D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsCharged Particles in Uniform Electric Fields
D2.3 D2.5About 75 minutes
Learning goals- I can find the acceleration of a charged particle in a uniform field, a = qE/m, and show that gravity is usually negligible.
- I can use energy (qΔV = ½mv²) to find the speed of a particle accelerated through a potential difference.
- I can analyse a charged particle that enters a uniform field at right angles, using projectile-motion methods.
- I can explain how electron guns, deflection plates and inkjet printers use these ideas.
Words to know
electron gun
A device that makes a narrow beam of fast electrons by boiling electrons off a hot cathode and accelerating them toward a positive anode.
accelerating potential difference
The potential difference ΔV through which a charged particle is accelerated; it gains kinetic energy qΔV.
deflection
The sideways change in direction of a moving charged particle caused by a field at an angle to its motion.
cathode-ray tube (CRT)
A vacuum tube in which an electron beam, steered by electric or magnetic fields, hits a screen. Used in older TVs and oscilloscopes and in J. J. Thomson's 1897 discovery of the electron.
electrostatic precipitator
A pollution-control device that charges smoke or dust particles and collects them on oppositely charged plates.
Steering electrons
In 1897, J. J. Thomson bent beams of "cathode rays" with electric and magnetic fields and showed that they were made of tiny negative particles, much lighter than atoms: electrons. For most of the 20th century, televisions and computer screens worked by steering an electron beam with fields to paint pictures on a glowing screen. Today the same ideas run electron microscopes, inkjet printers, X-ray tubes, mass spectrometers and particle accelerators.
The good news: you already know how to solve these problems. A charge in a uniform electric field feels a constant force, just like a ball in the near-Earth gravitational field. So everything from SPH3U kinematics and projectile motion (the unit Dynamics in this course reviews it) works here, with a = qE/m in place of g.
Acceleration in a uniform field
Newton's second law with F = qE:
a = qE/m
For an electron in a modest field of 1.0 × 10⁴ N/C, a = (1.60 × 10⁻¹⁹)(1.0 × 10⁴)/(9.11 × 10⁻³¹) ≈ 1.8 × 10¹⁵ m/s², about 10¹⁴ times g. That is why we can ignore gravity for electrons, protons and ions in almost any electric field. (Millikan's oil drops in the next lesson are the famous exception: they are heavy enough that gravity matters.)
The direction: a positive charge accelerates along E; a negative charge accelerates against E.
Speeding up: the energy method
When a particle with charge q starts from rest and moves through a potential difference ΔV, it gains kinetic energy:
|q|ΔV = ½mv² so v = √(2|q|ΔV/m)
This works in any field, uniform or not, because it only uses the start and end potentials. An electron gun uses it: a hot metal cathode releases electrons, and a positive anode with a small hole, a few hundred to many thousands of volts higher, accelerates them into a beam.
Worked exampleExample 1: Speed from an electron gun
An electron is accelerated from rest through 2.0 kV. Find its final speed. Repeat for a proton.
- Electron: v = √(2eΔV/m) = √(2(1.60 × 10⁻¹⁹)(2.0 × 10³)/(9.11 × 10⁻³¹)) = √(7.03 × 10¹⁴) = 2.65 × 10⁷ m/s.
- Proton: v = √(2(1.60 × 10⁻¹⁹)(2.0 × 10³)/(1.67 × 10⁻²⁷)) = √(3.83 × 10¹¹) = 6.19 × 10⁵ m/s.
Electron ≈ 2.7 × 10⁷ m/s (about 9% of the speed of light). Proton ≈ 6.2 × 10⁵ m/s. Both gain the same energy (2.0 keV), but the proton is 1836 times heavier, so it is √1836 ≈ 43 times slower.
A speed limit ahead
v = √(2qΔV/m) is Newtonian. Once an electron's speed gets above roughly 10% of the speed of light (accelerating voltages above a few kilovolts), the true speed is noticeably less than this formula predicts. You will learn the correct relativistic treatment in the unit Special Relativity.
Bending a beam: the projectile method
Now send a particle horizontally between two parallel plates, with the field perpendicular to its velocity. Just like a ball thrown horizontally off a cliff:
- Along the plates (x): no force, so constant speed v₀. Time between the plates: t = L/v₀.
- Across the plates (y): constant acceleration a = qE/m from rest. Sideways displacement: y = ½at²; sideways velocity: v_y = at.
The path between the plates is a parabola. After it leaves the plates, there is no force, so it travels in a straight line at angle θ, where tan θ = v_y/vₓ.
Worked exampleExample 2: Deflecting an electron beam
An electron enters the region between two plates moving horizontally at 2.0 × 10⁷ m/s. The plates are 5.0 cm long and the field between them is 2.0 × 10⁴ N/C, pointing down. Find (a) the electron's acceleration, (b) its vertical displacement at the end of the plates, (c) its velocity as it leaves.
- (a) a = eE/m = (1.60 × 10⁻¹⁹)(2.0 × 10⁴)/(9.11 × 10⁻³¹) = 3.51 × 10¹⁵ m/s², upward (opposite to E, because the electron is negative).
- (b) Time between plates: t = L/v₀ = 0.050/(2.0 × 10⁷) = 2.5 × 10⁻⁹ s. y = ½at² = ½(3.51 × 10¹⁵)(2.5 × 10⁻⁹)² = 1.10 × 10⁻² m.
- (c) v_y = at = (3.51 × 10¹⁵)(2.5 × 10⁻⁹) = 8.78 × 10⁶ m/s. vₓ = 2.0 × 10⁷ m/s.
- v = √((2.0 × 10⁷)² + (8.78 × 10⁶)²) = 2.18 × 10⁷ m/s. θ = tan⁻¹(8.78 × 10⁶/2.0 × 10⁷) = 23.7° above the horizontal.
(a) 3.5 × 10¹⁵ m/s² upward; (b) 1.1 cm upward (so the gap must be wider than that or the electron hits the plate); (c) 2.2 × 10⁷ m/s at 24° above horizontal.
As the particle leaves the plates, its horizontal velocity is unchanged and it has gained a vertical velocity v_y = at. The angle comes from tan θ = v_y/vₓ.Worked exampleExample 3: Stopping an electron
An electron moving at 3.0 × 10⁶ m/s enters a uniform field of 500 N/C that points in the same direction as its velocity. How far does it travel before stopping, and what potential difference does that correspond to?
- The force on the electron is opposite to E, so it slows down: a = eE/m = (1.60 × 10⁻¹⁹)(500)/(9.11 × 10⁻³¹) = 8.78 × 10¹³ m/s².
- Kinematics: v² = v₀² − 2ad, so d = v₀²/(2a) = (3.0 × 10⁶)²/(2 × 8.78 × 10¹³) = 0.051 m.
- Energy check: ½mv₀² = ½(9.11 × 10⁻³¹)(3.0 × 10⁶)² = 4.10 × 10⁻¹⁸ J = 25.6 eV, so ΔV = 25.6 V; and Ed = 500 × 0.0512 = 25.6 V ✓.
It stops after about 5.1 cm, having moved through about 26 V. Both methods agree.
Technologies that use these ideas (D1.1)
- Inkjet printers (continuous type): tiny ink droplets are given a charge as they form, then pass between deflection plates. The size of each droplet's charge decides how far it is deflected, which places it on the page. Uncharged droplets go straight into a gutter and are recycled.
- Cathode-ray oscilloscopes: two pairs of plates steer an electron beam left-right and up-down, drawing a graph of a voltage on the screen. (Modern oscilloscopes are digital.)
- Electrostatic precipitators: smoke from a factory chimney passes high-voltage wires that charge the particles, which are then pulled onto collecting plates. They remove most of the fine particles before the gas leaves the stack.
- Electron microscopes and X-ray tubes start with an electron gun. In an X-ray tube, electrons accelerated through tens of thousands of volts slam into a metal target and produce X-rays.
Just for funAn electron in a field of 10⁴ N/C accelerates about 10¹⁴ times faster than a dropped ball. If cars could do that, 0 to 100 km/h would take about 2 × 10⁻¹⁴ s.
Hands-on activityProjectile motion, electric style (simulation plus a kitchen model)
Simulation: search for a free online "charged particle in an electric field" simulation that lets you launch a charge between parallel plates (several physics-teaching sites offer one). If you have neither, use the kitchen model only.
- Set up a uniform field (a row of + charges above and − charges below, or parallel plates). Launch a charge horizontally and sketch its path.
- Double the field. Does the deflection at the end of the plates double?
- Double the launch speed. What happens to the deflection? (Predict first: the time between the plates halves.)
Kitchen model: tilt a large baking sheet slightly by putting a book under one long edge. Roll a marble across it parallel to the raised edge (from one short end to the other). The slope pulls the marble sideways with a constant force, like E acting on a charge.
- Mark the start and end points. Measure the sideways drift.
- Roll faster. Then add a second book (stronger "field").
What to look for: the path is a curve (parabola). Doubling the field doubles the deflection (y ∝ a). Doubling the speed makes the deflection one quarter as large, because y = ½a(L/v₀)², and time is halved. Keep marbles away from small children.
Your turn
1Find the acceleration of a proton in a uniform field of 5.0 × 10⁴ N/C. How many times larger is it than g?
a = eE/m = (1.60 × 10⁻¹⁹)(5.0 × 10⁴)/(1.67 × 10⁻²⁷) = 4.8 × 10¹² m/s², about 4.9 × 10¹¹ times g. Gravity is negligible.
2An electron starts from rest and crosses a gap with 100 V across it. Find its final speed.
v = √(2(1.60 × 10⁻¹⁹)(100)/(9.11 × 10⁻³¹)) = 5.9 × 10⁶ m/s.
3An alpha particle (q = +2e, m = 6.64 × 10⁻²⁷ kg) is accelerated from rest through 1.0 kV. Find its kinetic energy in eV and its speed.
Ek = qΔV = 2e × 1000 V = 2.0 keV = 3.2 × 10⁻¹⁶ J. v = √(2(3.2 × 10⁻¹⁶)/6.64 × 10⁻²⁷) = 3.1 × 10⁵ m/s.
4An electron gun accelerates electrons from rest through 2.0 kV. They then pass between plates 4.0 cm long and 1.0 cm apart, with 50 V across them. Find the vertical displacement at the end of the plates.
v₀ = 2.65 × 10⁷ m/s (Example 1). E = 50/0.010 = 5.0 × 10³ N/C, a = eE/m = 8.78 × 10¹⁴ m/s². t = 0.040/2.65 × 10⁷ = 1.51 × 10⁻⁹ s. y = ½at² = 1.0 × 10⁻³ m (1.0 mm).
5In the previous question, at what angle does the beam leave the plates?
v_y = at = (8.78 × 10¹⁴)(1.51 × 10⁻⁹) = 1.33 × 10⁶ m/s. tan θ = 1.33 × 10⁶/2.65 × 10⁷ = 0.050, so θ = 2.9°.
6Why is the path of a charge between the plates a parabola, but a straight line after it leaves?
Between the plates there is a constant force perpendicular to the initial velocity, as for a projectile in uniform gravity: constant vₓ and steadily growing v_y give a parabola. Outside the plates the field is (nearly) zero, so no force acts and the particle moves in a straight line (Newton's first law).
More practice: warm-up
1Write the acceleration of a charge q of mass m in a field E.
2Which way does an electron accelerate in a field pointing east?
3An electron is accelerated through 500 V. How much kinetic energy does it gain in eV?
4In a deflection problem, which velocity component does not change between the plates?
The component parallel to the plates (vₓ).
5Why can we usually ignore gravity for electrons in electric fields?
The electric force is many orders of magnitude larger than the electron's weight.
More practice: on your own
1A continuous inkjet droplet (m = 1.0 × 10⁻¹⁰ kg, q = 1.5 × 10⁻¹³ C) moves at 20 m/s between deflection plates 1.0 cm long where E = 1.2 × 10⁶ N/C. Find its sideways deflection at the end of the plates.
a = qE/m = (1.5 × 10⁻¹³)(1.2 × 10⁶)/(1.0 × 10⁻¹⁰) = 1.8 × 10³ m/s². t = 0.010/20 = 5.0 × 10⁻⁴ s. y = ½at² = 2.3 × 10⁻⁴ m (0.23 mm). It drifts farther on the way to the paper (v_y = 0.90 m/s).
2Electrons from rest cross 1.0 cm between plates with 100 V across them. How long does the crossing take?
E = 100/0.010 = 1.0 × 10⁴ N/C; a = 1.76 × 10¹⁵ m/s². d = ½at², t = √(2d/a) = √(2(0.010)/1.76 × 10¹⁵) = 3.4 × 10⁻⁹ s.
3A beam of electrons is deflected by plates. Describe how the deflection changes if (a) the plate voltage is doubled, (b) the electrons' speed is doubled.
(a) E doubles, so a doubles and the deflection doubles. (b) The time between the plates halves; y = ½at² becomes one quarter as large.
4An electron at 3.0 × 10⁶ m/s moves directly toward a negative plate. What minimum potential difference between its starting point and the plate will stop it before it arrives?
½mv² = eΔV, so ΔV = mv²/(2e) = (9.11 × 10⁻³¹)(3.0 × 10⁶)²/(2 × 1.60 × 10⁻¹⁹) = 26 V.
5Explain how an electrostatic precipitator removes smoke particles, and name one environmental benefit and one cost.
A high voltage gives the particles a charge; a field then pushes them onto oppositely charged collecting plates, which are shaken or washed to remove the dust. Benefit: much less fine particulate pollution (better air quality and health). Cost: it uses electricity, and the collected ash must be disposed of safely.
Challenge
1A particle of charge q and mass m is accelerated from rest through ΔV₁ and then enters deflecting plates of length L with field E. Show that its deflection y = EL²/(4ΔV₁) does not depend on q or m.
From the gun: ½mv₀² = qΔV₁, so v₀² = 2qΔV₁/m. Between the plates: y = ½(qE/m)(L/v₀)² = ½(qE/m)L² × m/(2qΔV₁) = EL²/(4ΔV₁). q and m cancel. (Check with Your turn question 4: (5.0 × 10³)(0.040)²/(4 × 2000) = 1.0 × 10⁻³ m ✓.) So electrons, protons and ions from the same gun follow the same path through the plates.
2Electrons from rest are accelerated through 15 kV. Find their speed using v = √(2eΔV/m) and compare with c. Should you trust this answer?
v = √(2(1.60 × 10⁻¹⁹)(1.5 × 10⁴)/9.11 × 10⁻³¹) = 7.3 × 10⁷ m/s ≈ 0.24c. At about a quarter of the speed of light, relativistic effects change the answer by a few percent, so the Newtonian result is only approximate. (See Special Relativity.)
3A proton is fired at 4.0 × 10⁵ m/s at 30° above the horizontal into a region where the field is 1.0 × 10³ N/C downward. Find the maximum height it rises and its horizontal range when it returns to the starting level.
a = eE/m = (1.60 × 10⁻¹⁹)(1.0 × 10³)/(1.67 × 10⁻²⁷) = 9.58 × 10¹⁰ m/s² downward (proton moves along E). v_y = 4.0 × 10⁵ sin 30° = 2.0 × 10⁵ m/s; vₓ = 3.46 × 10⁵ m/s. Max height = v_y²/(2a) = (2.0 × 10⁵)²/(1.916 × 10¹¹) = 0.21 m. Time of flight = 2v_y/a = 4.18 × 10⁻⁶ s. Range = vₓt = 1.4 m.
Quiz
1. A proton and an electron are released from rest in the same uniform field. Compared with the proton, the electron has
2. An electron is accelerated from rest through 400 V. If the voltage is increased to 1600 V, its final speed becomes
3. Between deflecting plates, the path of a charged particle that enters at right angles to the field is
4. The time a particle spends between deflecting plates of length L depends on
5. Which device uses deflection of charged droplets by electric fields?
6. Why is the gravitational force on an electron usually ignored in electric-field problems?
Big idea (D2.3, D2.5): a uniform electric field gives a charge a constant acceleration a = qE/m, so the familiar kinematics of projectiles applies. Energy (qΔV = ½mv²) gives speeds quickly, even in non-uniform fields.
How to help: Ask your teen to explain how a deflection problem is like throwing a ball horizontally off a cliff, and what plays the role of g. Have them predict, before calculating, what happens to the deflection if the speed doubles.
Common difficulties:
- Getting the force direction wrong for electrons (opposite to E).
- Using the plate gap instead of the plate length for the time t = L/v₀.
- Forgetting to square the time in y = ½at², or mixing up the energy and kinematics methods.
- Using the Newtonian speed formula for high voltages without noting it is only approximate.
Safety: use marbles only away from small children.