# Charged Particles in Electric and Magnetic Fields > How electric fields speed charges up and bend their paths into parabolas, and how magnetic fields bend them into circles. From HSC Physics on WhizzNotes, free study notes for NSW students: https://whizznotes.com/physics/charged-particles-in-fields/. Updated 2 November 2026. Year 11 and Year 12 boxes go deeper than the Year 10 text around them. ### Story time: The glowing screen Before flat screens, every television had a heavy glass tube at the back. At the narrow end, a hot wire boiled off electrons. A high voltage shot them down the tube at millions of metres per second, and they lit up a dot where they hit the screen. But a screen needs a picture, not a single dot. So coils around the tube made magnetic fields that swept the beam left and right, top to bottom, hundreds of times a second. Your eye saw a moving picture. Every one of those televisions used the two ideas in this chapter: electric fields push charges, and magnetic fields bend moving charges. **In short:** A charge in an electric field feels a force $F = qE$, so it speeds up or curves like a projectile. A moving charge in a magnetic field feels a force $F = qvB\sin\theta$ 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: $$F = qE$$ - 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. ### Worked example: Field between plates Plates 2.0 cm apart have 500 V across them. $E = 500 \div 0.020 = 25\,000$ V/m. An electron between them feels a force $F = qE = 1.602 \times 10^{-19} \times 25\,000 = 4.0 \times 10^{-15}$ N. ## Speeding up a charge: work and energy When a charge $q$ moves through a potential difference $V$, the field does work on it: $$W = qV = \Delta K$$ All this work becomes kinetic energy. An electron accelerated from rest through 100 V gains $1.602 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}$ J. Physicists often use the **electron volt** (eV): the energy one electron gains through 1 V. $1 \text{ eV} = 1.602 \times 10^{-19}$ 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 $a = \dfrac{qE}{m}$. The path is a **parabola**. This is the same maths as a ball thrown horizontally, with $qE/m$ 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 $B$: $$F = qvB\sin\theta$$ - $B$ is in **tesla** (T). - $\theta$ 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**: $$qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB}$$ - 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. ### Worked example: A proton in a field A proton ($m = 1.673 \times 10^{-27}$ kg) moves at $2.0 \times 10^6$ m/s at right angles to a 0.50 T field. $$r = \frac{1.673 \times 10^{-27} \times 2.0 \times 10^6}{1.602 \times 10^{-19} \times 0.50} = 0.042 \text{ m}$$ ## Comparing the three fields | | Gravitational field | Electric field | Magnetic field | |---|---|---|---| | Acts on | mass | charge | **moving** charge | | Force | $mg$ | $qE$ | $qvB\sin\theta$ | | Path when fired across a uniform field | parabola | parabola | circle | | Changes speed? | yes | yes | **no** | **Mind map** ```text 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 11: The 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, $qV = \frac{1}{2}mv^2$, so $v = \sqrt{\dfrac{2qV}{m}}$. An electron through 200 V reaches $\sqrt{\dfrac{2 \times 1.602 \times 10^{-19} \times 200}{9.109 \times 10^{-31}}} \approx 8.4 \times 10^6$ m/s. **Deflection between plates.** The time between the plates is $t = \dfrac{L}{v}$ (length over speed along the plates). Across the plates, $a = \dfrac{qE}{m}$ and the sideways distance is $y = \frac{1}{2}at^2$. 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: $qE = mg$, so $m = \dfrac{qV}{gd}$. This is how Millikan measured the charge on the electron. **Released from rest between plates.** With $a = \dfrac{qE}{m}$ and distance $s$ to the plate, the time is $t = \sqrt{\dfrac{2s}{a}}$. ### Year 12: Crossed 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: $qE = qvB \quad\Rightarrow\quad v = \frac{E}{B}$ 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.** $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$. It does not depend on speed. A particle in a magnetic field for half a circle spends $\dfrac{\pi m}{qB}$ 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, $r = \dfrac{mv}{qB}$ gives an alpha radius $\dfrac{4}{2} = 2$ times the proton's. If instead both were accelerated through the same voltage, use $K = qV$ 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 $qE \times$ (distance moved across the field), not $qV$ for the whole gap. **Identifying a particle.** From $r = \dfrac{mv}{qB}$, the ratio $\dfrac{q}{m} = \dfrac{v}{rB}$. Compare it with known particles: about $9.6 \times 10^7$ C/kg is a proton, and $1.76 \times 10^{11}$ C/kg is an electron. **Same voltage, then a field.** After acceleration through $V$, $v = \sqrt{\dfrac{2qV}{m}}$, so $r = \dfrac{mv}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$. An alpha particle has $\frac{m}{q}$ twice a proton's, so its radius is $\sqrt{2}$ times larger. If a second particle in a field $2B$ at speed $2v$ follows the same path, it has the same $\dfrac{m}{q}$ as the first. ## Quick summary - Electric field: $F = qE$. Between parallel plates $E = \dfrac{V}{d}$. - Moving through a voltage: $W = qV$ becomes kinetic energy. - Fired across plates, a charge follows a parabola, like a projectile with $a = \dfrac{qE}{m}$. - Magnetic force $F = qvB\sin\theta$ acts only on moving charges, at right angles to $v$ and $B$. - In a magnetic field a charge moves in a circle of radius $r = \dfrac{mv}{qB}$, at constant speed. ## Quiz Try these questions from the chapter quiz. 1. The force on a charge in an electric field is given by: - A. F = qvB - B. F = qE - C. F = mg - D. F = V/d - Answer: B. F = qE. Electric force = charge × field strength. 2. An electron is placed in an electric field that points to the right. The force on it is: - A. To the right - B. To the left - C. Up - D. Zero - Answer: B. To the left. Electrons are negative, so they are pushed against the field direction. 3. Two parallel plates 0.050 m apart have 200 V across them. The field between them is: - A. 10 V/m - B. 4000 V/m - C. 250 V/m - D. 0.00025 V/m - Answer: B. 4000 V/m. E = V ÷ d = 200 ÷ 0.050 = 4000 V/m. The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/physics/charged-particles-in-fields/#quiz