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02/10/2016 02/10/2016 Forces on charged particles Read Summary Notes, page 13 and 14. Fill in the worked examples on pages 15 and 16. Page 15 Definition.

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Presentation on theme: "02/10/2016 02/10/2016 Forces on charged particles Read Summary Notes, page 13 and 14. Fill in the worked examples on pages 15 and 16. Page 15 Definition."— Presentation transcript:

1 02/10/2016 02/10/2016 Forces on charged particles Read Summary Notes, page 13 and 14. Fill in the worked examples on pages 15 and 16. Page 15 Definition of potential difference and the volt Potential difference (p.d.) is defined to be a measure of the work done in moving one coulomb of charge between two points in an electric field. Potential difference (p.d.) is often called voltage. This gives the definition of the volt. There is a potential difference of 1 volt between two points if 1 joule of energy is required to move 1 coulomb of charge between the two points, 1 V = 1 J C −1 This relationship can be written mathematically: Where E w _____________________________, Q ____________________________ and V_____________________________________. E w = QV is energy (work done) in joules (J)is the charge in coulombs (C) is the potential difference (p.d.) in volts (V) If the small positive charge, above, is released there is a transfer of energy to kinetic energy, i.e. the charge moves. Again, using the conservation of energy means that; E w = E K QV = ½mv 2

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3 (a) Q = V = E w = ? (a)Calculate the electric potential energy gained by the charge–field system. 3.0  C = 3.0 × 10 −6 C 2.0 kV = 2.0 × 10 3 V E w = QV E w = 3.0 × 10 −6 × 2.0 × 10 3 E w = 6.0 × 10 −3 J (b) The electric field is uniform so the charge experiences a constant unbalanced force. The charge accelerates uniformly towards the negative plate A (c)By conservation of energy, E K = E w = 6.0 × 10 −3 J (d) m = E K = v = ? 5.0 mg = 5.0 × 10 −6 kg 6.0 × 10 −3 J E K = ½ mv 2 6.0 × 10 −3 = 0.5 × 5.0 × 10 −6 × v 2 v 2 = 2.4 × 10 3 v = 49 m s −1

4 Problems…Class Questions…p.7, nos. 1 to 9.


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