Charged Particles in Electric and Magnetic Fields

In short: A charge in an electric field feels a force $F = q E$ , so it speeds up or curves like a projectile. A moving charge in a magnetic field feels a force $F = q v B sin θ$ at right angles to its motion, so it moves in a circle at a steady speed.

Electric fields

An electric field is a region where a charge feels a force. The field strength $E$ is the force on each coulomb of charge:

$F = q E$

A positive charge is pushed along the field lines.
A negative charge (like an electron) is pushed against the field lines.
Field lines point from positive to negative.

Parallel plates

Two flat metal plates with a voltage $V$ across them, a distance $d$ apart, make a uniform field between them (the same strength everywhere):

$E = \frac{V}{d}$

The unit of $E$ is V/m, which is the same as N/C.

Speeding up a charge: work and energy

When a charge $q$ moves through a potential difference $V$ , the field does work on it:

$W = q V = Δ K$

All this work becomes kinetic energy. An electron accelerated from rest through 100 V gains $1.602 \times 1 0^{- 19} \times 100 = 1.6 \times 1 0^{- 17}$ J.

Physicists often use the electron volt (eV): the energy one electron gains through 1 V. $1 eV = 1.602 \times 1 0^{- 19}$ J.

Bending in an electric field: a parabola

A charge fired sideways into the field between plates feels a constant force towards one plate. Its motion is just like a projectile:

Along the plates: no force, so constant velocity.
Across the plates: constant force, so constant acceleration $a = \frac{q E}{m}$ .

The path is a parabola. This is the same maths as a ball thrown horizontally, with $q E / m$ in place of $g$ .

Magnetic fields and moving charges

A magnetic field only pushes on charges that are moving. For a charge $q$ moving at speed $v$ across a field $B$ :

$F = q v B sin θ$

$B$ is in tesla (T).
$θ$ is the angle between the velocity and the field. If the charge moves along the field, $sin 0° = 0$ and there is no force.
The force is always at right angles to both the velocity and the field.

Finding the direction: right-hand palm rule

For a positive charge: point your fingers along the field $B$ , your thumb along the velocity $v$ . Your palm pushes in the direction of the force.

For a negative charge, the force is the opposite way.

Circles in a magnetic field

Because the magnetic force is always at right angles to the motion, it changes direction but never speed. That is exactly a centripetal force, so the charge moves in a circle:

$q v B = \frac{m v ^{2}}{r} \Rightarrow r = \frac{m v}{q B}$

Faster or heavier particles make bigger circles.
A stronger field or a bigger charge makes smaller circles.
The magnetic force does no work, so the kinetic energy does not change.

Comparing the three fields

	Gravitational field	Electric field	Magnetic field
Acts on	mass	charge	moving charge
Force	$m g$	$q E$	$q v B sin θ$
Path when fired across a uniform field	parabola	parabola	circle
Changes speed?	yes	yes	no

Charges in Fields

Electric field

F = qE
Plates: E = V / d
+ along field, − against

Energy

W = qV = ΔK
1 eV = 1.602 × 10⁻¹⁹ J

Across plates

Like a projectile
a = qE / m
Parabola

Magnetic field

F = qvB sin θ
Only moving charges
Right-hand palm rule

Circles

r = mv / qB
No work done
Speed stays the same

Year 11The right-hand palm rule in practice

Diagrams show fields going into the page with crosses (×), like the tail of an arrow, and out of the page with dots (•), like its tip.

An electron moving right through a field into the page: for a positive charge, fingers into the page and thumb right would push the palm up. The electron is negative, so it is pushed down.

A charge moving parallel to the field feels no magnetic force, so it carries on in a straight line. A straight path through a region does not prove there is no magnetic field there.

Speed from a voltage. For a charge starting from rest, $q V = \frac{1}{2} m v^{2}$ , so $v = \frac{2 q V}{m}$ . An electron through 200 V reaches $\frac{2 \times 1.602 \times 1 0 ^{- 19} \times 200}{9.109 \times 1 0 ^{- 31}} \approx 8.4 \times 1 0^{6}$ m/s.

Deflection between plates. The time between the plates is $t = \frac{L}{v}$ (length over speed along the plates). Across the plates, $a = \frac{q E}{m}$ and the sideways distance is $y = \frac{1}{2} a t^{2}$ . To get through without hitting a plate, $y$ must be less than the gap available, which sets a minimum speed.

Hanging still. A charged oil drop floats when the electric force balances its weight: $q E = m g$ , so $m = \frac{q V}{g d}$ . This is how Millikan measured the charge on the electron.

Released from rest between plates. With $a = \frac{q E}{m}$ and distance $s$ to the plate, the time is $t = \frac{2 s}{a}$ .

Year 12Crossed fields, time in a field and comparing particles

Velocity selector. Put an electric field and a magnetic field at right angles. A charge goes straight through when the two forces balance:

$q E = q v B \Rightarrow v = \frac{E}{B}$

Only particles with exactly this speed pass undeflected, whatever their charge or mass. Thomson used this idea to measure the speed of cathode rays (see Structure of the Atom).

Time for one circle. $T = \frac{2 π r}{v} = \frac{2 π m}{q B}$ . It does not depend on speed. A particle in a magnetic field for half a circle spends $\frac{π m}{q B}$ inside it.

Comparing particles. An alpha particle has 4 times the mass and 2 times the charge of a proton. At the same speed in the same field, $r = \frac{m v}{q B}$ gives an alpha radius $\frac{4}{2} = 2$ times the proton's. If instead both were accelerated through the same voltage, use $K = q V$ first to find each speed.

Work done across plates. If a charge enters between plates and leaves at a different position across the gap, the work done is $q E \times$ (distance moved across the field), not $q V$ for the whole gap.

Identifying a particle. From $r = \frac{m v}{q B}$ , the ratio $\frac{q}{m} = \frac{v}{r B}$ . Compare it with known particles: about $9.6 \times 1 0^{7}$ C/kg is a proton, and $1.76 \times 1 0^{11}$ C/kg is an electron.

Same voltage, then a field. After acceleration through $V$ , $v = \frac{2 q V}{m}$ , so $r = \frac{m v}{q B} = \frac{1}{B} \frac{2 mV}{q}$ . An alpha particle has $\frac{m}{q}$ twice a proton's, so its radius is $2$ times larger. If a second particle in a field $2 B$ at speed $2 v$ follows the same path, it has the same $\frac{m}{q}$ as the first.

Quick summary

Electric field: $F = q E$ . Between parallel plates $E = \frac{V}{d}$ .
Moving through a voltage: $W = q V$ becomes kinetic energy.
Fired across plates, a charge follows a parabola, like a projectile with $a = \frac{q E}{m}$ .
Magnetic force $F = q v B sin θ$ acts only on moving charges, at right angles to $v$ and $B$ .
In a magnetic field a charge moves in a circle of radius $r = \frac{m v}{q B}$ , at constant speed.

Quiz

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The force on a charge in an electric field is given by:
1. A F = qvB
2. B F = qE
3. C F = mg
4. D F = V/d
Show answer
B: F = qE
Electric force = charge × field strength.
An electron is placed in an electric field that points to the right. The force on it is:
1. A To the right
2. B To the left
3. C Up
4. D Zero
Show answer
B: To the left
Electrons are negative, so they are pushed against the field direction.
Two parallel plates 0.050 m apart have 200 V across them. The field between them is:
1. A 10 V/m
2. B 4000 V/m
3. C 250 V/m
4. D 0.00025 V/m
Show answer
B: 4000 V/m
E = V ÷ d = 200 ÷ 0.050 = 4000 V/m.