Coaching Centre
Charged particles in electric fields
The uniform electric field between parallel plates, E = V/d, the force and acceleration on a charged particle, work done by the field and the electron volt, and the parabolic path of a charge fired between plates, compared with a projectile.
Every electron in the Australian Synchrotron at Clayton starts its trip at almost zero speed, boiled off a hot cathode in an electron gun. A few centimetres and a few nanoseconds later it is moving at a large fraction of the speed of light. The thing doing the pushing is nothing more exotic than the electric field between two charged metal surfaces.
The field between parallel plates
Connect two flat, parallel metal plates to a power supply and one becomes positive, the other negative. Away from the edges, the electric field between them is uniform: the same strength and direction everywhere. Field lines are drawn straight, parallel and evenly spaced, pointing from the positive plate to the negative plate (the way a positive test charge would be pushed).
The field strength depends on the potential difference across the plates and how far apart they are:
$$E = \frac{V}{d}$$
with E in volts per metre (V m−1), V in volts and d in metres. One volt per metre is the same as one newton per coulomb, because the field strength is also the force per unit charge, $E = F/q$. Double the voltage and the field doubles; move the plates twice as far apart and it halves.
Force and acceleration
A charge q in the field feels a force
$$F = qE$$
which is constant everywhere between the plates. A constant force means a constant acceleration, $a = F/m = qE/m$. For electrons and protons this acceleration is enormous, because their masses are so small. Gravity still acts, but it is so much weaker that we ignore it: for an electron in a 104 V m−1 field the electric force is about 1014 times its weight.
Two plates 2.0 cm apart have a potential difference of 400 V across them. Find the field strength, and the force on and acceleration of an electron between the plates. ($q_e = -1.602 \times 10^{-19}$ C, $m_e = 9.109 \times 10^{-31}$ kg)
$E = \dfrac{V}{d} = \dfrac{400}{0.020} = 2.0 \times 10^{4}$ V m−1, from the positive to the negative plate.
$F = qE = (1.602 \times 10^{-19})(2.0 \times 10^{4}) = 3.2 \times 10^{-15}$ N, towards the positive plate (the electron is negative, so the force is opposite to the field).
$a = \dfrac{F}{m} = \dfrac{3.2 \times 10^{-15}}{9.109 \times 10^{-31}} = 3.5 \times 10^{15}$ m s−2 towards the positive plate.
Work, energy and the electron volt
When the field moves a charge from one plate to the other, it does work on it. Work is force times displacement, so
$$W = qEd = qV$$
If nothing else acts on the particle, all of that work becomes kinetic energy (the work–energy theorem): $\Delta K = qV$. For a particle that starts from rest,
$$qV = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{\frac{2qV}{m}}$$
Notice that the plate separation has dropped out. Closer plates give a stronger field and a bigger force, but the force acts over a shorter distance, so the work, and the final speed, depend only on the voltage. That is why electron guns are rated by their accelerating voltage.
Physicists often measure particle energies in electron volts (eV): the kinetic energy an electron gains when it is accelerated through 1 V. $1\ \text{eV} = 1.602 \times 10^{-19}$ J. An electron accelerated through 1000 V gains 1000 eV, or 1.602 × 10−16 J. An alpha particle (charge +2e) accelerated through the same 1000 V gains 2000 eV.
A proton starts from rest and is accelerated through a potential difference of 5.0 kV. Find its final speed. ($m_p = 1.673 \times 10^{-27}$ kg)
$W = qV = (1.602 \times 10^{-19})(5.0 \times 10^{3}) = 8.01 \times 10^{-16}$ J $= 5.0$ keV.
$\tfrac{1}{2}mv^2 = 8.01 \times 10^{-16}$, so $v = \sqrt{\dfrac{2(8.01 \times 10^{-16})}{1.673 \times 10^{-27}}} = 9.8 \times 10^{5}$ m s−1.
That is about 0.3% of the speed of light, so we don't need relativity. An electron through the same voltage would move about 43 times faster ($\sqrt{m_p/m_e} \approx 43$), at 4.2 × 107 m s−1, or 14% of c, where relativistic effects start to show.
The path of a charge fired between plates
Now fire a charged particle into the field at right angles to the field lines, like the beam in an old cathode ray oscilloscope. Treat the motion in two perpendicular directions, exactly as you did for projectiles in Module 5:
- Parallel to the plates there is no force, so the velocity is constant: $x = u_x t$.
- Across the plates there is a constant force qE, so a constant acceleration $a = qE/m$: $y = \tfrac{1}{2}at^2$ and $v_y = at$ (if it entered with no velocity across the plates).
Constant velocity one way and constant acceleration at right angles gives a parabola. Once the particle leaves the plates there is no force (ignoring gravity), so it travels in a straight line at the angle it had when it left.
| Projectile near Earth's surface | Charged particle between plates | |
|---|---|---|
| Field | Uniform gravitational field, g = 9.8 N kg−1 | Uniform electric field, E = V/d |
| Force | F = mg, always towards Earth | F = qE, along or against the field depending on the sign of q |
| Acceleration | g, the same for every mass | qE/m, depends on the charge-to-mass ratio |
| Path | Parabola | Parabola (while between the plates) |
| Work done by the field | mgh | qV |
The one important difference is that every projectile has the same acceleration, but charged particles do not. Two particles fired in at the same speed are deflected according to their charge-to-mass ratio, q/m. Electrons, with a huge q/m, swing round much more sharply than protons.
An electron moving at 2.0 × 107 m s−1 enters midway between two plates 5.0 cm long and 1.0 cm apart, travelling parallel to them. The potential difference across the plates is 150 V. Does the electron get through?
$E = \dfrac{150}{0.010} = 1.5 \times 10^{4}$ V m−1, so $a = \dfrac{qE}{m} = \dfrac{(1.602 \times 10^{-19})(1.5 \times 10^{4})}{9.109 \times 10^{-31}} = 2.64 \times 10^{15}$ m s−2.
Horizontal velocity is constant: $t = \dfrac{x}{u_x} = \dfrac{0.050}{2.0 \times 10^{7}} = 2.5 \times 10^{-9}$ s.
$y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(2.64 \times 10^{15})(2.5 \times 10^{-9})^2 = 8.2 \times 10^{-3}$ m $= 8.2$ mm.
It starts only 5.0 mm from each plate, so it hits the positive plate before reaching the end. Since y is proportional to V, the largest voltage that lets it through is $150 \times \dfrac{5.0}{8.2} \approx 91$ V.
- Deflect mode, electron, accelerating voltage 1000 V, plate voltage 200 V, separation 2.0 cm. Check the sim's deflection at exit (6.25 mm) using $y = \tfrac{1}{2}at^2$ with the readouts.
- Switch to a proton, then an alpha particle, without changing anything else. The deflection doesn't change. Show algebraically that $y = \dfrac{V L^2}{4 d V_{acc}}$, so the charge and mass cancel when the particle was accelerated by a voltage.
- Raise the plate voltage until the beam just hits the plate. Predict the value first: the beam starts 1.0 cm from each plate.
- Accelerate mode, electron, 1000 V. Note the final speed. Halve the separation: what changes, and what doesn't? Then try 4000 V. By what factor does the speed increase?
Test yourself
The potential difference across two parallel plates is doubled and their separation is halved. The electric field between them
E = V/d: doubling V doubles E, halving d doubles it again, so E is 4 times larger.
An electron and a proton are each accelerated from rest through the same potential difference. Which statement is correct?
ΔK = qV and both have charge of magnitude e. The electron ends up much faster because its mass is smaller.
An alpha particle is accelerated from rest through 2.0 kV. Its kinetic energy is
An alpha particle has charge +2e, so it gains 2 × 2000 = 4000 eV.
A proton enters the uniform field between two horizontal plates moving horizontally. While it is between the plates, its horizontal velocity
The force is vertical, so nothing changes the horizontal component, just as for a projectile.
Why is the gravitational force usually ignored for an electron in an electric field?
The weight (about 9 × 10−30 N) is negligible compared with forces of order 10−15 N.
Label the parallel-plate diagram.
Sort the statements.
The field between parallel plates is and points from the positive plate to the plate. Its strength in volts per metre equals the voltage divided by the plate . The work done on a charge q moved through a potential difference V is qV, and it becomes energy. A charge fired across the field follows a path. The energy an electron gains through 1 V is one electron .
An electron is accelerated from rest through a potential difference of 2.5 kV and then enters, at right angles, the uniform field between two plates 4.0 cm long. (a) Calculate the speed of the electron as it enters the plates. (b) Explain why the electron's path between the plates is parabolic.
(a) The work done by the field becomes kinetic energy: qV = ½mv². v = √(2qV/m) = √(2 × 1.602 × 10−19 × 2500 ÷ 9.109 × 10−31) = 3.0 × 107 m s−1.
(b) Between the plates the field is uniform, so the electron feels a constant force F = qE directed towards the positive plate, at right angles to its initial velocity. There is no force parallel to the plates, so that velocity component stays constant and the electron covers equal distances along the plates in equal times. At right angles to the plates the constant force produces a constant acceleration, so the sideways displacement grows with t². Constant velocity in one direction combined with constant acceleration perpendicular to it gives a parabolic path, the same as a horizontally launched projectile.
Explain why the final speed of a particle accelerated from rest between two parallel plates does not depend on the separation of the plates, provided the potential difference is fixed.
Moving the plates closer increases the field strength, since E = V/d, so the force F = qE on the particle is larger. However, that larger force acts over a proportionally shorter distance. The work done is W = Fd = qEd = q(V/d)d = qV, which depends only on the charge and the potential difference. All of this work becomes kinetic energy, so ½mv² = qV and the final speed √(2qV/m) is the same for any separation.