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- 1. Chapter 8
- 2.  Electric charge is a fundamental quantity – we don’t really know what it is ◦ But we can describe it, use it, control it ◦ Electricity runs motors, lights, heaters, A/C, stereos, TV’s, computers, etc.  Electric Forces – at the microscopic level they hold atoms and molecules together ◦ Electric Forces hold matter together ◦ Gravitational Forces hold the universe together  Magnetism is also closely associated with electricity Intro
- 3.  Experimental evidence leads us to conclude that there are two types charges ◦ Positive (+) ◦ Negative (-)  All matter is composed of atoms, which in turn are composed of subatomic particles ◦ Electrons (-) ◦ Protons (+) ◦ Neutron (neutral) Section 8.1
- 4.  Note that the Proton and Neutron each have about 1000x more mass than the Electron  If the atom has the same number of protons and electrons it is electrically neutral Section 8.1
- 5.  Named after a French scientist – Charles Coulomb (1736-1806)  Note that the charge on a single electron (-) or proton (+) is 1.60 x 10-19 C (very small!)  q usually designates electric charge ◦ Excess of positive charges +q ◦ Excess of negative charges -q Section 8.1
- 6.  An electric force exists between any two charged particles – either attractive or repulsive Section 8.1
- 7.  Law of Charges – like charges repel, and unlike charges attract ◦ Two positives repel each other ◦ Two negatives repel each other ◦ Positive and negative attract each other  The force between two charged bodies is directly proportional to the product of the two charges & inversely proportional to the square of their distance apart. Section 8.1
- 8.  Force = Section 8.1 constant x charge 1 x charge 2 (distance between charges)2 λ In equation form F = kq1q2 / r2 λ k is a constant = 9.0 x 109 Nm2 /C2
- 9.  An object with an excess of electrons is said to be negatively charged  An object with a deficiently of electrons is said to be positively charged  Photocopier (xerography) – practical and widespread use of electrostatics Section 8.1
- 10. Two negative charges repel Two positive charges repel One negative and one positive attract Repulse Repulse Attract Section 8.1
- 11.  Force of attraction/repulsion between two charged bodies is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them  F = (kq1q2) / r2 ◦ F = force of attraction or repulsion ◦ q1 = magnitude of the first charge ◦ q2 = magnitude of the second charge ◦ r = distance between charges ◦ k = constant = 9.0 x 109 N-m2 /C2 Section 8.1
- 12.  Equations look similar ◦ F = kq1q2 / r2 & F = Gm1m2/ r2  Both depend on r2  Coulomb’s law can describe either an attractive or repulsive force – gravity is always positive  Electrical charges are much stronger than gravitational forces Section 8.1
- 13.  I = charge/time = q/t ◦ I = electric current (amperes) ◦ q = electric charge flowing past a point (coulombs) ◦ t = time for the charge to pass point (seconds)  1 ampere (A) = flow of 1 Coulomb per second  Rearrange equation above: ◦ q = It or 1 coulomb = 1 ampere x 1 second  Therefore, 1 coulomb is the amount of charge that flows past a given point in 1 second when the current is 1 ampere Section 8.2
- 14.  Electrical conductor – materials in which an electric charge flows readily (most metals, due to the outer, loosely bound electrons)  Electrical insulator – materials that do not conduct electricity very well due to very tight electron bonding (wood, plastic, glass)  Semiconductor – not good as a conductor or insulator (graphite) Section 8.2
- 15.  A wire carries a current of 0.50 A for 2 minutes. a) how much (net) charge goes past a point in the wire in this time? b) how many electrons make up this amount of charge?  GIVEN: I = 0.50 A, t = 2 minutes (120 seconds)  WANTED: q (charge) & n (number of electrons)  (a) q = It = (0.50 A)(120 s) = 60 C (coulombs) Section 8.2
- 16.  A wire carries a current of 0.50 A for 2 minutes. a) how much (net) charge goes past a point in the wire in this time? b) how many electrons make up this amount of charge?  To solve for (b), we know that each electron has a charge of 1.6 x 10-19 C and we know the total charge from part (a) = 60 Coulombs  n = 60/(1.6 x 10-19 C / electron) = 3.8 x 1020 electrons Section 8.1
- 17.  When work is done to separate positive and negative charges, we have electric potential energy Section 8.2
- 18.  Instead of measuring electric potential energy, we measure the potential difference, or voltage  Voltage – the amount of work it would take to move a charge between two points, divided by the value of the charge  Voltage = work / charge = V = W/q  Measured in volts (V) = 1 joule/Coulomb  When we have electric potential energy, this may be used to set up an electrical current Section 8.2
- 19.  Whenever there is an electrical current, there is resistance (R) within the conducting material ◦ R is due to atomic/subatomic collisions  Georg Ohm (1787-1854) – formulated a simple relationship between voltage, current, and resistance  Ohm’s Law  V = IR ◦ V = voltage in volts, I = current in amperes, and R = resistance in ohms  1 ohm = 1 volt/1 ampere (R=V/I) Section 8.2
- 20.  Electrons flow from negative terminal to positive terminal (provided by the chemical energy of the battery) -- negative to positive  Open switch – not a complete circuit and no flow of current (electrons)  Closed switch – a complete circuit and flow of current (electrons) exists  Closed Circuit Required – to have a sustained electrical current Section 8.2
- 21. The light bulb offers resistance. The kinetic energy of the electric energy is converted to heat and radiant energy. Section 8.2
- 22. Section 8.2
- 23.  Recall: P = W/t (Power = work/time) – Ch. 4  V = W/q (Voltage = work/charge) – Ch. 8  Rearrange V=W/q  W = qV  Substitute W = qV into P = W/t equation  Result  P = qV/t  Recall that I = q/t (Current = charge/time)  ∴ P = IV (Electric power = current x voltage)  also P = I2 R (substitute in V = IR)  P in watts, I in amperes, R in ohms, V in volts Section 8.2
- 24.  Find the current and resistance of a 60-W, 120-V light bulb in operation.  Given: P = 60W, V = 120 V  Find: I (current in amperes), R (resistance in ohms)  Since P = IV  I = P/V = 60 W/120 V = 0.50 A  Since V = IR  R = V/I = 120 V/0.50A = 240 Ω  Since P = I2 R  R = P/I2 = 60 W/(0.50 A)2 = 240 Ω (same answer as above) Section 8.2
- 25.  A coffeemaker draws 10 A of current operating at 120 V. How much electrical energy does the coffeemaker use each second.  Given: I = 10 A, V = 120 V  Find: P (electrical energy in watts)  P = IV  P = (10 A) x (120 V) = 1200 W or 1200 J/s Section 8.2
- 26.  Current: I = q/t ◦ I = current (amperes), ◦ q = electric charge (coulombs), ◦ t = time (seconds)  Coulomb’s Law: F = kq1q2/r2  Voltage: V = W/q [Work is in joules (J)]  Ohm’s Law: V= IR  Electrical Power: P = IV = I2 R Section 8.2
- 27.  Direct Current (dc) – the electron flow is always in one direction, from (-) to (+) ◦ Used in batteries and automobiles  Alternating Current (ac) – constantly changing the voltage from positive to negative and back ◦ Used in homes. ◦ 60 Hz (cycles/sec) and Voltage of 110-120 V Section 8.3
- 28.  Plugging appliances into a wall outlet places them in a circuit – either in series or in parallel  In a series circuit, the same current (I) passes through all the resistances. ◦ The total resistance is simply the sum of the individual resistances. Section 8.3
- 29.  Rs = R1+R2+R3+…  If one bulb were to burn out the circuit would be broken, and all the lights would go out Section 8.3 Same current all the way
- 30.  In a parallel circuit, the voltage (V) across each resistance is the same, but the current (I) through each may vary. Section 8.3
- 31. Current Divides, Voltage is the same Section 8.3 I = I1+I2+I3+… for R in parallel: 1/Rp = 1/R1+1/R2+… for 2 R in parallel: Rp = (R1R2)/(R1+ R2)
- 32.  Three resistors have values of R1=6.0Ω, R2=6.0Ω, and R3=3.0Ω. What is their total resistance when connected in parallel, and how much current will be drawn from a 12-V battery if it is connected to the circuit?  1/Rp = 1/R1+1/R2+1/R3 = 1/6 + 1/6 + 1/3  1/Rp = 1/6 + 1/6 + 2/6 = 4/6 Ω  Rp = (6 Ω)/4 = 1.5 Ω = Total Resistance  I = V/Rp = (12 V)/(1.5 Ω) = 8.0 A = current drawn Section 8.3
- 33.  Wired in parallel independent branches any particular circuit element can operate when others in the same circuit do not. Section 8.3
- 34.  Wires can become hot as more and more current is used on numerous appliances.  Fuses are placed in the circuit to prevent wires from becoming too hot and catching fire.  The fuse filament is designed to melt (and thereby break the electrical circuit) when the current gets too high.  Two types of fuses: Edison and S-type  Circuit Breakers are generally now used. Section 8.3
- 35.  A dangerous shock can occur if an internal ‘hot’ wire comes in contact with the metal casing of a tool.  This danger can be minimized by grounding the case with a dedicated wire through the third wire on the plug. Section 8.3
- 36. Section 8.3
- 37. Section 8.3
- 38.  Closely associated with electricity is magnetism.  In fact electromagnetic waves consists of both vibrating electric and magnetic fields. These phenomena are basically inseparable.  A bar magnet has two regions of magnetic strength, called the poles.  One pole is designated “north,” one “south.” Section 8.4
- 39.  The N pole of a magnet is “north-seeking” – it points north.  The S pole of a magnet is “south-seeking” – it points south.  Magnets also have repulsive forces, specific to their poles, called …  Law of Poles – Like poles repel and unlike poles attract ◦ N-S attract ◦ S-S & N-N repel Section 8.4
- 40. All magnets have two poles – they are dipoles Section 8.4
- 41.  Magnetic field - a set of imaginary lines that indicates the direction in which a small compass needle would point if it were placed near a magnet  These lines are indications of the magnetic force field.  Magnetic fields are vector quantities. Section 8.4
- 42.  The arrows indicate the direction in which the north pole of a compass would point. Section 8.4
- 43.  The source of magnetism is moving and spinning electrons.  Hans Oersted, a Danish physicist, first discovered that a compass needle was deflected by a current- carrying wire. ◦ Current open  deflection of compass needle ◦ Current closed  no deflection of compass needle  A current-carrying wire produces a magnetic field: stronger current  stronger field  Electromagnet – can be switched on & off Section 8.4
- 44.  Most materials have many electrons going in many directions, therefore their magnetic effect cancels each other out  non-magnetic  A few materials are ferromagnetic (iron, nickel, cobalt) – in which many atoms combine to create magnetic domains (local regions of magnetic alignment within a single piece of iron)  A piece of iron with randomly oriented magnetic domains is not magnetic. Section 8.4
- 45.  The magnetic domains are generally random, but when the iron is placed in a magnetic field the domains line up (usually temporarily). Section 8.4
- 46.  A simple electromagnet consists of an insulated coil of wire wrapped around a piece of iron.  Stronger current  stronger magnet  Electromagnets are made of a type of iron that is quickly magnetized and unmagnetized – termed “soft.” Section 8.4
- 47.  Materials cease to be ferromagnetic at very high temperatures – the “Curie temperature” (770o C for iron)  Permanent magnets are made by permanently aligning the many magnetic domains within a piece of material.  One way to create a permanent magnet is to heat the ferromagnetic material above the Curie temperature and then cool the material under the influence of a strong magnetic field. Section 8.4
- 48.  This planet’s magnetic field exists within the earth and extends many hundreds of miles into space.  The aurora borealis (northern lights) and aurora australis (southern lights) are associated with the earth’s magnetic field.  Although this field is weak compared to magnets used in the laboratory, it is thought that certain animals use it for navigation. Section 8.4
- 49.  Similar to the pattern from a giant bar magnet being present within the earth (but one is not present!) Section 8.4
- 50.  The origin of the earth’s magnetic field is not completely understood. ◦ Probably related to internal currents of electrically charge particles in the liquid outer core of the earth, in association with the earth’s rotation  The temperatures within the earth are much hotter than the Curie temperature, so materials cannot be ferromagnetic.  The positions of the magnetic poles are constantly changing, suggesting “currents.” Section 8.4
- 51.  Magnetic declination – the angle between geographic (true) north and magnetic north  The magnetic declination varies depending upon one’s location on Earth. ◦ In the northern hemisphere the magnetic North Pole is about 1500 km from the geographic North Pole.  Therefore, a compass does not point to true north, but rather magnetic north. ◦ An adjustment (magnetic declination) must be made to determine true north from a compass. Section 8.4
- 52. Section 8.4
- 53.  Resistance in Series: Rs = R1+R2+R3+…  Resistance in Parallel: 1/Rp = 1/R1+1/R2+…  For 2 R in parallel: Rp = (R1R2)/(R1+R2) Section 8.4
- 54.  Electromagnetism – the interaction of electrical and magnetic effects  Two basic principles: 1) Moving electric charges (current) give rise to magnetic fields (basis for an electromagnet). 2) A magnetic field will deflect a moving electric charge (basis for electric motors and generators). Section 8.5
- 55.  Telephone transmission involves the conversion of sound waves into varying electric current  Sound waves at the transmitting end vibrate a diaphragm that puts pressure on carbon grains  The varying pressure on the carbon, results in varying amounts of resistance in the circuit  By Ohm’s law (V=IR), varying resistance results in varying current Section 8.5
- 56.  The varying current gives an electromagnet at the receiving end varying magnetic strengths  The electromagnet is drawn toward/away a permanent magnet as a function of the electric current in the line  The electromagnet is attached to a diaphragm that vibrates, creating sound waves that closely resemble the original sound waves Section 8.5
- 57. Section 8.5
- 58.  When a moving charge (a current) enters a magnetic field, the moving charge will be deflected by the magnetic field  The magnetic force (Fmag) is perpendicular to the plane formed by the velocity vector (v) of the moving charge and the magnetic field (B) Section 8.5
- 59.  Electrons entering a magnetic field experience a force Fmag that deflects them “out of the page” Section 8.5
- 60.  A beam of electrons is deflected downward due to the introduction of the magnetic field of a bar magnet Section 8.5
- 61.  Since a magnetic field deflects a moving charge, a current-carrying wire loop will experience a force Section 8.5
- 62. Section 8.5
- 63. Section 8.5
- 64.  Electric motor – a device that converts electrical energy into mechanical work  When a loop of coil is carrying a current within a magnetic field, the coil experiences a torque and rotates Section 8.5
- 65.  The inertia of the spinning loop carries it through the positions where unstable conditions exist  The split-ring commutator reverses the loop current every half-cycle, enabling the loop to rotate continuously Section 8.5
- 66.  When a wire is moved perpendicular to a magnetic field, the magnetic force causes the electrons in the wire the move – therefore a current is set up in the wire Section 8.5
- 67.  Generator – a device that converts mechanical energy into electrical energy  Generators operate on the principle of electromagnetic induction  Electromagnetic induction was discovered by the English scientist, Michael Faraday in 1831 ◦ When a magnet is moved toward a loop or coil of wire, a current is induced in the wire ◦ The same effect is obtained if the magnetic field is stationary and the loop is rotated within it Section 8.5
- 68.  A current is induced in the circuit as the magnet is moved toward and into the coil Section 8.5
- 69.  When the loop is rotated within the magnetic field a current is induced in the loop  The current varies in direction every half-cycle and is termed alternating current (ac) Section 8.5
- 70.  Transformers serve to step-up or step-down the voltage of electric transmissions  A transformers consists of two insulated coils of wire wrapped around an iron core ◦ The iron core serves to concentrate the magnetic field when there is a current in the input coil  The magnetic field resulting from the primary coil goes through the secondary coil and induces a voltage and current Section 8.5
- 71.  Since the secondary coil has more windings, the induced ac voltage is greater than the input voltage.  The factor of voltage step-up depends on the ratio of the windings on the two coils. Section 8.5
- 72.  The main reason is to reduce the amount of current going through the lines ◦ As the current is reduced, the amount of resistance is also reduced ◦ If the amount of resistance is reduced, the amount of heat (called ‘joule heat’ or ‘I2 R’ losses)  For example, if the voltage is stepped-up by a factor of 2, the resulting I2 R losses would be reduced four-fold Section 8.5
- 73.  The voltage change for a transformer is given by the following:  V2 = (N2/N1)V1  V1 = primary voltage  V2 = secondary voltage  N1 = Number of windings in primary coil  N2 = Number of windings in secondary coil Section 8.5
- 74.  A transformer has 500 windings in its primary coil and 25 in its secondary coil. If the primary voltage is 4400 V, find the secondary voltage.  Use equation V2 = (N2/N1)V1  Given: N1 = 500, N2 = 25, V1 = 4400 V  V2 = (25/500)(4400 V) = 220 V  This is typical of a step-down transformer on a utility pole near homes Section 8.5
- 75. Voltage is dramatically stepped-up at the generating plant to minimize joule heat loss during long-distance transmission. The voltage must then be stepped-down for household use. Section 8.5
- 76.  F = kq1q2/r2 Coulomb’s Law  I = q/t Current  V= W/q Voltage  V= IR Ohm’s Law  P= IV = I2 R Electric Power Review
- 77.  Rs = R1 + R2 + R3… Resistance in Series  1/Rp = 1/R1 + 1/R2 + 1/R3 …. Resistance in Parallel  Rp = (R1R2)/(R1 + R2) Two Resistances in Parallel  V2 = (N2/N1)V1 Transformer (Voltages and Turns) Review

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