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# 8.2: Charged Particle in an Electric Field

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

There is really very little that can be said about a charged particle moving at nonrelativistic speeds in an electric field $$\textbf{E}$$. The particle, of charge q and mass $$m$$, experiences a force $$q\textbf{E}$$, and consequently it accelerates at a rate $$q\textbf{E}/m$$. If it starts from rest, you can calculate how fast it is moving in time t, what distance it has travelled in time $$t$$, and how fast it is moving after it has covered a distance $$x$$, by all the usual first-year equations for uniformly accelerated motion in a straight line. If the charge is accelerated through a potential difference $$V$$, its loss of potential energy $$qV$$ will equal its gain in kinetic energy $$\frac{1}{2} m v^2$$ . Thus $$v = \sqrt{2qV/m}$$.

Let us calculate, using this nonrelativistic formula, the speed gained by an electron that is accelerated through 1, 10, 100, 1000, 10000, 100,000 and 1,000,000 volts, given that, for an electron, $$e/m = 1.7588 \times 10^{11} \text{C kg}^{−1}$$. (The symbol for the electronic charge is usually written $$e$$. You might note here that that's a lot of coulombs per kilogram!). We'll also calculate $$v/c$$ and $$v^2 /c 2$$.

\begin{array}{c c c c} \nonumber
V \text{ volts} & \nu \text{ m s}^{-1} &\nu /c & \nu^2/c^2 \\
\hline
1 & 5.931\times 10^5 & 1.978\times 10^{-3} & 3.914\times 10^{-6} \\
10 & 1.876\times 10^6 & 6.256\times 10^{-3} & 3.914\times 10^{-5}\\
100 & 5.931\times 10^6 & 1.978\times 10^{-2} & 3.914\times 10^{-4}\\
1000 & 1.876\times 10^7 & 6.256\times 10^{-2} & 3.914\times 10^{-3} \\
10000 & 5.931\times 10^7 & 1.978\times 10^{-1} & 3.914\times 10^{-2} \\
100000 & 1.876\times 10^8 & 6.256\times 10^{-1} & 3.914\times 10^{-1} \\
1000000 & 5.931\times 10^8 & 1.978 & 3.914\\
\end{array}

We can see that, even working to a modest precision of four significant Figures, an electron accelerated through only a few hundred volts is reaching speeds at which $$v^2 /c^2$$ is not quite negligible, and for less than a million volts, the electron is already apparently moving faster than light! Therefore for large voltages the formulas of special relativity should be used. Those who are familiar with special relativity (i.e. those who have read Chapter 15 of Classical Mechanics!), will understand that the relativistically correct relation between potential and kinetic energy is $$qV = (\gamma-1)m_0c^2$$ , and will be able to calculate the speeds correctly as in the following table. Those who are not familiar with relativity may be a bit lost here, but just take it as a warning that particles such as electrons with a very large charge-to-mass ratio rapidly reach speeds at which relativistic formulas need to be used. These Figures are given here merely to give some idea of the magnitude of the potential differences that will accelerate an electron up to speeds where the relativistic formulas must be used.

\begin{array}{c c c c} \nonumber
V \text{ volts} & \nu \text{ m s}^{-1} &\nu /c & \nu^2/c^2 \\
\hline
1 & 5.931\times 10^5 & 1.978\times 10^{-3} & 3.914\times 10^{-6} \\
10 & 1.875\times 10^6 & 6.256\times 10^{-3} & 3.914\times 10^{-5}\\
100 & 5.930\times 10^6 & 1.978\times 10^{-2} & 3.912\times 10^{-4}\\
1000 & 1.873\times 10^7 & 6.247\times 10^{-2} & 3.903\times 10^{-3} \\
10000 & 5.845\times 10^7 & 1.950\times 10^{-1} & 3.803\times 10^{-2} \\
100000 & 1.644\times 10^8 & 5.482\times 10^{-1} & 3.005\times 10^{-1} \\
1000000 & 2.821\times 10^8 & 0.941 & 0.855\\
\end{array}

If a charged particle is moving at constant speed in the $$x$$-direction, and it encounters a region in which there is an electric field in the $$y$$-direction (as in the Thomson $$e/m$$ experiment, for example) it will accelerate in the $$y$$-direction while maintaining its constant speed in the $$x$$-direction. Consequently it will move in a parabolic trajectory just like a ball thrown in a uniform gravitational field, and all the familiar analysis of a parabolic trajectory will apply, except that instead of an acceleration g, the acceleration will be $$q/m$$.