Motion of Charged Particles in Electric Fields
A charged particle in a uniform electric field E experiences a constant force F = qE and so a constant acceleration a = qE ÷ m, which means that the particle moves along a straight line if it starts at rest or along the field, and along a parabola if it enters perpendicular to the field, like a projectile in a gravitational field. In brief: in the direction of the field, apply v = u + at and s = ut + ½at2; in the perpendicular direction, the velocity is constant. The energy gained in moving through a potential difference V is qV. Applications include cathode-ray tubes, inkjet printers, particle accelerators and the Millikan oil-drop experiment.
Oscilloscopes and old TV tubes deflected electron beams with electric fields, and printers use the same principle to steer droplets of ink. This guide applies Newton’s second law and kinematics to charges in uniform electric fields, compares the motion in electric and gravitational fields, works through numerical examples of acceleration and deflection of an electron, and corrects common mistakes.
What happens when a charge moves along the field?
The force on a charge q in a field E is F = qE, so the acceleration is a = qE ÷ m, in the direction of E for positive charges and opposite for negative ones. In a uniform field (between parallel plates, E = V ÷ d), the acceleration is constant, and the kinematic equations apply. A charge released from rest moves in a straight line. An alternative approach uses energy: the work done by the field in moving the charge through a potential difference V is qV, which becomes kinetic energy: ½mv2 = qV, so v = √(2qV ÷ m). Lighter particles reach higher speeds: electrons (m = 9.11 × 10−31 kg) reach speeds about 43 times higher than protons for the same voltage. At speeds close to the speed of light (accelerating voltages above about 10 kV for electrons), relativistic effects must be included.
What happens when a charge enters perpendicular to the field?
When a charge enters a region of uniform field with a velocity vx perpendicular to the field direction (the y direction), the motion in the two directions is independent:
- horizontally (x): no force, so constant velocity, x = vxt;
- vertically (y): constant acceleration a = qE ÷ m, so y = ½at2 and vy = at.
Eliminating t: y = (qE ÷ 2mvx2) x2, a parabola, just like a ball thrown horizontally in a gravitational field. After leaving the plates, the particle moves in a straight line at the angle θ with tan θ = vy ÷ vx. In the Millikan experiment, an oil drop of charge q and mass m is held stationary between horizontal plates when the electric force balances the weight: qE = mg, which allows q to be found.
What are the applications?
- Cathode-ray tube (CRT) oscilloscopes and old televisions: an electron gun accelerates electrons through a few kV, and plates deflect the beam.
- Inkjet printers: droplets of ink are charged and deflected to the position on the paper.
- Electrostatic precipitators: charged particles of smoke are attracted to plates.
- Particle accelerators: electric fields accelerate protons and electrons (for example, the linear accelerators).
- Mass spectrometers: ions are accelerated by an electric field and sorted by mass in a magnetic field.
- Photocopiers and laser printers use electric fields to move toner.
Electric and gravitational fields compared
| Feature | Uniform electric field | Uniform gravitational field |
|---|---|---|
| Force | qE | mg |
| Acceleration | qE ÷ m (depends on q/m) | g (independent of m) |
| Path for perpendicular entry | Parabola | Parabola |
| Direction | Either, depending on the sign of q | Always toward the mass |
Worked example: an electron between plates
An electron enters horizontally with a speed of 2.0 × 107 m/s into a region between two horizontal plates that are 5.0 cm long, where a uniform vertical field of 5,000 N/C is present. (e = 1.602 × 10−19 C, m = 9.11 × 10−31 kg.) Ignore gravity: it is negligible, since the electric force of 8.0 × 10−16 N is about 1014 times larger than the weight of the electron (8.9 × 10−30 N).
- Acceleration: a = eE ÷ m = 1.602 × 10−19 × 5,000 ÷ 9.11 × 10−31 = 8.8 × 1014 m/s2, directed opposite to the field (the electron is negative).
- Time in the field: t = L ÷ vx = 0.050 ÷ 2.0 × 107 = 2.5 × 10−9 s.
- Deflection at the exit: y = ½at2 = ½ × 8.8 × 1014 × (2.5 × 10−9)2 = 2.7 × 10−3 m = 2.7 mm.
- Vertical velocity: vy = at = 8.8 × 1014 × 2.5 × 10−9 = 2.2 × 106 m/s. The angle of the exit velocity: tan θ = 2.2 × 106 ÷ 2.0 × 107 = 0.11, so θ = 6.3°.
- Speed at the exit: v = √((2.0 × 107)2 + (2.2 × 106)2) = 2.01 × 107 m/s.
- Oil drop: a drop of mass 1.0 × 10−14 kg is held stationary in a field of 3.0 × 105 V/m. q = mg ÷ E = 1.0 × 10−14 × 9.8 ÷ 3.0 × 105 = 3.3 × 10−19 C, about 2 elementary charges (2 × 1.602 × 10−19 = 3.2 × 10−19 C).
How was this studied?
J. J. Thomson deflected cathode rays with electric and magnetic fields in 1897 and found that the rays consisted of negatively charged particles much lighter than atoms, the electrons. Robert Millikan and Harvey Fletcher measured the elementary charge with oil drops in 1909 to 1913. Karl Ferdinand Braun invented the cathode-ray tube in 1897, and it was the basis of television displays for much of the 20th century.
Which mistakes are common?
Mistake 1: Using the direction of the field for the force on an electron
The force on a negative charge is opposite to the field.
Mistake 2: Including gravity for electrons and ions
It is negligible for elementary particles. For oil drops, it may matter.
Mistake 3: Using constant speed in the direction of the field
The particle accelerates along the field. The velocity component perpendicular to the field is constant.
Mistake 4: Applying the nonrelativistic formula at very high energy
At speeds above about 10 percent of the speed of light, use relativistic formulas.