30  Charged Particles in Fields

30.1 Syllabus inquiry question

  • How do charged particles behave in electric and magnetic fields?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 1:

The motion of charged particles in electric and magnetic fields underlies almost all of modern technology—from cathode ray tubes to particle accelerators. The beauty is that two simple force laws explain it all.

30.2 Learning Objectives

  • Apply electric field relationships: \(E = V/d\), \(F = qE\)
  • Calculate work and energy changes: \(W = qV\), \(K = \frac{1}{2}mv^2\)
  • Analyse magnetic force on moving charges: \(F = qvB\sin\theta\)
  • Compare trajectories in electric vs magnetic fields
  • Derive circular motion radius in magnetic fields

30.3 Content

30.3.1 Uniform Electric Field Between Parallel Plates

A uniform electric field exists between parallel plates with potential difference \(V\) and separation \(d\):

\[E = \frac{V}{d}\]

where: - \(E\) = electric field strength (N/C or V/m) - \(V\) = potential difference (V) - \(d\) = plate separation (m)

Key Features
  • Field is uniform (constant magnitude and direction)
  • Field lines are parallel, equally spaced
  • Force on a charge is constant throughout the field

30.3.2 Interactive: Charged Particle in Electric Field

Visualise the motion of charged particles between parallel plates:

30.3.3 Force and Acceleration in Electric Fields

A charged particle experiences a force:

\[F = qE\]

By Newton’s second law:

\[a = \frac{qE}{m}\]

Trajectory Comparison

In a uniform electric field, charged particles follow parabolic paths—identical to projectile motion in a gravitational field. The key difference: \(a = qE/m\) instead of \(g\).

30.3.4 Work and Energy

Work done on a charge moving through potential difference:

\[W = qV = qEd\]

This equals the change in kinetic energy:

\[qV = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2\]

For a particle starting from rest:

\[v = \sqrt{\frac{2qV}{m}}\]

30.3.5 Interactive: Energy Gain Across Plates

30.3.6 Magnetic Force on Moving Charges

A charged particle moving through a magnetic field experiences a force:

\[F = qvB\sin\theta\]

where: - \(F\) = magnetic force (N) - \(q\) = charge (C) - \(v\) = velocity (m/s) - \(B\) = magnetic field strength (T) - \(\theta\) = angle between \(\vec{v}\) and \(\vec{B}\)

Critical Understanding
  • The force is perpendicular to both velocity and field
  • No force when \(v \parallel B\) (θ = 0° or 180°)
  • Maximum force when \(v \perp B\) (θ = 90°)
  • The magnetic force does no work (always perpendicular to motion)

30.3.7 Interactive: Charged Particle in Magnetic Field

30.3.8 Circular Motion in Magnetic Fields

For motion perpendicular to \(B\), the magnetic force provides centripetal force:

\[qvB = \frac{mv^2}{r}\]

Solving for radius:

\[r = \frac{mv}{qB}\]

Key Relationships
  • Larger mass → larger radius (heavier particles curve less)
  • Larger velocity → larger radius (faster particles curve less)
  • Larger field → smaller radius (stronger field bends more)
  • Radius is independent of entry point (depends only on \(m\), \(v\), \(q\), \(B\))

30.3.9 Comparing Electric and Magnetic Fields

Property Electric Field Magnetic Field
Force direction Along field (+ charge) Perpendicular to v and B
Force on stationary charge Yes (\(F = qE\)) No
Work done Yes (\(W = qV\)) No
Trajectory Parabolic Circular (if v ⊥ B)
Speed change Yes No

30.4 Worked Examples

30.4.1 Example 1: Electric field between plates

Parallel plates are separated by 0.015 m with a potential difference of 600 V. Find the electric field strength.

Solution:

  1. Use \(E = V/d\)

  2. \(E = 600/0.015\)

  3. \(E = 4.0 \times 10^4\ \text{N/C}\)

30.4.2 Example 2: Electron acceleration

An electron starts from rest and accelerates through 1200 V. Find its final speed. (\(m_e = 9.11 \times 10^{-31}\) kg, \(q_e = 1.6 \times 10^{-19}\) C)

Solution:

  1. Use energy conservation: \(qV = \frac{1}{2}mv^2\)

  2. \(v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.11 \times 10^{-31}}}\)

  3. \(v = \sqrt{4.22 \times 10^{14}} = 2.05 \times 10^7\ \text{m/s}\)

30.4.3 Example 3: Magnetic force on a proton

A proton moves at \(2.0 \times 10^6\) m/s perpendicular to a 0.30 T magnetic field. Find the magnetic force. (\(q_p = 1.6 \times 10^{-19}\) C)

Solution:

  1. Use \(F = qvB\sin\theta\) with \(\theta = 90°\)

  2. \(F = 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times 0.30 \times 1\)

  3. \(F = 9.6 \times 10^{-14}\ \text{N}\)

30.4.4 Example 4: Circular path radius

Find the radius of the proton’s circular path in Example 3. (\(m_p = 1.67 \times 10^{-27}\) kg)

Solution:

  1. Use \(r = mv/(qB)\)

  2. \(r = \frac{1.67 \times 10^{-27} \times 2.0 \times 10^6}{1.6 \times 10^{-19} \times 0.30}\)

  3. \(r = \frac{3.34 \times 10^{-21}}{4.8 \times 10^{-20}} = 0.070\ \text{m}\) (7.0 cm)

30.4.5 Example 5: Deflection in electric field

An electron enters a uniform electric field (\(E = 3.0 \times 10^4\) N/C) with horizontal velocity \(5.0 \times 10^6\) m/s. The field region is 0.10 m long. Find the vertical deflection.

Solution:

  1. Time in field: \(t = d/v_x = 0.10/(5.0 \times 10^6) = 2.0 \times 10^{-8}\) s

  2. Vertical acceleration: \(a = qE/m = (1.6 \times 10^{-19} \times 3.0 \times 10^4)/(9.11 \times 10^{-31}) = 5.27 \times 10^{15}\ \text{m/s}^2\)

  3. Vertical deflection: \(y = \frac{1}{2}at^2 = 0.5 \times 5.27 \times 10^{15} \times (2.0 \times 10^{-8})^2 = 0.0011\ \text{m}\) (1.1 mm)

30.5 Common Misconceptions

Common Misconceptions
  • Misconception: Magnetic force can speed up a charged particle. Correction: Magnetic force is always perpendicular to velocity, so it changes direction but not speed. No work is done.

  • Misconception: Electric field strength depends on the test charge. Correction: Field strength \(E = V/d\) is independent of any test charge. The force depends on charge, but the field does not.

  • Misconception: Force on an electron in an electric field is opposite to the field. Correction: This is correct! Electrons (negative charge) accelerate opposite to the field direction.

  • Misconception: Circular motion in a magnetic field requires continuous energy input. Correction: The magnetic force provides centripetal force but does no work. Kinetic energy stays constant.

  • Misconception: A stationary charge experiences magnetic force in a magnetic field. Correction: Magnetic force requires motion: \(F = qvB\sin\theta\). If \(v = 0\), then \(F = 0\).

30.6 Practice Questions

30.6.1 Easy (2 marks)

Calculate the electric field between plates separated by 0.020 m with a potential difference of 400 V.

  • Use \(E = V/d\) correctly (1)
  • Correct answer: \(E = 400/0.020 = 20000\) N/C = \(2.0 \times 10^4\) N/C (1)

Answer: \(2.0 \times 10^4\) N/C

30.6.2 Medium (4 marks)

An electron is accelerated from rest through a potential difference of 500 V. Calculate its final speed and the kinetic energy gained.

  • Use \(qV = \frac{1}{2}mv^2\) correctly (1)
  • \(v = \sqrt{2 \times 1.6 \times 10^{-19} \times 500 / 9.11 \times 10^{-31}} = 1.33 \times 10^7\) m/s (1)
  • Kinetic energy \(= qV = 1.6 \times 10^{-19} \times 500 = 8.0 \times 10^{-17}\) J (1)
  • Or 500 eV (1)

Answer: \(v = 1.3 \times 10^7\) m/s, \(K = 8.0 \times 10^{-17}\) J (or 500 eV)

30.6.3 Hard (5 marks)

A proton enters a magnetic field of 0.25 T perpendicular to the field with velocity \(3.0 \times 10^6\) m/s. Calculate the radius of its circular path and the time for one complete revolution.

  • Use \(r = mv/(qB)\) correctly (1)
  • \(r = (1.67 \times 10^{-27} \times 3.0 \times 10^6)/(1.6 \times 10^{-19} \times 0.25) = 0.125\) m (1)
  • Period formula: \(T = 2\pi r/v\) (1)
  • \(T = 2\pi \times 0.125/(3.0 \times 10^6) = 2.6 \times 10^{-7}\) s (1)
  • Alternatively: \(T = 2\pi m/(qB)\) gives same answer (1)

Answer: \(r = 0.125\) m (12.5 cm), \(T = 2.6 \times 10^{-7}\) s (260 ns)

30.7 Multiple Choice Questions

Test your understanding with these interactive questions:

30.8 Summary

Key Takeaways
  • Electric field between plates: \(E = V/d\)
  • Force on charge in E-field: \(F = qE\), acceleration \(a = qE/m\)
  • Energy gained: \(qV = \frac{1}{2}mv^2\) (particle from rest)
  • Magnetic force: \(F = qvB\sin\theta\) (perpendicular to both v and B)
  • Circular motion radius: \(r = mv/(qB)\)
  • Electric fields: do work, change speed, parabolic paths
  • Magnetic fields: do no work, constant speed, circular paths

30.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: