Lesson 6 of 6Gravitational, Electric, and Magnetic Fields

Unit 3 · Lesson 6 of 6

Charged Particles in Uniform Fields: Deflection and e/m

How an electron gun speeds charges up, why a charge crossing between parallel plates follows a parabola just like a thrown ball, and how J. J. Thomson combined electric and magnetic fields to measure the charge-to-mass ratio of the electron.

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Key things to remember

1Electron gun: $q Δ V = \frac{1}{2} m v^{2}$ , so $v = 2 q Δ V / m$ for a charge starting from rest.
2Between parallel plates a charge has constant velocity along the plates and constant acceleration $a = qE / m = q Δ V / (m d)$ across them, so its path is a parabola.
3Treat deflection like a horizontal projectile: $t = L / v_{x}$ , $y = \frac{1}{2} a t^{2}$ , and $tan θ = v_{y} / v_{x}$ on exit.
4Velocity selector: crossed fields pass undeflected only the particles with $v = E / B$ .
5Thomson's method: $\frac{e}{m} = \frac{v}{B r} = \frac{E}{B ^{2} r}$ ; the accepted value is $1.76 \times 1 0^{11} C/kg$ .
6Electric and gravitational fields do work and bend paths into parabolas; a magnetic field does no work and bends paths into circles.

In 1897 nobody knew what the glowing 'cathode rays' inside an evacuated glass tube actually were. Some physicists thought they were a kind of light; others thought they were particles. J. J. Thomson settled it by steering the rays with electric and magnetic fields and measuring how sharply they bent. The answer was a particle far lighter than any atom: the electron. The same steering tricks went on to power television picture tubes, oscilloscopes, industrial printers and the ion engines on deep-space probes. Everything in this lesson uses physics you already have: the energy from Electric Potential Energy and Potential, the magnetic force from Magnetic Fields and Forces, and the projectile motion from Unit 1.

Accelerating charges: the electron gun

An electron gun starts with a heated metal filament, the cathode. Heat gives some electrons enough energy to leave the metal, a process called thermionic emission. A nearby anode with a hole in it is held at a high positive potential, so the electrons accelerate toward it, and many shoot straight through the hole as a narrow beam. Because the electric force is conservative, you do not need to know the distance or the shape of the field: the electric potential energy lost, $q Δ V$ , all becomes kinetic energy. For electrons, this simple energy equation works well up to a few tens of kilovolts. Beyond that the speed becomes a large fraction of $c$ , and the relativistic energy of Unit 5 is needed instead.

q Δ V = \frac{1}{2} m v^{2} \Rightarrow v = \frac{2 q Δ V}{m} e = 1.60 \times 1 0^{- 19} C m_{e} = 9.11 \times 1 0^{- 31} kg

For a charge starting from rest. Use the magnitude of the charge and the size of the accelerating potential difference.

Sideways deflection: projectile motion with electricity

Fig 1An electron enters the gap between two charged plates moving horizontally. The constant upward force bends its path into a parabola; after leaving the plates it travels in a straight line.

Now send the beam horizontally between two parallel plates with a potential difference across them. Inside, the field $E = Δ V / d$ is uniform and points straight across the gap, so the particle feels a constant force at right angles to its initial velocity. That is exactly the situation of a ball thrown horizontally off a cliff. Nothing acts along the plates, so the horizontal velocity stays constant and the time between the plates is $t = L / v$ . Across the gap there is a constant acceleration $a = qE / m$ , so the sideways displacement grows as $\frac{1}{2} a t^{2}$ and the path is a parabola. Once the particle leaves the plates the force stops, and it travels in a straight line at the angle it had on exit. Gravity is ignored, and rightly: for an electron in a typical field the electric acceleration is around $1 0^{14}$ times $g$ .

t = \frac{L}{v _{x}} a_{y} = \frac{qE}{m} = \frac{q Δ V}{m d} y = \frac{1}{2} a_{y} t^{2} tan θ = \frac{v _{y}}{v _{x}} = \frac{a _{y} t}{v _{x}}

L is the length of the plates, d their separation, and theta the angle of the path when it leaves the plates.

ABall thrown horizontally

Constant horizontal velocity $v_{x}$
Constant downward acceleration $g$ from gravity
Acceleration is the same for every mass
Parabolic path while gravity acts

BCharge between parallel plates

Constant velocity $v_{x}$ along the plates
Constant acceleration $qE / m$ across the gap
Acceleration depends on the charge-to-mass ratio $q / m$
Parabola between the plates, then a straight line

A neat shortcut for gun-plus-plates problems

If a particle is accelerated from rest through

Δ V_{a cc}

and then crosses plates with field

E

, substituting

v^{2} = 2 q Δ V_{a cc} / m

into

y = \frac{1}{2} (qE / m) (L / v)^{2}

gives

y = \frac{E L ^{2}}{4Δ V _{a cc}}

. The charge and the mass cancel. An electron and a negative ion fired from the same gun follow the same path, which is why a cathode-ray tube's image does not depend on exactly what the beam is made of, only on the two voltages.

Thomson's e/m experiment

Thomson could not measure the electron's charge or mass separately, but he could measure their ratio. First he applied an electric field and a magnetic field at right angles, both across the beam, arranged so the electric force $e E$ and the magnetic force $e v B$ pointed in opposite directions. He adjusted them until the beam went straight. This is a velocity selector: balance means $e E = e v B$ , so the beam's speed is $v = E / B$ , whatever the particles are. Then he switched off the electric field. The magnetic force alone bent the beam into a circular arc of radius $r = m v / (e B)$ , and measuring that radius gave the ratio. Thomson's own values were rough, but they were already about a thousand times larger than the ratio for a hydrogen ion; the modern value, $1.76 \times 1 0^{11} C/kg$ , is about 1840 times larger. Either the particle carried an enormous charge or it had a tiny mass. When Millikan later measured the charge $e$ directly, the mass followed: $m_{e} = e / (e / m) = 9.11 \times 1 0^{- 31} kg .$

e E = e v B \Rightarrow v = \frac{E}{B} r = \frac{m v}{e B} \Rightarrow \frac{e}{m} = \frac{v}{B r} = \frac{E}{B ^{2} r}

The velocity selector fixes v; the radius in the magnetic field alone then gives the charge-to-mass ratio.

Technologies and a comparison of the three fields

Technology	How fields steer the charges
Analog oscilloscope (cathode-ray tube)	An electron gun makes the beam; two pairs of deflecting plates move the spot sideways and up and down in proportion to the voltages being measured.
Older televisions and computer monitors	Also cathode-ray tubes, but most used magnetic coils to sweep the beam across the screen line by line.
Continuous inkjet printers (industrial coding)	Tiny ink droplets are given an electric charge, then deflected by charged plates onto the right spot; uncharged drops fly straight into a gutter and are recycled.
Ion thrusters on spacecraft	Xenon ions are accelerated through a potential difference of the order of a kilovolt, leaving at tens of kilometres per second, many times faster than chemical rocket exhaust.
Mass spectrometers	A velocity selector picks one speed, then a magnetic field separates ions by their charge-to-mass ratio.

Uniform field	Force on the particle	Path when entering at right angles to the field	Does the field change the speed?
Gravitational ( $g$ )	$m g$ , along the field, same acceleration for every mass	Parabola	Yes: gravity does work
Electric ( $E$ )	$qE$ , along the field (opposite it for negative charges)	Parabola	Yes: the field does work $q Δ V$
Magnetic ( $B$ )	$q v B$ , perpendicular to both $v$ and $B$	Circle of radius $m v / (qB)$	No: the force is always perpendicular to the motion, so it does no work

Motion in uniform fields. Gravitational and electric fields act alike; the magnetic field is the odd one out.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

An electron starts from rest and is accelerated through a potential difference of

2.0 kV

in an electron gun. Find its kinetic energy and its speed.

Example 2

The electron from the previous example (

v = 2.65 \times 1 0^{7} m/s

) enters midway between two horizontal plates

5.0 cm

long and

1.0 cm

apart, with a potential difference of

150 V

across them. Find its sideways deflection on leaving the plates and the angle of its path.

Example 3

In a Thomson-style experiment, a beam passes undeflected through crossed fields of

E = 3.0 \times 1 0^{4} N/C

and

B = 2.0 \times 1 0^{- 3} T

. With the electric field switched off, the beam bends in an arc of radius

4.27 cm

. Find the speed of the particles and their charge-to-mass ratio.

Example 4

An ion thruster accelerates singly charged xenon ions (

m = 2.18 \times 1 0^{- 25} kg

) from rest through a potential difference of

1.10 kV

. Find the exhaust speed and compare it with a chemical rocket's exhaust speed of about

4.5 km/s

Where marks get lost

Using the plate separation $d$ as the distance travelled along the plates. The time between the plates comes from the plate length $L$ and the horizontal speed; $d$ only sets the field $E = Δ V / d$ .
Letting the horizontal speed change between the plates. The electric force is perpendicular to the plates, so the component along them stays constant, exactly as $v_{x}$ does for a projectile.
Continuing the parabola after the particle leaves the plates. Outside the field there is no force, so the path is a straight line at the exit angle.
Writing $e / m = E / (B r)$ instead of $E / (B^{2} r)$ : the field $B$ appears once in $v = E / B$ and again in $r = m v / (e B)$ .
Expecting a magnetic field to speed a particle up or slow it down. The magnetic force is always perpendicular to the velocity, so it changes only the direction.

Flashcards 9

Electron gun

Thermionic emission

Acceleration between parallel plates

Path of a charge between plates

+ 5 more

Practice

An electron starts from rest and accelerates through a potential difference of

500 V

. What is its final speed, in units of

1 0^{7} m/s

? Give three significant figures.

× 10⁷ m/s

What potential difference is needed to accelerate an electron from rest to

3.0 \times 1 0^{7} m/s

, in kilovolts? Give three significant figures.

An electron moving at

2.0 \times 1 0^{7} m/s

enters a uniform field of

2.0 \times 1 0^{4} N/C

between plates

4.0 cm

long, moving parallel to the plates. How far is it deflected sideways by the time it leaves the plates, in millimetres? Give three significant figures.

A velocity selector uses

E = 4.5 \times 1 0^{4} N/C

and

B = 3.0 \times 1 0^{- 3} T

. What speed do the undeflected particles have, in km/s?

km/s

A beam passes undeflected through

E = 2.4 \times 1 0^{4} N/C

and

B = 1.5 \times 1 0^{- 3} T

. With

E

switched off, the beam curves with radius

6.07 cm

. What is the particles' charge-to-mass ratio, in units of

1 0^{11} C/kg

? Give three significant figures.

× 10¹¹ C/kg

An electron enters the region between two charged parallel plates moving parallel to them. What is the shape of its path while it is between the plates?

A proton and an electron enter the same pair of deflecting plates with the same speed. How do their deflections compare?

In a velocity selector, a particle moves faster than

E / B

. What happens to it?

J. J. Thomson's 1897 cathode-ray experiment measured what property of the electron?

Put the steps of Thomson's e/m measurement in order.

1Calculate e/m = E/(B²r)
2Switch off the electric field and measure the radius of the curved beam
3Adjust the fields until the beam passes straight through, so v = E/B
4Accelerate electrons from a hot cathode into a beam
5Apply crossed electric and magnetic fields

Match each technology to how it uses fields to control charged particles.

Oscilloscope

Continuous inkjet printer

Ion thruster

Mass spectrometer

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