2.6 Electricity & Magnetism

1. 2.6 Electricity & Magnetism 6 credits - External

2. Static Electricity

3. Electrical Charge An electrical charge (q) can be positive or negative. Like charges repel, unlike charges attract. Charge is measured in Coulombs, C

4. Electrical Charge e.g An electron holds 1.6 x 10-19C of charge. How many electrons are contained in these charges? a.) -2.24 x 10-18 C b.) -3.6 mC c.) -1 C 14electrons 2.25 x 1016electrons 6.25 x 1018electrons Charge is measured in Coulombs, C

5. Conductors and Insulators Conductors transport charges freely. Charges rapidly distribute over the surface of a conductor. Insulators do not readily transport charge. Charges on an insulator tend to stay where they are put. Both conductors and insulators can store charge.

6. Conductors and Insulators Because Earth is so large, it can be considered an infinite ‘sink’ for charges – charges spread out over the Earth’s surface and it remains effectively uncharged. Symbol: earth ground

7. Electroscope A simple tool used to detect the presence of small electric charges.

8. Charging an Electroscope An electroscope gives an indication of the amount of charge on an object. An electroscope can be charged: A. By contact Charged object is touched against the electroscope. B. By induction 1. Charged object is placed near the electroscope. 2. Electroscope is earthed. 3. Earth is removed. 4. Charged object is removed.  

9. Coulomb’s Law Coulomb’s Law gives the force between two point charges. It is similar to Newton’s Law of gravity. FE = kq1q2 r2 . FE= electrostatic force, N k = Coulomb constant, 9.0 x 109 N.m2.C-2 q = charge, C r = distance between charges, m.

10. Coulomb’s Law Coulomb’s Law gives the force between two point charges. It is similar to Newton’s Law of gravity. FE = kq1 q2 r2 . FE= electrostatic force, N k = Coulomb constant, 9.0 x 109 N.m2.C-2 Q = charge, C r = distance between charges, m.

11. Coulomb’s Law e.g.Three protons are separated from a single electron by a distance of 1x10-6 m. Find the electrostatic force between them. Is this force attractive or repulsive? q1 = 3 protons = 3(1.6x10-19) = 4.8 x 10-19C FE = kq1 q2 r2 . q2 = 1 electron = -1.6x10-19C = (9.0 x 109)(4.8 x 10-19)(-1.6x10-19) (1x10-6)2 . = -6.912 x 10-16 N FE= electrostatic force, N k = Coulomb constant, 9.0 x 109 N.m2.C-2 Q = charge, C r = distance between charges, m. (attractive)

12. Electric Fields An electric field is a region of space where a charge will experience an electric force. Electric field lines: • go from positive to negative • don’t cross • closer lines indicates stronger the electric fields. A gravitational field is a region of space where a mass will experience a force - - - - - - - - - - - - -

13. Electric Fields E = F Electric Field strength (E) is the force on each Coulomb of test charge at that location q E = electric field strength, N.C-1 F = force on a charge, N q = charge, C.

14. Electric Fields q A 35μC test charge is placed inside an electric field with a strength, E, of 170 NC-1 at that location. Calculate the electric force on this charge. E = F e.g. F = Eq = (170 NC-1)(35 x 10-6 C) = 5.95 x 10-3 N E = electric field strength, N.C-1 F = force on a charge, N q = charge, C.

15. Electric Field of a Point Charge Substituting FE = kq1q2 r2 into E = F/q, we get: E = kq1q2 ÷ q r2 E = kq r2 E = the electric field due to a point charge, N.C-1 k = Coulomb’s constant, 9.0 x 109 q = point charge, C r = distance from the charge, m.

16. Uniform Electric Field A uniform electric field has a constant strength and direction. Uniform gravitational field field lines flow from positive to negative

17. V + + + + - - - - - - - - d Uniform Electric Field A uniform electric field has a constant strength and direction. If a voltage is applied across two metal plates that are facing each other, an electric field is formed between the two plates.

18. V + + + + - - - - d Uniform Electric Field A uniform electric field has a constant strength and direction. E = V d - - - - E = electric field strength, N.C-1 (or V.m-1) V = voltage difference between the plates, V d = distance between the plates, m. d

19. Electrical Potential Energy Work is done to move a charge against an electric field This work is stored as electric potential energy. If the charge is then allowed to move, the electric potential energy is released as kinetic energy. work is done to move a mass against a gravitational field W = ΔEp = Eqd Since W = F ×d and F = E×q ΔEp = change in electric potential energy, J E = electric field strength, N.C-1 F = force on a charge, N q = charge, C d = distance the charge is moved, m.

20. V= 2134 V - - + + 123 mm Uniform Electric Fields e.g. A +2.00μC charge moves between two plates, separated by a distance of 123 mm, with a voltage of 2134 V across it. Find: a.) the electric field strength (E) between the plates b.) The electric force (F) on the charge. c.) The loss in electric potential energy ΔEp of the charge in moving the full distance between the plates E = V d E = 2134/123 x 10-3 = 17.3 x 103Vm-1 E = F q F = Eq = (17.3 x 103)(2 x 10-6)= 0.0346 N W = ΔEp = Eqd d = ΔEp = Eqd = (17.3 x 103)(2 x 10-6)(123 x 10-3)= 4.26 x 10-3 J

21. E = F q Uniform Electric Fields e.g. Two metal plates are separated by a distance of 68.4 mm, connected to a battery of voltage 597 V and an open switch. A charge of -34.7μC is placed between the plates. a.) When the switch is closed which way will conventional positive charge flow through the battery? b.) Which of the two plates will become negatively charged? c.) Calculate the electric field strength between the plates. What is the direction of the electric field d.) What is the magnitude and direction of the electric force? E = V d negative to positive terminal bottom plate E = 597/68.4x10-3 = 8.73 x 103 Vm-1 + + + + + - - 0.303 N F = Eq = (8.73 x 103)(-34.7 x 10-6)= - - - - (towards the top plate)

22. Uniform Electric Fields e.g. Two metal plates are separated by a distance of 68.4 mm, connected to a battery of voltage 597 V and an open switch. A charge of -34.7μC is placed between the plates. e.) Calculate the loss in electric potential energy of the charge as it moves a distance of 13.4 mm between the plates ΔEp = Eqd = (8.73 x 103)(-34.7 x 10-6)(0.0134) W = ΔEp = Eqd = 4.06 x 10-3 J + + + + + - - - - - -

23. Current Electricity

24. Current Current (I) is a measure of how many charges are going past a point every second. Conventional current is the movement of positive charge. + - Current is measured in Amperes or Amps, A.

25. Current + I = Q t - I = current, A Q = charge, C t = time, s. Ammeters are placed in series.

26. Electrical Energy Electrical energy (Ep) is carried by charges. Electrical energy is measured in Joules, J. V = Ep q Voltage (V) is the amount of electrical energy gained or lost by a coulomb of charge as it travels through an energy user.. Voltage is measured in Volts, V.

27. Voltage The energy lost or gained for every Coulomb of charge e.g.Calculate the voltage lost if 600J of energy is released when 6 C of charge passes a resistor. V = Ep q Voltage V = 600 6 = 100 V Ep= change in electric potential energy, J V =Voltage (potential), JC-1 or V q = charge, C

28. Voltage Voltages are measured in parallel with a circuit component. e.g. Calculate the energy change when 2.0 x 108electrons are moving across a potential difference, ΔV, of 30 V. V = Ep q ΔEp = Vq = (30V)(2.0 x 108 x 1.6 x 10-19 C) = 9.6 x 10-10 J ΔEp = change in electric potential energy, J V =Voltage (potential) is work per Coulomb, JC-1 or V q = charge, C

29. Circuit Diagrams • Circuit diagrams need to have: • clear, ruled lines • connected lines (no gaps). + -

30. Circuit Diagrams You need to be familiar with the following circuit symbols: + +

31. Ohm’s Law The resistance of a circuit controls the size of the current through the circuit. Ohmicresistors obey Ohm’s Law: if the voltage across the resistor is increased, the current through the resistor will increase proportionately. Voltage V Slope = rise run = V / I = Resistance Current I

32. Ohm’s Law The resistance of a circuit controls the size of the current through the circuit. Ohmicresistors obey Ohm’s Law: if the voltage across the resistor is increased, the current through the resistor will increase proportionately. V = voltage, V I = current, A R = resistance, W. V = IR

33. Resistances in series & parallel

34. Resistors in Series A series circuit is a circuit with only one conducting pathway. For resistors in series, the total resistance is the sum of the individual resistances. RT= total resistance, Ω Rn = individual resistance, Ω

35. Resistors in Parallel A parallel circuit is a circuit with more than one conducting pathway. For resistors in parallel, use the reciprocal rule: RT= total resistance, Ω Rn = individual resistance, Ω

36. Electromagnetism

37. Magnetic Fields A magnetic field is a region of space where a magnetic force can be detected. Like poles repel, unlike poles attract. Magnetic field lines go from North to South.   Magnetic fields exist around magnetised objects, e.g. a bar magnet conductors carrying a current.

38. Magnetic Fields Magnetic fields can be: strong weak uniform (in strength and direction) directed into the page x xxxx x xxxx x xxxx directed out of the page ● ● ● ● ● ● ● ● ● ● ● ●

39. F q v Magnetic Force on a Moving Charge The direction of the magnetic force experienced by a charge moving in an electric field is also given by the right-hand slap rule. + fingers = magnetic field thumb = direction positive charge is moving in (velocity vector) slap = force on the charge

40. Magnetic Fields

41. Magnetic Force on a Moving Charge The direction of the magnetic force experienced by a charge moving in an electric field is also given by the right-hand slap rule (positive charge) left-hand slap rule (negative charge – electrons) • The magnetic force experienced by the charge is: • F = Bqv • F = force on the current-carrying wire, N • B = magnetic field strength, T • q = charge, C • v = velocity, m.s-1.

42. Aurora Australis The southern lights often seen at the lower latitudes of New Zealand on a clear dark night. The phenomenon begins at the sun where a particularly large coronal outburst creates a strong stream of charged particles known as the solar wind. The solar wind interacts with the Earth's magnetic field lines at the altitude called the magnetopause and the charged positive and negative charges are then associated with the Earth's magnetic field, creating a large charged electric field concentrated at the poles. In the southern hemisphere the field lines are streaming out of the Southpole which is like a north pole of a magnet. Current flows between these electrical polarities, through the ionosphere, across the polar region and back. The streams of electrons which are the negatively charged particles are flowing - spiraling, along the magnetic field lines and intercepting the oxygen and nitrogen atoms in the upper atmosphere. These excite the oxygen and nitrogen atoms causing them to fluoresce, just as they do in a conventional neon fluorescent light tube.

43. Aurora Australis Excited oxygen atoms emit a greenish white light, nitrogen molecules which are excited by more energetic electrons are made to emit the red to pinkish colours. Ionised nitrogen gives off a blue violet light. The light emission or fluorescence from atoms is caused when electromagnetic energy strikes an atom raising the electron orbital level to a set, higher energy level, which may then collapse back again emitting a photon of light of a particular wavelength. This determines its colour.