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Chapter 6

Chapter 6. Motion of Charged Particles in Electric Fields. Electric potential difference. Electric Potential Energy. A mass in a gravitational field that is free to move will accelerate down, gaining kinetic energy as it does

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Chapter 6

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  1. Chapter 6 Motion of Charged Particles in Electric Fields

  2. Electric potential difference

  3. Electric Potential Energy • A mass in a gravitational field that is free to move will accelerate down, gaining kinetic energy as it does • This energy comes from the mass having energy due to it’s position in the field, we call this gravitational potential energy

  4. Electric Potential Energy • In a similar manner, a charge free to move in an electric field will accelerate, gaining kinetic energy • Relating this to a mass in a gravitational field, we can deduce that: Any charge placed in an electric field has energy due to its position in the field • This is called electric potential energy

  5. Electric Potential Energy • Whether a charge gains or loses electric potential energy depends on the direction it moves in the field, and the type of charge (positive or negative) • If the charge accelerates freely, it loses potential energy • If the charge moves against the force acting on it by the field, it gains potential energy

  6. Electric Potential Energy • If a positive charge moves in the direction of the electric field it loses potential energy • If a positive charge moves opposite to the electric field, it gains energy

  7. Electric Potential Energy • If a negative charge moves in the direction of the electric field it gains potential energy • If a negative charge moves opposite to the electric field, it loses energy

  8. Electric Potential Difference • The electric potential difference between 2 points is the work done per unit charge in moving a positive (test) charge between the 2 points, provided all other charges remain undisturbed Potential difference = work done charge ΔV = W (units are Volts (V) or J.C-1) q

  9. Energy Changes • If a charge is moved against the force exerted by the electric field, it gains potential energy equal to the work done: W = qΔV = |ΔPE|

  10. Energy Changes • If a charge moves freely due to the force exerted by the electric field, it loses potential energy and gains kinetic energy: |ΔPE| = qΔV = |ΔKE|

  11. The Electron Volt • A joule is a large unit of energy to use when describing interactions of subatomic particles • A smaller unit of energy is more convenient – the electron volt

  12. The Electron Volt • Definition: One electron volt (1eV) is the work done when an electron moves through a potential difference of one volt OR One electron volt is the energy gained or lost by an electron moving through a potential difference of one volt

  13. The Electron Volt • When an electron moves through one volt: Work done = ΔPE = qΔV = 1.60x10-19 x 1 = 1.60x10-19J = 1eV

  14. The Electron Volt • Note: • The electron volt is not the standard unit of energy • The electron volt is interchangeable with other units of energy • The electron is a very small unit

  15. Acceleration in a uniform electric field

  16. Acceleration in an electric field • The force on a particle in an electric field can be given by: F = Eq • Using F = ma gives: ma = Eq Therefore, a = Eq m

  17. Electric field strength in a uniform electric field

  18. Formula for electric field strength • Since the electric field is uniform, the force on a charge is constant everywhere, as F = qE

  19. Formula for electric field strength • Work is done to move a charge against the electric field, given by W = ΔPE = qΔV • The work done can also be found by W = Fs = Eqd (as F = Eq)

  20. Formula for electric field strength • Equating these gives us: Eqd = qΔV • Therefore: E = ΔV d

  21. Motion of Particles parallel to an electric field

  22. Motion in a gravitational field • The acceleration due to gravity near Earth’s surface is 9.8 m.s-1, this treated as uniform • Objects in this field can have motion in the direction of the gravitational field (vertical) or undergo projectile motion (parabolic) • The object is subject to a constant force due to the uniform gravitational field

  23. Motion in a gravitational field • For motion in Earth’s gravitational field, the acceleration always had a fixed value of 9.8m.s-1 downwards • For particles in electric fields the acceleration depends on the electric field strength, the magnitude of the charge and the mass of the charged body

  24. Motion of an initially stationary particle in an electric field The particle will accelerate in a direction determined by the direction of the field and charge of the particle: • Positive charges will accelerate in the direction of the field • Negative charges will accelerate in the opposite direction to that of the field

  25. Motion of an initially stationary particle • Once the acceleration has been determined, the problem can be treated as a straight line problem and solved in one of two ways: • Kinematic – dynamic approach using equations of motion, i.e. ΔV = Ed, v = u + at, KE = ½mv2, etc. • Energetic approach using energy changes, i.e. ΔPE = qΔV = ΔKE

  26. Particles initially moving parallel or anti-parallel to the field A proton moving against an electric field behaves like a mass moving against a gravitational field, i.e. upwards: • It will reach a maximum “height” before moving back in the opposite direction • The force on the proton is constant and vectorially opposite the initial velocity

  27. Particles initially moving parallel or anti-parallel to the field A proton moving against an electric field behaves like a mass moving against a gravitational field, i.e. upwards: • The acceleration of the proton is constant and vectorially opposite the initial velocity • The proton’s initial motion is described as being anti-parallel (opposite) to the field

  28. Particles initially moving parallel or anti-parallel to the field If the charged particle is initially moving in the same direction as the force acting on it, then: • Its motion is said to be parallel to the field • Its motion is equivalent to throwing a mass vertically downwards towards Earth, where the mass moves in the direction of the force it experiences due to gravity

  29. Example problems Example 4 – Particle initially stationary Two large flat parallel plates have a potential difference of 20V applied to them. The plates are 2.0cm apart. A proton, initially adjacent to the positive plate, accelerates towards the negative plate. Find: • The force on the proton • The acceleration of the proton • The time the proton takes to move between the plates • The final velocity of the proton (just before it hits) • The final kinetic energy of the proton The worked solution to this problem is on p. 132 of the textbook (2nd edition)

  30. Example problems Example 5 – Particle initially moving parallel to the field Two large flat parallel plates 2cm apart have a potential difference of 20V between them. A proton is fired upwards towards the positive plate through a hole in the negative plate. The Initial speed of the proton is 10,000m.s-1. Calculate: • The force on the proton • The acceleration of the proton • The time the proton takes to come to rest (time of flight) • The distance that the proton moves upwards in the space between the plates The worked solution to this problem is on p. 134 of the textbook (2nd edition)

  31. Projectile Motion of Particles electric fields

  32. Motion of a particle entering an electric field at right angles • A charged particle entering an electric field at right angles to the direction of the field is equivalent to the case of horizontal projection in a gravitational field

  33. Motion of a particle entering an electric field at right angles • Since the particle’s velocity perpendicular to the field remains constant, the time taken to traverse the field can be found using the relation: time = length of plates initial velocity

  34. Motion of a particle entering an electric field at right angles • For the example shown, the vertical deflection can be determined using the time taken to traverse the field and the vertical acceleration of the particle by using the formula s = ut + ½at2 • Since the initial vertical velocity is zero, this relation becomes s = ½at2

  35. Example problem Example 6 – Particle initially moving perpendicular to the electric field Two large parallel plates 15cm long and 5cm apart have a potential difference of 600V between them. In an electron gun, electrons are accelerated through a potential difference of 2000V. On leaving the electron gun, these electrons enter the field between the plates at right angles to the field. Find the: • Time taken for the electrons to traverse this field • Deflection of the electrons on leaving the field • The final velocity of the electrons on leaving the field The worked solution to this problem is on p. 137 of the textbook (2nd edition)

  36. Class problems Conceptual questions: 1-5, 9-12 Descriptive questions: 1-3, 5, 8 Analytical questions: 1-9, 12, 14, 17

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