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 Starter: 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. mass of a proton = 1.67 x 10-27 kg charge on a proton = 1.6 x 10-19 C Starter If a 15kV voltage is applied across a 0.2 mm gap, calculate: a.) The electric field strength b.) The force on a proton inside the gap c.) The work done on the proton d.) The proton’s final speed as it travels across the gap E = V d 7.5 x 107 Vm-1 E = F q 1.2 x 10-11 N W = ΔEp = Eqd 2.4 x 10-15 J ΔEp = Ek = ½ mv2 1.7 x 106 m/s

24. Electric potential, V, is the same as Voltage Starter • V =ΔEp q Calculate the energy needed to move a 5 mC charge across a potential difference (ΔV) of 30 V. The potential at point X is 24 V. It needs 28 J of work to move a 3.5 Coulomb charge from X to Y. What is the potential of Y? 0.15 J (150mJ) ΔEp = qV (5 x 10-3)(30) = (28)/(3.5) = 8 V V =ΔEp/ q Pot. at Y = 24 + 8 = 32 V

25. Current Electricity

26. 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.

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

28. Current e.g.A voltmeter reads that the voltage loss across a light bulb is 6.8 V and an ammeter is reading that the current (I) flowing through the bulb filament is 0.72 A. Calculate the charge, q, that flows through the bulb filament in 2.0 minutes. I = q t Charge q = I t = (0.72 A)(120 s) = 86.4 C (3 sf) Calculate the electrical potential energy loss Ep of the charge q Ep =V q V = Ep q = (6.8 V)(86.4 C) = 588 J (3 sf)

29. 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.

30. 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

31. 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

32. Starter Volts Voltage is measured in _______,V. It is measured using a voltmeter placed in parallel with a circuit component. Current is measured using an ________ and is placed in _______ . Current is measured in Amperes, A. Conventional current is the movement of ______charge. It flows from the _______ to the _______ terminal. Electrons flow in the opposite direction. Electrical energy is carried around a circuit by the _________. Ammeter series positive negative positive electrons

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

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

35. Ohm’s Law The resistance of a circuit controls the size of the current through the circuit. Ohmic resistorsobey 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

36. Ohm’s Law The voltage loss V across a constant temperature conductor is proportional to the current I flowing through it. V = voltage, V I = current, A R = resistance, Ω. V = IR

37. Resistances in series & parallel

38. Measuring Voltage and Current across a resistor An ideal voltmeter has infinite resistance, so nocurrent goes through the voltmeter. An ideal ammeter has zero resistance, so there is no voltage drop across the ammeter. V A R

39. Series Circuits Have just one current pathway. Currents in a series circuit are the same at all points. Voltages across series resistors add up to the supply voltage. Vs = V1 + V2 Vs + + V A V = Ep q V1 V2 + + V V R1 R2 Rs = R1 + R2 Energy is converted into heat as it passes through the resistors

40. Series Circuits Have just one current pathway. Currents in a series circuit are the same at all points. The battery supplies the power input Ps for the circuit. The resistors use this power to give out heat energy. Ps = P1 + P2 Vs I I + I I V1 V2 R1 R2 Rs = R1 + R2 Energy is converted into heat as it passes through the resistors

41. Parallel Circuits Currents through parallel branches add up to the supply current. Have more than one current pathway. Voltages across parallel resistors are the same as the supply voltage. + V Vs = V1 = V2 Is Vs + A + Is V1 + V I1 1 1 1 RP R1 R2 I1 + = + A R1 V2 Is = I1 + I2 I2 + V I2 + A R2

42. Parallel Circuits Voltages across parallel resistors are the same as the supply voltage. Have more than one current pathway. For any circuit the poweroutput of each resistor must equal the power input of the battery (or electrical mains). + V Is Ps = P1 + P2 Vs + A + Is V1 + V I1 1 1 1 RP R1 R2 I1 + = + A R1 V2 I2 + Is = I1 + I2 V I2 + A R2

43. 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

44. 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

45. 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

46. Starter Ohmic conductors have a _________ resistance (at constant temperature). The V-I graph is a straight line through the _________. Resistors are ohmic conductors within their rated current. constant V origin I Non-ohmic conductors have a resistance which varies for different currents. e.g. A lamp is non-ohmic. It’s resistance _______ as the current through it increases V increases I

47. e.g.Calculate the following quantities for this circuit: a.) The total circuit resistance b.) The total supply current c.) The current through each resistor 10 V Is + A + Is I1 I1 Rs = 3Ω + 9Ω 3 Ω 12 Ω 9 Ω = 12Ω I2 I2 6 Ω

48. e.g.Calculate the following quantities for this circuit: a.) The total circuit resistance b.) The total supply current c.) The current through each resistor 10 V Is + Is I1 1 3 RP 12 1 1 1 RP 12 6 1 1 1 RP R1 R2 I1 = + = + = 12 Ω RP = 4Ω I2 I2 6 Ω 4 Ω

49. e.g.Calculate the following quantities for this circuit: b.) The total supply current c.) The current through each resistor 10 V Is + Is = V / R Is = 10 / 4 = 2.5 A 4 Ω

50. e.g.Calculate the following quantities for this circuit: c.) The current through each resistor I1 = V / R 10 V = 10 / 12 2.5 A + = 0.83 A Is I1 I1 I2 = 10 / 6 12 Ω = 1.67 A I2 I2 6 Ω