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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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**Chapter 6**Motion of Charged Particles in Electric Fields**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**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**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**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**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**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**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|**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|**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**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**The Electron Volt**• When an electron moves through one volt: Work done = ΔPE = qΔV = 1.60x10-19 x 1 = 1.60x10-19J = 1eV**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**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**Formula for electric field strength**• Since the electric field is uniform, the force on a charge is constant everywhere, as F = qE**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)**Formula for electric field strength**• Equating these gives us: Eqd = qΔV • Therefore: E = ΔV d**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**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**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**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**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**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**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**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)**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)**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**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**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**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)**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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