Notes for Phy 122 College Physics 2 e' Section 1: Electric Forces & Fields ELECTRIC CHARGE: The particles making up an atom are charged: Entire objects become charged by gaining or loosing electrons: Electrons "rub off" fur onto rod. Two kinds of charge (pos. & neg.): Likes repel, opposites attract. Charge, q, is measured in coulombs: Charge on one electron = -1.60 x 10-19 C (Charge is to electricity what mass is to gravity.) Force = How hard something pushes or pulls. (and direction.) Units: pounds or newtons. See unit table in formula handout. m for meter, C for coulomb, N for newton (1 N ¼ lb.) Coulomb's law: Force between two pointlike charges, in a vacuum: Example 1-a: Find the force (including direction) on the -3 μc charge. (1μC = 10-6 C) (Taking notes yourself, rather than getting printed ones, focuses your attention on what you re writing. You don t need to learn the question, so I printed that. But I saved the solution to write on the board.)
-2- Notice I had to convert μc to C and cm to m for the units to cancel properly: Put standard units in to get standard unit out Also notice: Complete answer includes unit. Example 1-b: Convert that into pounds. To convert units: Multiply by a fraction where numerator equals denominator, set up so unwanted unit cancels out. (To decide whether to multiply or divide by the conversion factor.) yes: no: mi 60min mi min 70 4200 2 hr 1hr hr Note: The word per means division: Miles per hour (speed = distance time), Newtons per coulomb (E = F q), and so on. Problem 1-1: Find the force (including direction) on: a. the 2 microcoulomb charge. b. the 3 microcoulomb charge. c. What is this in pounds? (It s better to do this in your own notebook rather than on this handout. That way, if you want to do it again later for practice, you can look at the question without having to see the solution.) Answers: a. 5.39 N left, b. I don t want to give this part away, c. 1.21 lb Vectors vs. Scalars: Vectors have both magnitude and direction. Examples: Force, velocity, Scalars have only a magnitude. Examples: Electric charge, time, Vectors names include an arrow: F = 23 N, 14 above horizontal Without arrow means the magnitude: F = 23 N (Or, F = 23 N) Charged objects create an invisible influence around themselves called an ELECTRIC FIELD. One way to describe its strength is Electric field vector: E = F q E 's direction: That of the force on a positive charge. (Away from + and toward -.)
-3- Significant Figures: I used to try to cover this, but eventually decided there wasn't room in the course. All I expect now is that you won t round too far from the correct answer. For example, don't round 3.46 to 3. If you are more than 2% from the correct answer on a test, I will treat it like an arithmetic error. Otherwise, I will treat incorrect rounding like incorrect spelling. Spelling physics "fizzyx" won t cost you credit either. But for those of you who prefer not to look quite that ignorant, here is how you ought to round your numbers: The basic idea is not to mislead your reader about your accuracy with more decimal places than you actually know. For example, say you measured the width of a piece of paper as 8.0 +.3 cm. If you then calculate one third of that, your calculator tells you 8.0/3 = 2.666666667. But, using the rules for experimental uncertainty from PHY 121(which I won't go through here) this is only accurate to +.1 cm. So, most of those figures, or digits, are meaningless. The 7 on the end, for example, might in fact be a 3 or a 9 or anything. The significant figures are the ones about which you have some information. The last of those would be the first 6, because based on the +.1 it might really be a 5 or a 7, but at least you know it isn't anything else. So, usually you should round after the decimal place which is somewhat, but not completely, uncertain: 2.7 +.1 cm. When you don't calculate an uncertainty, round using the idea that the answer can't be any more accurate than the least accurate number that went into it. However, exactly how you define "accuracy" depends on what operation you are performing: If adding or subtracting: Round to the same number of DECIMAL PLACES as the number with the fewest. Example: 1234.5 + 6.78 6.78 goes to the hundredths place, but 1234.5 only goes to the tenths place. Round 1241.28 on your calculator to 1241.3. The reason: You don't know what is in the hundredths place of 1234.5, so you don't know what to add to the 8, so you don't know the hundredths place of the answer. If multiplying or dividing: Round to the same number of SIGNIFICANT FIGURES as the number with the fewest. Example: (1234.5)(6.78) 1234.5 has five significant figures, but 6.78 has only three. Round 8369.91 on your calculator to 8370. (Or, in a case like this, the number of significant figures is clearer written as 8.37 x 10 3.) The reason: Let? stand for a figure you don t know. (1234.5)(6.78?) = (1234.5)(6) + (1234.5)(.7) + (1234.5)(.08) + (1234.5)(.00?). The last term might be as big as (1234.5)(.009) 11 or as small as (1234.5)(.000) = 0. Since you are adding in something that could be anywhere between 0 and 11, you have no idea what s in the last place before the decimal. Other operations: Rounding by the number of significant figures, like when multiplying, is usually ok. A couple of final notes: - Unless specifically told otherwise, assume that all numbers given by the text or me are good to three significant figures. (If I write 7, I'm being lazy and really mean 7.00.) - When solving a problem involves several steps, wait until the end to round. If you keep rounding along the way, it sometimes adds up to a considerable difference in the answer.
-4- Example 1-c: Find E (including direction) at the location of the -3.0 µc charge in example a. Problem 1-2: a. Find E (including direction) at the location of the 2.0 µc charge in problem 1. b. What would E at this point be if an electron was located there instead? Combining E = F/q with F = kqq /r 2 (One q gets canceled): E = kq r 2 Electric field around a point charge, q, in a vacuum. To find the force produced by a field, re-arrange E = F/q: F = qe Electric field lines (or "lines of force"): show direction of the force on a positive test charge. Note: Where field is stronger, lines are closer together. Problem 1-3: In each case, draw an arrow at point P showing the direction of the electric field there.
-5- Superposition: With several charges, F on a charge = Add F vectors from all charges at other points. E at a point = Add E vectors from all charges at other points. Example 1-d: Find the force on the 3 μc. Problem 1-e: Find the electric field (including direction) at P, an empty point in space. Problem 1-4: Find the force (including direction) on A, and on B. Ans: a) 4.49 N right
-6- Section 2: Potential Another way to describe the strength of an electric field is in terms of work & energy instead of force. Work: Example: Pushing a piano down the hall: W = F Δx (Assuming F parallel to x.) Energy = the ability to do work. unit: 1 newton meter = 1 joule Examples: - Kinetic energy (Work an object could do due to its motion.) - Chemical Energy - Heat - Others Potential Energy: PE = The work an object could do because of its position. (Example from PHY 121: An elevated object can do work as gravity pulls it downward.) A charge in an electric field has electric potential energy, just as a mass in a gravitational field has gravitational potential energy. The strength of an electric field can be described by the potential energy per unit charge. (How much work is needed to move a unit of charge through it.) Potential: V = PE q Potential difference (between two points): V = PE q Unit: 1 joule coulomb = 1 volt Δ means the amount of difference; ΔV = V 2 V 1 Example 2-a: a. Find the energy delivered by each coulomb of charge flowing from a 12 V battery. b. Repeat for a 120 V battery. Problem 2-1: How much does an electron s energy change if it moves from the ( ) to the (+) terminal of a 12 V battery? Ans: decreases 1.92 x 10-18 J
-7- Potential is higher at (+) and lower at ( ). (Electric field lines, which run + to, point toward lower potential just as a gravitational field points toward lower elevation.) Relationship between E and ΔV: E = ΔV d (Units: It can be shown that N/C is equivalent to V/m.) Problem 2-2: 800 V is applied to a pair of metal plates 7.00 mm apart. Find: a. The strength of the electric field between the plates. b. The force on an electron there. Ans: 1.14 x 10 5 V/m, 1.83 x 10-14 N Capacitors: Capacitors store charge. (Connect a pair of metal plates to a battery. Going into the one plate lets electrons from the battery s negative terminal get farther from each other. (They repel.) They are also attracted by positive charge on the other plate. Current flows until what's on the plates repels charge in the wires as strongly as the battery does.) A higher voltage, though, can push more charge in: Capacitance: Charge stored per volt. C = q / V Unit: 1 C/V = 1 farad Special case: Parallel-plate capacitor, nothing between plates. C 0 = ε 0A d ε 0 = 8.85 x 10-12 C 2 /N m 2 A & d = area & distance as above. Problem 2-b: 1000 V is applied across these plates. Find a. their capacitance, and b. the charge on each.
-8- Dielectric: An insulating material placed between capacitor plates. Becomes polarized and pulls more charges into capacitor from battery. (In a sense, the presence of matter amplifies the field, so it pulls more charge in.) Therefore, capacitor can store more charge. Dielectric constant, κ: C = κc 0 C 0 = capacitance with nothing between plates C = capacitance with dielectric between plates Example 2-c: Find the capacitance in 2-b if the space between the plates is filled with polystyrene. Problem 2-3: Two 1.0 m 2 plates are 2.0 mm apart. The space between them is filled with thick paper. Find: a. the capacitance. b. the charge this holds if connected to a 12 V battery. Ans:.0164 μf,.197 μc Dielectric strength: The value of E at which the material "breaks down" - allows a spark to jump. Ex. 2-4: Find the potential difference which will cause a spark to jump 1 cm through air, between parallel plates. Ans: 30 000 V
-9- Capacitor combinations: Parallel: Series: V is same across each one. Q tot = Q 1 + Q 2 + Q 3 + Q tot /V = Q 1 /V + Q 2 /V + Q 3 /V + C eq = C 1 + C 2 + C 3 + Same Q on each. V tot = V 1 + V 2 + V 3 + V tot /Q = V 1 /Q + V 2 /Q + V 3 /Q + 1/C eq = 1/C 1 + 1/C 2 + 1/C 3 + Example 2-d: What is the equivalent capacitance? Problem 2-5: What is the equivalent capacitance? Ans: 2.67 μf
-10- Section 3: Conductors (examples: salt water, metals, acids): Contain many charged particles which are free to move. Insulators (rubber, glass, etc.): very few free charges. An electric current is a flow of charge. An electric field in a conductor pushes on the free charges, making them flow. In an insulator, there are no charges free to flow when pushed on. def: Electric current: I = q t q = charge, t = time unit: 1 coulomb = 1 ampere sec Problem 3-1: A steady 5.0 amps flows through a wire. How much charge passes one particular point in 10 sec? Problem 3-2: 1.0 x 10 21 electrons flow through a headlight in 30 seconds. Find the current. Ans: 5.34 A I is conventionally thought of as a flow of positive charge: Big voltage is on big side of battery symbol: Higher V on that side since a + charge would "flow downhill" from + to. Electrical Resistance: (Analogous to friction: changes electrical energy into heat.) Ohm's Law: V = IR V = Potential difference across resistor. I = Current through resistor. R = Resistance. Unit: 1 V/A = 1 Ohm = 1 Ω
-11- Comparing electrical charge to water, wires to pipes, and so on: A battery's "voltage" is related to how hard it pushes the charges in a wire, so it is like the amount of pressure from a pump. Current (amps) is the rate that charge flows through the wire. Coulombs per second through a wire would be like gallons per second of water through a pipe. Resistance involves factors like how big around the pipe is. Problem 3-3: Taking the ground to be at V = 0, what is the potential at points A and B? Ans: b) 11 V Power = The rate work is done, or the rate energy is delivered. P = PE t ΔPE = work done, or energy delivered, t = time units: 1 joule/sec = 1 watt 550 ft lb/sec = 1 horsepower ex: You run up a flight of stairs. Your friend, who has the same mass, walks up. Compare work done. Compare power. Ans: Same work: Same weight was lifted through the same height. Your power is greater: You did the work at a faster rate. Electrical Power: From the definition, P = ΔPE/t, it can be shown that P = VI P = I 2 R P = V 2 /R Problem 3-4: A hair dryer is labeled "1200 watts, 120 V (meaning it consumes 1200 W if 120 V is applied). a) Find how much current it draws. b) Find its resistance. c) If you run it 5 minutes, how much charge flows through it? d) How much electrical energy is converted into heat?
-12- Ans: 10.0 A, 12.0 Ω, 3000 C, 360 kj House wiring is in parallel: Problem 3-5: How many 55 W 120 V light bulbs can be lit at once on a circuit with a 20.0 A circuit breaker? Ans: 43 Types of injury: - burns - stopping the heart or lungs Electrical safety: Injury depends on amount of current flowing through body. Precautions: - Be careful you're not grounded. - Keep one hand in your pocket (to keep current out of your chest.) - Be very careful touching someone else being shocked.
-13- Section 4: The circuit's "pump" (battery or generator) is called a source of EMF, electromotive "force". Definition: EMF = Work done by source, per unit charge. E = ΔPE/q (unit: Volts E : script E) (For example, with a 1.5 V flashlight battery, E is the 1.5 V.) Kirchhoff's Laws: -Loop Rule: The sum of the potential changes around a closed loop equals zero. An increase is positive, a decrease is negative: (Current, like water, flows "downhill.") Example: 1.5 -.9 -.6 = 0 -Point Rule (or junction rule ): Total current into a point = total current out Example: 1 + 2 + 3 = 6 Example 4-a: A 10 Ω resistor is placed in parallel with a 20 Ω resistor across a 1.5 V emf. Find the current through the battery. Problem 4-1: Find the current in this circuit: Ans:.020 A
-14- Note that in series, I tot = I 1 = I 2 = I 3 = (In parallel, I tot = I 1 + I 2 + I 3 + from the point rule.) Problem 4-2: Find I 1, I 2 and I 3. Ans: 2.47 A, 1.67 A,.800 A Problem 4-3: Find: (a) The current through the 10 Ω, (b) the current through the battery and (c) the emf of the battery. Ans:.20 A, 1.20 A, 5.0 V It can be shown from Kirchhoff's laws that In series: R eq = R 1 + R 2 + R 3 +... In parallel: 1/R eq = 1/R 1 + 1/R 2 +... Problem 4-4: What is the equivalent resistance? Ans: 23.3 Ω
-15- Review of Sections 1-4: My exams have five parts, each worth 25 points. The best four you do are counted. So, it s best to do all five, and let me drop the worst one. How to study: Be on intimate terms with your notes. Know the basic concepts and how to apply them. Review the sample problems. If you ve kept up with the weekly work, just going through the notes to refresh your memory and fill in any gaps is probably all you need. If you haven t been keeping up, you should still focus on the notes, especially on solving problems. If it turns out you can t learn a month of material in one night, do more each week in the future. Don t just memorize quiz solutions without getting a feel for the concepts. If necessary, use Mastering Physics and/or work more problems from the text in areas where you feel weak. As always, you can come to my office for help. The questions below are not so much a review for the test as one more look at some points you might need help with. Questions on the actual test vary from year to year, and won t necessarily be similar to these. You need to study everything we covered. 1. The resistor on top has a resistance of 12.0 ohms. The one on the bottom consumes 12.0 watts of power. The current through the circuit is 3.00 amps. What is the voltage of the battery? Ans: 40.0 V 2. Find the electric field vector at point P. State whether its direction is right or left. Ans: 212 kn/c (I won t give away the direction.) 3. Find the current through the 15 V battery. Ans: 1.25 A 4. Each of these identical capacitors has a capacitance of C. The equivalent capacitance of the whole group is 6.00 μf. Find C. Ans: 9.0 μf
-16-5. Short answer, 5 points each: a. E 1 < E 2 < E 3. Which has more volts across it: The 5 Ω, the 15 Ω, or are they the same? b. Which is at the higher potential: A battery s negative terminal or its positive terminal? c. If a proton was at point P, what would the direction of the force on it be? (Give an approximate value for θ, where 0 means right, 90 means up, 180 means left, etc.) d. A 1500 W space heater runs for three minutes. How many joules of heat does it give off? e. If you increase the voltage applied to a resistor, what happens to its resistance? (Increases? Decreases? Stays the same? Assume its temperature is not affected.)