\frac{q}{m}$ from the slope, is unchanged. Computing $\frac{q}{m}$ from a single point instead, as $\frac{v^2}{2\,\Delta V}$, would come out too high, which is another reason to use the slope.

Where students lose points

On the exam

Questions on this topic give a potential difference (as numbers, a graph of $V$, or an equipotential map) and ask for a speed, a kinetic energy, a stopping point, or how one of these changes when the charge, mass or potential difference is scaled. Justifications should name the principle ("only the electric force does work, so mechanical energy is conserved") and state the system. Experimental questions use the same relation $\frac{1}{2}mv^2 = q\,\Delta V$ and expect you to choose quantities to graph that make it a straight line, then read the physical quantity from the slope.