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Unformatted text preview: MasteringPhysics: Assignment Print View http://session.masteringphysics.coni/myct/assignmentPrint?assignmen... Assigiment Display Mode: 1 MW} PHYSICSBCF08
{Same rs and Camera? time a: ‘EEﬁtgtz/a at: wanders, i‘éovembes 3‘ 25% View Grading Details Description: Introduces the concept of capacitance, and the basic formula for a ﬁlled parallelplate capacitance Leaming Goal: To understand the meaning of capacitance and ways of calculating capacitance When a positive charge r; is placed on a conductor that is insulated from ground, an electric ﬁeld emanates from the conductor to ground, and the conductor will have a nonzero potential 5/" relative to ground. If more charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance (I? of this conductor: (.7 (git? . i Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the ﬁrst (they
E are often called electrodes), and the relevant voltage is the voltage between these two electrodes. ‘ This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallelplate capacitor, which consists of two plates each of area A separated by a small distance {i with air or vacuum in between. In ﬁguring out the capacitance of this conﬁguration ofconductors, it is important to keep in mind that the voltage difference is the line integral of the electric ﬁeld betweenthe plates. 2 E
z ANSWER: Z ability to conduct electric current
ability to distort an external electrostatic ﬁeld
. ability to store charge
ability to store electrostatic energy Capacitance is a measure of the ability of a system of two conductors to store electric charge and enery. Capacitance is deﬁned as (f m: This ratio rermins constant as long as the system retains its geometry and the ir amount of dielectric does not change. Capacitors are special devices designed to combine a large capacitance with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has sonic capacitance. Even an isolated electrode has a mall capacitance. That is, if a charge rig is placed on it, its potential V with respect to ground will change, and the ratio is its capacitance (3. Part B
Assume that charge w t; is placed on the top plate, and t; is placed on the bottom plate. What is the magnitUde of the electric ﬁeld between the plates? Part 3.1 How do you find the magnitude of the electric ﬁeld?
What is the easiest way to obtain 15‘? ANSWER: . 9 Use Gauss's law and the fact that a; t) outside the capacitor. Use Gauss's law and the symmetry of the lower plate.
Use Coulomb's law integrating over all charge on the bottom plate.
Use Coulomb's law integrating over all charge on both plates. What Is the electnc ﬂux integral due to the electric ﬁeld?
Apply Gauss's law to a small box whose top side is just above the lower plate and whose bottom is just below it1 where 3;” 8. Start by ﬁnding the electric ﬂux integral (ﬁg. Express this integral in terms of the area :3 of the top side of the box and the magnitude 32} of the electric ﬁeld between the plates. ANSWER: gig; : gig :Part 8.3 Find the enclosed charge
Find the amount of charge gm} enclosed in a small box whose top side isjust above the lower plate and whose bottom is just below it, where [3‘ m 01 E;
i:
a
'3 ; 10f19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen... Part B.3.a Find the surface charge What is o, the charge per unit area on the lower plate? Express at in terms of any necessary constants and quantities given in the introduction. ANSWER: Vs: Express the enclosed charge in terms of the crosssectional area of the box a and otherquantities given in the introduction. ANSWER: ﬁrm2 : Hint BA Recall Gauss's law Gauss's law states that @ Express E in temts of x} and other quantities given in the introduction, in addition to ti, and any other constants needed ANSWER: g I; z What is the voltage 1’“ between the plates of the capacitor? [[mt CI The electric ﬁeld is the derivative of the potential The voltage difference is the integral of the electric ﬁeld from one plate to the other; in symbols, V w» E v ti. Express t’ in terms of the quantities given in the introduction and any required physical constants. 3'
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2 ANSWER: Increase the charge on the capacitor.
Decrease the charge on the capacitor.
Increase the spacing between the plates of the capacitor.
. Decrease the spacing between the plates of the capacitor.
Increase the length of the wires leading to the capacitor plates. Part F
Consider a charged parallelplate capacitor. Which combination of changes would quadruple is capacitance? Double the charge and double the plate area.
Double the charge and double the plate separation
Halve the charge and double the plate separation.
Halve the charge and double the plate area. a Halve the plate separation; double the plate area.
Double the plate separation; halve the plate area. A Simple Network of Capacitors
Description. Given the capacitance of three capacitors connected in a network and the ﬁnal charge on one of the capacitors, calculate the ﬁnal charge on the other two capacitors and the potential diﬁerence for the network. i In the ﬁgure are shown three capacitors with capacitances (‘13 {35):} “a (‘3 x 3‘60 gm? {‘3 me {am 331". The capacitor network is connected to an applied potential C
2 §
§ as, Aﬁer the charges on the capacitors have reached their ﬁnal values, the charge {:32 on the second capacitor is 40.0 1:50. »‘; j
E . 20f19 10/31/200811221AM MasteringPhysics: Assignment Print View http://sessi0n.masteringphysics.com/myct/assignmentPrint?assignmen... . Part A
What is the charge {3} on capacitor C‘, 7 ‘ Hint AJ How to approach the problem
; ‘ Consider only the initial section of the network (ﬂ'om a to d). Develop a relation between the capacitance and charge for each capacitor in that section. Use the capacitances Cg and C2 and the ﬁnal charge Q3 on (33 to calculate
E‘the charge Q: on Cg. Part A.2 Series or parallel? 3Are capacitors C; and (,3 connected in series or in parallel? ANSWER: series
‘ parallel ' the two capacitors have correspondinguplates connected to the same side of the potential diéerence in the ﬁrst section, that is, the leﬁ plates are both connected to pomt a, and the right plates are both connected to point
~ d they are connected in parallel. Part A.3 Calculate the potential difference across the second capacitor Calculate the potential difference {is across the second capacitor. . Hint A.3.a Equation for capacitance Q  Recall that for any capacitor one has the relation C’ ‘ 17;, where is the charge on each plate of the capacitor and V is the potential difference across the capacitor (not the difference across the whole network). Express your answer in volts to three significant figures. ANSWER: p3: = 333 V ' *5! Pan AA Calculate the potential difference across the ﬁrst capacitor, V} Calculate the potential difference V3 across the first capacitor. int A.4.a Parallel capacitors and potential difference f two capacitors are connected in parallel, how are the potential differences across each capacitor related? Note that the left plate and right plate are connected to points a and 11, respectively, for bath capacitors.
Express your answer in volts to three signiﬁcant ﬁgures. ANSWER: V3 2 mg; 3? Since (liltliand {lg"are connected in parallel, the potential difference will sameufor hoth capacitors, t What is the charge on capacitor Cg? > Hint 3.1 How to approach the problem
' Determine whether the third capacitor, (3: , is connected in series or in parallel with the first section of the network (ﬁrm a to d). Calculate the total charge of the initial section of the network, {3%. Use the result to ﬁnd the
5 charge on the third capacitor, egg. ’ Part B.2 Series or parallel? Is the third capacitor, (1‘3, connected in series or in parallel with the ﬁrst section of the network (froma to d)? ANSWER: ‘ series parallel § Since the two capacitors in the initial section are connected between points a and d, and the third capacitor (33 is connected between points d and b, the initial section and the third capacitor are connected together in series. 1% Him 113 Capacitors in series . For capacitors connected in series, the positive plate of one capacitor is connected to the negative plate of the other capacitor. Since the charge on the two plates must come from separation of charges in the initially neutral plates
(charge is never created or destroyed), the two plates must have charges of equal magnitude and opposite signs. Therefore, the charge must be the same for the two capacitors. in this problem, the network between a and d is in 3 ofl9 10/31/200811121AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... series with capacitor (:25. Therefore, the total charge on that network must equal the charge on (3’3. ix .Part 8.4 Calculate the total charge in the initial section
' Calculate {35.4, the total charge stored in the initial section Hint B.4.a How to find the charge in the initial section
You already know the charge on the first two capacitors, ﬁg and Q3, in the initial section between points a and d. The total charge must therefore be the sum of the charges in each capacitor. Express your answer in microeouiomhs to three signiﬁcant ﬁgures. ANSWER: M = {39 ﬁg: i Express your amwer in microcoulombs to three signiﬁcant figures. ANSWER: ‘ : Egg} #0. \ Part C What is the applied voltage, mi? s
i Hint C.l How to approach the problem
i Reduce the network of capacitors to a single equivalent capacitor connecting points a and b directlyi Determine the capacitance (315* and charge {34% on the equivalent capacitor. Then calculate the total potential diﬁerence V6,». Calculate the equivalent capacitance I f 1’0 Part (2.2
3 Calculate the equivalent capacitance if“, of the network
3 Pan C.2.a 1 section Calculate the equivalent capacitance {3‘ of the ca acitors in the initial section between points a and d.
as P Calculate the equivalent capacitance of the ini Hint C.2.a.i Capacitance for capacitors in parallel The total capacitance for two capacitors with capacitances C?" and (,2; connected in parallel is given by CM“; 0“ {I . Express your answer in microfarads to three signiﬁcant figures. ANSWER:  (Tag : Qbm 3&7 V Hint C.2.b Capacitance for capacitors in series : . . . . , . . i 1 . 1
j The total capacitance for two capacitors in series, {3, and f9, is given by FEW 3;» a Remember to solve for Swat and not its reciprocal!
amt » a «a 3 Express your answer in microfarads to three signiﬁcant figures. ANSWER: {to = 3:21 a? To pm C,3 Find the total charge ind the total charge (3 stored in the equivalent capacitor. Hint C.3.a Finding the charge of the equivalent capacitor
Remember that since the section between points a and d and the section between points d and b are in series, the charge on the equivalent capacitor will be the same as the charge stored in each section. Using the value of
for the third capacitor in the network, as calculated in Part B, you can show that Q“ Q“ :25; (3,3) 2 Q3. ’ Express your answer in microcoulonibs to three signiﬁcant ﬁgures. ANSWER: 53“ = gm 3ng Express your answer in volts to three signiﬁcant figures. ANSWER: V ﬁgs = 37;; x: Capacitor With a Dielectric Description: For a capacitor with a dielectric at a given electric ﬁeld strength, calculate the charge per unit area on the conducting plates and dielectric surfaces, as well as the total electricﬁeld energy stored in the capacitor. Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of It: The resultant electric field in the E
t
E
E i dielectric is $38 x 1% volts per meter. Part A
3 Compute the magnitude of the charge per unit area a» on the conducting plate 5 Hint A.l How to approach the problem g Use the dielectric constant to detemiine what the ﬁeld strength would be in the capacitor if the dielectric were not there; then use this to ﬁnd the charge density on the capacitor plates.
i 4 of19 10/31/2008 11:21 MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assignmentPrint?assignmen.. , Part A.2 Calculate the electric ﬁeld Without the dielectric
Calculate the magnitude of the electric ﬁeld 1425 if the dielectric were not present between the conducting plates. Hint A.2.a Electric ﬁeld With a dielectric If a capacitor has a dielectric added between the plates, the new electric ﬁeld strength will be related to the electric ﬁeld strength without the dielectric by ~ The dielectric will decrease the electric ﬁeld by a factor of ‘ it” because the dielectric will become polarized, creating a ﬁeld in the opposite direction to the ﬁeld created by the charges on the capacitor plates. Express your answer in volts per meter to three signiﬁcant ﬁgures. ANSWER‘ 23;, = more“ Vg’m i
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g: 752 Him A,3 Charge density in a parallelplate capacitor
3 For a capacitor (without a dielectric), the electric ﬁeld is related to the charge density on the plates at by .52.; afar If you want to prove this for yourself, use a Gaussian surface that is a right cylinder with ends of area AC, : with one end parallel to the capacitor plates between the two plates and the other end inside one of the plates (again parallel to the plates). Since the side walls are parallel to the ﬁeld, the ﬂux through the side walls is zero. Also,
v since one end is inside the conducting plate, the ﬂux through that end is also zero. Therefore, since the enclosed charge inside the Gaussian cylinder is mic, and since the end between the plates is perpendicular to the ﬁeld, Gauss's law gives Eric omits. or E m gym. 3 Express your answer in coulombs per square meter to three signiﬁcant ﬁgures ANSWER‘ ; rs = 3.52.30“ ‘33:)” Part B
Compute the rmgnitude of the charge per mil area r}; on the surfaces of the dielectric. Hint 3.1 How to approach the problem For a capacitor with a dielectric, the magnitude of the electric ﬁeld is given by m amiss, where if: is the mgnitude of the ﬁeld with the dielectric inserted, and ﬂm is the net charge density at the surface where the plate and 3the dielectric meet, The net charge density is decreased by the presence of the dielectric because the electric ﬁeld induces an opposite signed charge on the surface of the dielectric to oppose the charges on the capacitor plates. Part 32 Find the net surface charge density The result If gym,ng holds for any two charged plates. In the case of the capacitor with a dielectric, you essentially have a pair of charged plates, where each "plate" consists of the charge on the capacitor plate plus the net charge induced on the surface of the dielectric next to the plate. So, knowing the electric ﬁeld, you can detemiine the surface charge density new of the effective plate created by the capacitor plate and the surface of the dielectric.
What is am here? xpress your answer in coulombs per square meter to three signiﬁcant figures. ANSWER: ma : tigers'3; Cm"? Nouticeihat at,“ is less than 3‘} that you calculated for the capacitor plate. This rmkes sense, because the induced surface charge density a; on the dielectric has the opposite of surface chargevr‘on the
capacitor plate, and at,” Express your answer using three signiﬁcant ﬁgures. i
g ANSWER= ; a, = more Cine Note that the charges on the dielectric will be polarized to counteract the charges (and electric ﬁeld) created by the capacitor. For example, near the positive surface of the capacitor the dielectric will have a negative charge.
However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. W \ Part C Find the total electricﬁeld energy 3; stored in the capacitor. ‘Hint C.l How to approach the problem
A Calculate the electricﬁeld energy density in the dielectric; then use the volume in the dielectric to calculate the total electricﬁeld energy stored in the capacitors 1"III Part C_2 Calculate the electricﬁeld energy density
alculate the electricﬁeld energy density is in the capacitor with the dielectric. Hint C. .a Equation {or the electri ﬁeld eneigy deity ' The electricﬁeld energy density for a capacitor with a dielectric is given by a a: éh’rﬁﬁg. v : Express your answer in joules per cubic meter to three signiﬁcant ﬁgures. ANSWER: a = 32,9 Sim“ 1? Hint Q3 Volume where the electric ﬁeld exists For parallelplate capacitors, you assume that the electric ﬁeld is uniform in the space between the plates and is zero outside. This is a good approximation, as long as the plates are mmh larger than the distance separating them,
as in this problem Thus, use the volume of the space between the plates 1/ an go, where if is the volume of the space between the plates, at is the area of one of the plates, and it is the distance between the plates. Recall that it is just the formula for the volume of a prism, which you should be able to see is the shape of the space between the plates. Express your answer in joules to three signiﬁcant ﬁgures. 5 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmeiL.. t,
E
a
t ANSWER f l?=3.t12t§tz“y‘ .i Because of the dielectric, there is le‘ssvenergy stored than if there dielectric Since already know P gajlf, it follows electric energy stored inthe capacitor will be reduced by a factor of A" from the value if the dielectric were not present. of a Ca acitor Description: Two questions about the change in emery stored in a capacitor. First, a dielectric is inserted while no external voltage source is connected. Second, the dielectric is removed with an external voltage source connected
across the capacitor. parallel—plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is removed, and then a dielectric material with
ielectric constant A” is inserted into the capacitor, ﬁlling the space betweenthe plates Finally, the capacitor is fully discharged through a resistor (which is connected cross the capacitor terminals). Find ij}, the the energy dissipated in the resistor. Hint A.l How to approach the question
The energy dissipated in the resistor equals the energy stored in the dielectricﬁlled capacitor. Hint A.2 Energy of a capacitor
The electric potential energy of a capacitor with capacitance {f and charge Q is given by Hint A.3 Effects of the dielectric Inserting a dielectric into the capacitor in the manner described (with the charging battery removed) does not change the charge on the capacitor since its plate is isolated. However, the presence of the dielectric does increase the
capacitance by a factor of If . Express your awer in terms of {3' and other given quantities. ANSWER: ’ H i U, = m It" The energy of the capacitor in this case drops from to {s g k as the dielectric plate is inserted. This energy oss is associated with mutual attraction of the plate and the capacitor. As the plate goes into the capacitor, the g potential energy of the "capacitor and plate system" decreases, much like the potential energy ofa stretched Part B Consider the san‘e situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted. The battery is then disconnected and the capacitor is discharged. For this situation, what is 1);,
t the energy dissipated in the resistor? Hint 3.1 Energy of a capacitor
The potential energy of a capacitor with capacitance {’ and charge {:3 is given by Hint 13.2 Effects oi the dielectric
The battery maintains a constant voltage between the plates but, as before, the capacitance increases by a factor of It” . Express your answer in terms of if} and other given quantities. 6 of19 10/31/200811121AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigrunentPrint?assigmnen... ANSWER: : = g;}( In this case, the energy increase comes from the battery. The battery does positive work on the capacitor by moving more electrons ﬁ'om one plate to another to maintain the constant potential difference as the capacitance
'ncreases. M y , Problem 23.16 Description: (a) Three point charges +11 and a fourth, q are assembled to form a square of side a. Find an expression for the electrostatic energy of this charge distribution Part A
Three point charges r; and a fourth, m g are assembled to form a square of side gr. Find an expression for the electrostatic energy of this charge distribution Express your answer in terms of gt, a, and constant ANSWER: g = 0 Problem 23.32 i Description: (a) What is the equivalent capacitance of the combination shown in the ﬁgure ? (b) If a 100V battery is comected across the combination, what is the charge on each capacitor? (c) What is the voltage across each? , Part A What is the equivalent capacitance of the combination shown in the ﬁgure ‘2 Express your answer using two signiﬁcant ﬁgures. .L (:1 7‘": ct = (3.01pm (:3 0.02 as {7' : {one it? 3 Part B If a 100— V battery is connected across the combination, what is the charge on each capacitor? 1 Enter your answers numerically separated by commas, wing two signiﬁcant figures. ANSWER: as}, {gs = 3,2 0,41": use git“; Part C
i What is the voltage across each? Enter your answers numerically separated by commas, using two signiﬁcant figures. ANSWER: ' Vi, v.2, v.33: (so at) «to ‘v’ t Description: Three simple water "circuits" consisting of a pump (battery) and pipes (resistors) in series and in parallel. Simple multiplechoice questions elicit the analogy of voltage to pressure and water ﬂow to current. Learning Goal: To understand the analogy between water pressure, water ﬂow, voltage, and current i
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i i ’ As suggested by the fact that we call both currents, the ﬂow of charged particles through an electrical circuit is analogous in some ways to the ﬂow of water through a pipe. § When water ﬂows ﬁom a smll pipe to a large pipe, the ﬂow (measured, for instance, in gallons per minute) is the same in both pipes, because the amount of water entering one pipe must equal the amount leaving the other. If not,
: water would accumulate in the pipes. For the same reason, the total electric current I is constant for circuit elements in series. Water pressure is analogous to total electric potential (voltage), and a pump is analogous to a battery. Water ﬂowing through pipes loses pressure, just as current ﬂowing through a resistor falls to lower voltage. A pump uses
mechanical work to raise the water's pressure and thus its potential energy; in a battery, chemical reactions cause charges to ﬂow against the average local electric ﬁeld, ﬁ'orn low to high voltage, increasing their potential energy. PartA E Consider the following water circuit: water is continually pumped to high pressure by a pump, and then ﬁmnelled into a pipe that has lower pressure at its far end (else
the water would not ﬂow through the pipe) and back to the pump. Two such circuits are identical, except for one diﬁerence: the pipes in one circuit have a larger diameter than the pipes in the other circuit. Through which circuit is the ﬂow of water greater? 7of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://sessi0nmasteringphysics.com/myct/assignmentPrint?assignmen... zero pressure ,2 ANSWER: Small pipe
0 Large pipe The cross sectional area of the pipe is analogous to the area of a wire: the smaller the area the higher the resistance and the more the pipe/wire impedes ﬂow. If the change in pressure (proportional to the potential energy per
unit mass) of water traveling through two pipes is the same, the ﬂow will be less through the pipe with smaller cross sectional area. The electrical analog is Ohm's law I a where resistance R is inversely proportional o the area of a wire/resistor. Now consider a variant on the circuit. The water is pumped to high pressure, but the water then faces a fork in the pipe. Two pipes lead back to the pump: large pipe L and small pipe S. Since the water can ﬂow through either pipe,
the pipes are said to be in parallel: The overall ﬂow of water that enters the system before the fork is equal to high reignite we present * ‘ Hint 3.1 Water conservation Conserve water! A particular molecule of water ahead of the fork must go through one pipe or the other. ANSWER: V the ﬂow through pipe L.
' the sumofthe ﬂows throughLand S.
the average ofthe ﬂows through L and S. ? *3 Part c
5 What can you say about the drop in potential energy (per unit mass or volume) of water traveling through either pipe? ANSWER: j The drop is greater for pipe L.
The drop is greater for pipe 8.
g The drop is the same for both pipes. The pressure IS equal at the top of both pipes (to the pressure created by the pump) and at the bottom of (takenvto Vzero‘).vso thepressmevrlrop across each pipe is only ﬂows are different. This circuit IS
the analog of resistors in parallel, where the voltage is the same for both resistors, but the currents differ if the resistances are unequal. PartD Consider a new circuit: water is pumped to high pressure and fed into only one pipe. The pipe has two distinct segments of diﬁ'erent diameters; the second half of the pipe has a smaller diameter than the ﬁrst half:
Which of the following statements about the ﬂow and change in pressure through each segment is true? hi it pressure pressure ANSWER: The ﬂow through each segment is the same as the overall ﬂow; the change in pressure through each segment is the same as the overall change.
. The ﬂow through each segment is the same as the overall ﬂow; the sum of the changes in pressure through each segment equals the overall change.
The sum of the ﬂows through each segment equals the overall ﬂow; the change in pressm‘e through each segment is the same as the overall change.
The sum of the ﬂows through each segment equals the overall ﬂow; the sum of the changes in pressure through each segment equals the overall change. 80f19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen... Introduction to Electric Current Description: Mostly conceptual questions about electric current in metals Learning Goal: To understand the nature of electric current and the conditions under which it exists. Electric current is deﬁned as the motion of electric charge through a conductor. Conductors are materials that contain movable charged particles. In metals, the most commonly used conductors, such charged particles are electrons. The more electrons that pass through a cross section of a conductor per second, the greater the current. The conventional deﬁnition of current is where I is the current in a conductor and (9%,; is the total charge passing through a cross section of the conductor during the time interval Sat. E
i The motion of free electrons in metals not subjected to an electric ﬁeld is random: Even though the electrons move fairly rapidly, the net result of such motion is that QSM m i) (Let, equal numbers of electrons pass through the cross section in opposite directions). However, when an electric ﬁeld is imposed, the electrons continue in their random motion, but in addition, they tend to move in the direction of the force applied by the electric ﬁeld. A g In summary, the two conditions for electric current in a material are the presence of movable charged particles in the material and the presence of an electric ﬁeld. Quantitatively, the motion of electrons under the inﬂuence of an electric ﬁeld is described by the drift speed, which tends to be much smaller than the speed of the random motion of the electrons. The number of electrons passing
g through a cross section of a conductor depends on the drift speed (which, in turn, is determined by both the microscopic structure of the material and the electric ﬁeld) and the crosssectional area of the conductor. In this problem, you will he offered several conceptual questions that will help you gain an understanding of electric current in metals. i
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. PartA You are presented with several long cylinders made of different materials. Which of them are likely to be good conductors of electric current? E Check all that apply. ANSWER: . copper  aluminum glass quartz cork plywood table salt
. gold " i3 Part3 Metals are good conductors of electric current for which of the following reasons? ANSWER: ’ They possess high concentrations of protons.
' They possess low concentrations of protons. ‘ They possess high concentrations of free electrons. They possess low concentrations of ﬁ'ee electrons. Part C
Which of the following is the most likely driﬁ speed of the electrons in the ﬁlament of a light bulb? E
i ANSWER: i0J mféi
10% mfs Part D You are presented with several wires trade of the same conducting material. The radius and driti speed are given for each wire in terms of sonic unknown units r and 9. Rank the wires in order of decreasing electron current. Hint D.1 What the wires have in common Since the wires are rmde of the same material, the charge carriers and their densities are the same for all the wires. ' Hint 0.2 A formula for electric cinrent
Other conditions being equal, the current is proportional to the product of the cross‘sectional area of the wire and the drift velocity, that is, f m sigma
Where I is the current, ed is the driﬁ velocity, ,4 is the crosssectional area, a; is the density of charge carriers, and q is the charge on the carriers I Rank from most to least electron current. To rank items as equivalent, overlap them.
3 ANSWER: 9 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... . View Part E
The driﬁ speed of the electrons in a wire depends strongly on which of the following factors? ANSWER: the crosssectional area of the wire
the mass of the wire
the temperature of the wire
0 the internal electric ﬁeld inthe wire iv Part F
What quality must the charge density on the surface of a conducting wire possess if an electric ﬁeld is to act on the negatively charged electrons inside the wire? 2
3
% ANSWER: positive. negative. 3 The charge density must be .
. I nonuniform. Down To The Wire Description: Given the current in a wire and the characteristics of the material of the wire, ﬁnd the current density, drift velocity of the electrons in the wire and the average time between the collisions. current of I a: $13 A is ﬂowing in a typical extension cord oflength L a: 35X} m. The cord is made ofcopper wire with diameter :33 ,5} {M}. The charge of the electron is (3 34¢; X m» to (‘7‘ The resisitivity of copper is p 1.”? x it)”x i? ~ m. The concentration of free electrons in copper is 1: 85:3 28 if)?“ to”? Find the driﬂ velocity {36 ofthe electrons in the wire. 2
E Find the current density ﬁrst
Find 3', the magnitude ofthe current density inthis wire. Diameter and area sing the diameter of the wire, ﬁnd its crosssectional area. Express your answer in amperes per meter squared. Use two signiﬁcant figures. ANSWER: J = «ﬁrms A2115: Current density and the drift speed The current density is proportional to the drilt speed: J W n ‘ Express your answer in meters per second, to two signiﬁcant ﬁgures. ANSWER up 2 3.34:)” mfg 3
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3
; Note that this wire is carrying more current density than is carried by most household wiring in everyday use. With the given amount of current ﬂowing, the cord would be hot to the touch if it were under a rug or had otherwise
estricted air ﬂow around it. It would certainly be considered unsafe by standard electrical safety codes. Even though this wire is carrying a large amount of current for its size, the driﬁ velocity of the electrons is tiny (less than one millimeter per second) This reﬂects the fact that there is a huge number of free (mobile) electrons
n the wire. Let us illustrate this fact with a calculation. S ’3 Part B l The population of the Earth is roughly six billion people. If all free electrons contained in this extension card are evenly split among the humans, how many free electrons ( My) would each person get? z
3 Part 3.1 Find the volume ﬁrst
Find the volume V ofthe wire. Express your answer in cubic meters. Use two signiﬁcant figures. ANSWER: V V: Knows ms Use two signiﬁcant ﬁgures. 10 ofl9 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://sessi0n.masteringphysics.com/Inyct/assigmnentPrint?assignmen... Find the total number of collisions ( 3;.) that all free electrons in this extension cord undergo in one second. Hint C.l Consider a single electron How many collisions per second does each electron undergo per second? Part C.2 Find the time between collisions
Find the algebraic expression for the mean time between collisions 3. Express your answerin terms of p, the mass of the electron in, the charge of the electron and the concentration of the electrons rt. ANSWER: ' ' m
‘ pros? ANSWER: . A; = 13313”?
Note that 2' does not depend on the applied driﬁ speedihowever, does. Description: Derive a formula for power delivered to a resistor in terms of the resistance and the voltage across the resistor or current through the resistor. %
In this problem you will derive two diﬂ‘erent formulas for the power delivered to a resistor. Part A What is the power 1‘" supplied to a resistor whose resistance is {*2 when it is known that it has a voltage V across it? Part A.l Find an expression for power
What is the power I” supplied to any circuit element that has a voltage if across it and through which a current I ﬂows? Express the powerin terms of I and V. ANSWER: p = V1 Part AZ Find current in term of voltage and resistance Using Ohm‘s law, what is the current I through a resistor of resistance 3? that has a voltage 1" applied across it? 2 ANSWER: Express the power i” interms of If and V. , ANSWER: 7 i i I V2
I} : at“. I? ‘5‘! Part B
What is the power I" supplied to a resistor whose resistance is i? when it is known that it has a current ,i‘ ﬂowing through it? Part 3.1 Find an expression for power
What is the power 9‘ supplied to any circuit element that has a voltage V across it and through which a current 3 ﬂows? Express the power in terms of f and 1". ANSWER: p '=' g; Part 8.2 Find voltage in tern of current and resistance ' 2 Using Ohm's law, what is the voltage across a resistor of resistance If that has a current 3' ﬂowing through it? ANSWER: ; 1;; ' Express the power 13 in terms of 1% and I. ANSWER: p : pig? 110f19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignInen... Note that both expressions for power involve the square of some electric variable: t” or P}. This is typical of energy and power fortmilast.hey involve the squares of the amplitudes of the dynamic variables. Note also that there is no factor of 1/2 in these expressions as there is in many energy (but not power) formulas. “knife”; 24.30 Description: A 4.5W ﬂashlight bulb draws 750 mA (3) At what voltage does it operate? (b) What is its resistance?
A 4.5W ﬂashlight bulb draws 750 ﬁrst. Part A At what voltage does it operate? Express your answer using two signiﬁcant ﬁgures. ANSWER: : V: (H; is What is its resistance? Express your answer using two significant ﬁgures. ANSWER: ’ I ’3: 0 ; Description: Students are required to calculate the current through a hurmn body when subject to a low potential difference. 3 ‘ Most of us have experienced an electrical shock one way or another in our lives Most electrical shocks we receive are minor ones from wooly sweaters or from shoes. However, some shocks, especially ﬁ'om outlets or power g
mains, can be fatal. This question will show you how to estirmte the current through a human body when subject to an electrical shock. ‘ n'agine a situation in which a person accidentally touches an electrical socket with both hands. By modeling the arm and the chest to be a cylindrical tube with a total length l. iijl to, cross—sectional area ,4} m it} can“, and esistivity 5; a: 1,5, 01m; » m, you can calculate the current in amperes through the person when a potential difference of V w: iii) ‘5 is applied across the two hands. Assume that the current ﬂows only through the modeled
ylindrical tube. Part A
What is the current ﬂow through the body? Hint A.1 Calculating the resistance
To calculate the formula for resistance using resistivity, one needs the following formula: I.
Rm??? where I? is the resistance, ,a is the resistivity, L is the length of the resistor, and A is the crosssectional area of the resistor. Part A.2 Obtaining the expression for current
g Having worked out the resistance, apply Ohm's law to obtain an expression for I in terms of V, A, p, and Z,. ANSWER: Take care when calculating the numerical value for I . Remember that i m3 a to" are m tiresome“. Express your answer numerically to two signiﬁcant ﬁgures. ANSWER: ' f: 95337 A :l'henfollowing are‘the of current on humans. 0 1 mA 1 it?“ A or less: barely noticeable; I 1 to 8 mA: strong surprise;
' 8 to 15 mA: unpleasant, victims able to detach from source;
I 15 to 75mA: painful, dangerous;
0 75 mA or more: fatal.
These values vary according to sex, age, and weight. An Introduction to EMF and Circuits Description: Several basic conceptual and computational questions involving the concepts of emf, internal resistance, and simple circuits (batteryresistor). i '; Learning Goal: To understand the concept of electromotive force and internal resistance; to understand the processes in Oneloop circuits; to become familiar with the use of the ammeter and voltmeten order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, a closed path through which the charged particles can move without creating a "buildup." Such buildup, if it occurs, creates
ts own electric ﬁeld that cancels out the external electric ﬁeld, ultimately causing the current to stop. v 12 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrinﬂassignmen... However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a source of energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potential energy. In fact, electrostatic forces always push the particles in the direction that leads to a decrease in potential energy. At some point, each charged particle would reach the location in the circuit where it has the lowest possible potential energy.
g How can such a particle move toward a point where it would have a higher potential energy? Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higher potential energy despite the presence of electrostatic forces. In circuits, such forces exist inside a device commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the charged particles in continuous motion by increasing their potential energy through the action of some kind of nonelectrostatic force. The armunt of work that the battery does on each coulomb of charge that it "pushes through" is called (inappropriately) the electramotivefarce (pronounced "eeem—ef' and abbreviated emf or denoted by 5). Batteries are oﬁen
eferred to as sources of emf (rather than sources of energy, even though they are, ﬁindamentally, sources of energy). The emf of a battery can be calculated using the deﬁnition mentioned above: 5 m: Wfq. The units of emf are
oules per coulomb, that is, volts. The terminals of a battery are oﬁen labeled and for "higher potential" and "lower potential," respectively. The potential difference between the terminals is called the terminal voltage of the battery. If no current is running through a battery, the terminal voltage is equal to the emf of the battery: VS, 3‘. owever, if there is a current in the circuit, the terminal voltage is less than the emf because the battery has its own internal resistance (usually labeled 3‘). When charge r; passes through the battery, the battery does the amount of work Sc; on the charge; however, the charge also "loses" the amount of energy equal to [r ( I is the current through the circuit); therefore, the increase in potential energy is Sq m 41¢, and the terminal voltage is trig, mg “Ir. n order to answer the questions that follow, you should ﬁrst review the meaning of the symbols describing various elements of the circuit, including the amrneter and the voltmeter; you should also know the way the ammeter and the
oltmeter must be connected to the rest of the circuit in order to function properly. ‘ 3 Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol. It is important to keep in mind that this reSistor With reSistance r is actually inSide the battery.
; In all diagrams, 6.? stands for emf, r for the internal resistance of the battery, and f? for the resistance of the external circuit As usual, we'll assure that the connecting wires have negligible resistance. We will also assume that both 3
e ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance, and the voltmeter has a very large resistance. ( i Pan A ,, .. , ,,, E For the circuit shown in the diagram , which potential difference corresponds to the terminal voltage of the battery? ANSWER: between points K and L
j between points L and M
: . between points K and M Keep in mind that the "resistor" with resistance r is actually inside the battery. The next several questions refer to the four diagrams shown here labeled A, B, C, and D. \ 3? Part B
i In which diaganﬁs) (labeled A  D) does the ammeter correctly measure the current through the battery? Hint 3.1 How an ammeter works Recall that an ammeter works by the current that is being measured passing through the meter, which has a very low internal resistance compared to the rest of the circuit. Therefore, the ammeter does not have too great an effect
on the circuit ﬁ—om being a resistor placed in series (instead, one can consider it as just an extra section of wire). Enter the letter(s) of the comet diagranﬁs) in alphabetical order. For example if both A and C are correct enterAc. ANSWER: ; CD i ’5' Part C
I In which diagram is the current through the battery nearly zero? ; . Hint C.1 How to approach the problem 3 ' In order to determine the current through the battery we must follow the current loop through the circuit. Whichever loop has the highest resistance will have the lowest current. Keep in mind that the voltmeter has a very high
'nternal resistance. 13 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmert.. ANSWER: ‘ A § Dia am A is the only one in which the current through the battery is the same as the current through the voltmeter. Since the latter has a very large resistance, this current is essentially zero. *5? Part 1) In which diagram or diagrams does the arraneter correctly measure the current through the resistor with resistance R? Hint D.l How to approach the problem §
§ ' Note that current is conserved through a wire, and in order for an ammeter to measure the correct current passing through an element, it must be in series with that element. Enter the letteds) of the correct diagranﬁs) in alphabetical order. For example if both A and C are correct enterAc. ANSWER: V CD E
z 355' Part E
In which diagram does the voltmeter correctly measure the terminal voltage of the battery? Choose the best answer. A In diagrams A and B, the voltmeter readings would actually be quite close to the terrm‘nal voltage if the amineter has a very low resistance, and the voltmeter, a very high one. However, diagram C clearly shows the best way
to connect the voltmeter in order to measure the terminal voltage. f8? Part F
In which diagram does the voltmeter read almost zero? Enter the letterts) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enter AC. % The voltriiete‘ruin diagram D is connecédrto two points that are also by a wire presumably, low resistance. Therefore, charge ﬂowing throughthat wire will not lose appreciable amount I H §§ potential energy, and the potential difference (voltage) is ne 1 In which diagram or diagrams does the ammeter read almost zero? Enter the letteds) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enterAc. ANSWER: f. A]; In diagram A, the voltmeter is connected in series with the battery. Since the voltmeter has a very large resistance there is no (or nearly zero) current in the whole cireui Therefore, the arrrmeter reads no current. In diagram B, \
\ l S' the ‘tanc {the ltmeterisve lar thecun'entisnearl zero x the current through the ammeter is the same as the current through the 33m last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohrm. 1” Part H ' A voltmeter is connected to the terminals of the battery; the battery is not connected to any other external circuit elements. What is the reading of the voltmeter i2"? §
E
t
.: Express your awerinvolts. Use three signiﬁcant ﬁgures. ANSWER: : V = i333} \g‘ 39' Part I
The voltmeter is now removed and a 21 .0—ohm resistor is connected to the terminals of the battery. What is the current 1 throughthe battery? Express your answer in amperes. Use two signiﬁcant ﬁgures. ANSWER: I I= cm A is? Part J
In the situation described in Part I, what is the current I through the 21 .Ovohm resistor? 5 Express your answer in amperes. Use two significant figures. ANSWER . I= use a. é
2,
§
5 14 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/rnyct/assignmentPrint?assignmen... g Since the battery and the external resistor form one loop, the charge that passes through one must pass through another; therefore, the currents must be the same. How to approach the problem 2
I What is the potential difference across the 21.0ohm resistor from Part ['7
g ' Hint K.1 The best way to ﬁnd the potential difference V across a resistor 8 when a current I is ﬂowing is to use Ohm's law: ’ Express your arswer in volts. Use three signiﬁcant figures. ANSWER: y = “£3 3.: Kirchhoﬁ‘s voltage law
Kirchhoﬂ‘s voltage law states that the voltage difference across all the elements in a circuit (in this case just one resistor) is equal to the voltage at the temiinals fromthe source (in this case a battery). Express your answer in volts. Use three signiﬁcant figure ANSWER: i V = ms v % Since the ends of the resistor with resistance R are attached to the terminals of the battery, the voltage across the reSistor is the same as that between the terminals of the battery. ' is? Pm M How much work it” does the battery connected to the 21 .0<ohm resistor perform in one minute? Hint M.l How to approach the problem
Find the charge that passes through the battery, and then use the deﬁnition of emf. Part M.2 Find the charge
How much charge r; passes through the battery in one minute? Hint M.2.a Deﬁnition of current . Recall that current is deﬁned as the number of units of charge that pass through a wire per second. Express your answer in coulombs. Use three signiﬁcant ﬁgures. ANSWER: q : am <3 Express your answer in joules. Use three signiﬁcant ﬁgures. ANSWER: 3 a.“ : 3g} 3 Series And Parallel Connections Description: Several calculations of increasing complexity that help the students practice ﬁnding the equivalent resistance of the circuits combining series and parallel connections. Learning Goal: To learn to calculate the equivalent resistance of the circuits combining series and parallel connections. 1" Resistors are oﬁen connected to each other in electric circuits. Finding the equivalent resistance of combinations of resistors is a common and important task. Equivalent resistance is deﬁned as the single resistance that'can replace the given combination of resistors in such a manner that the currents in the rest of the circuit do not change. ‘ Finding the equivalent resistance is relatively straighforward if the circuit contains only series and parallel connections of resistors. £ An example of a series connection is shown in the diagram; For such a connection, the current is the same for all individual resistors and the total voltage is the sum of the voltages across the individual resistors. I Rt R3 12‘ E
’4‘ 1
WW WV «WW
% g An example of a parallel connection is shown in the diagram For resistors connected in parallel the voltage is the same for all individual resistors because they are all connected to the same two points (A and B on the diagram). The total current is the stirn of the currents through :
i the individual resistors. This shotild makes sense as the total current "splits" at points A and B. § R1 3 Using Ohm's law, one can show that. for a parallel connection, the reciprocal of the equivalent resistance is I, g it 5 g 15 of19 10/31/2008 11:21 MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... the sum of the reciprocals of the individual resistances. ) Mathematically, these relationships can be written as:
% vwnmnmamm 1 I: +13 ~§~ :5 «in ; “swunﬂnncusswﬂt g wi—ijsm‘m Rag“WMIM 5?; Rx 3;. ‘ Part A For the combination of resistors shown, ﬁnd the equivalent resistance between points A and B. ( Express your answer in Ohms. m 3m 43}
AH‘VX WNW“ ANSWER: ft“ = 5} g; § These resistors are connected in series; the current through each is the same Part B For the set—up shown, ﬁnd the equivalent resistance between points A and B. Express your answer in Ohms. 652 3S2 ANSWER: {gm = g 52 ,, I _ . . . , , . . V, V, " This 15 a parallel connection Since the voltage across each reststor IS the same. 19‘ Part C
K For the combination of resistors shown, ﬁnd the equivalent resistance between points A and B. 4f!
20 Kt
12$} Hint C.l How to approach the question
You cannot say that all three resistors are connected either in series or in parallel: this circuit has to be viewed as a combination of different connections. Find the equivalent resistance of the "40hm—12 Ohm" combination ﬁrst. : Part C.2 What kind of connection is this? :1
§
§ 16 0f19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen.. The 20hm resistor is connected: ANSWER: in series with the 4Ohm resistor
in series with the 12Ohm resistor
in series with the combination of the 4Ohm and the 12—Ohm resistors
in parallel with the 40hm resistor
in parallel with the lZOhm resistor
in parallel with the combination of the 40hm and the 12Ohm resistors Express your answer in Ohms. ANSWER: .' g” 9' H \ In this case, you cannot say that all three resistors are connected either in series or in parallel. You have a combination of a series and a parallel connection §
\
s Some circuits rmy contain a large number of resistors connected in various ways. To determine the equivalent resistance of such circuits, you have to take several steps, careﬁilly selecting the "subcombinations" of resistors
connected in relatively obvious ways. Good recordkeeping is essential here. § The next question helps you practice this skill. ‘8 Part D For the combination of resistors shown, ﬁnd the equivalent resistance between points A and B. Z
i
i me i Hint D.l How to approach the question Find separately the equivalent resistances of the top and the bottom branches of the circuit; then combine them Part D.2 Find It” for the "4612" combination What is the equivalent resistance for the "4 ohm  6 ohm  12 Ohm" combination? Express your answer in ohms. ANSWER: gm 9.3 Find am for the top branch 5 What is the equivalent resistance for the top branch of the circuit (between C and D)? Express your answer in ohms. ANSWER: gem”, = g; Q Part 0.4 Find Km for the bottom branch
What is the equivalent resistance for the bottom branch of the circuit (between E and F)? Express your answer in ohm. ANSWER: Rama». = 6? $2 Express your answer in Ohm. ANSWER: ; 3 gr The a circuit is to determine the voltages across and the through the various branches of the circuit. You will practice diatskill in the V 7 Of course, there are circuits that cannot possibly be represented as combinations of series and parallel connections. However, there are ways to analyze those, too. Problem 25.16 Description: Resistors R‘l and R_2 are connected in series, and this series combination is in parallel with R_3. This parallel combination is connected across a battery. (3) Choose the correct diagram representing this circuit. esistors m ard 1%: are connected in series, and this series combination is in parallel with R3. This parallel combination is connected across a battery. Part A 17 0f19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/Inyc‘r/assignmentPrint?assignmen... Choose the correct diagram representing this circuit. E
s
i ANSWER: \wl’roblem 25.48 Description: (a) In the circuit of the ﬁgure , ﬁnd the current supplied by the battery. (b) Find the current through the 6Omega resistor. Part A
In the circuit of the ﬁgure , ﬁnd the current supplied by the battery. g.
; ANSWER: : Ii: 3‘5? 2»; Part B
Find the current through the 69 resistor. ANSWER: > {ﬁg 2 m3 3311A : Problem 25.81 s . ....... .. , .......... .. . .................... .. . . . . . . . , i . . . . . . . . . . . . . .. . .. ......... .. . . i , , , . . . . . . . . . . . . . . . . . . . . . . . . ., . .. .. ...... ,, . . ,, . . ................................................. “1
Description: You have a dilemrm. You're building a small circuit. You need a 3.44(k)0mega resistor. You have three resistors, 3.30 ( k)0mega , 4.70 ( k)0mega , and 1.50 ( k)0rmga . (a) How can you connect them to get the ' required 3.44 ( k)Ornega ‘2 You have a dilemma. You're building a srmll circuit. You need a 3.44%!) resistor. You have three resistors, 3.30 3d} , 4.70 51$ 2 . and 1.50 M2 . Part A
i How can you connect them to get the required 3.44 kg} 7 ANSWER: § 29., ﬁg and m are parallel.
ﬁx, I33 and 133 are put in series. 18 of19 10/31/2008 11:21 AM MasteringPhysics: Assignment Print View http://sessi0n.masteringphysics.com/myct/assigmnentPrint?assignmen... {£3 and [£3 are parallel, R; is added in series to this combination.
(2‘; and ﬁg are parallel, RE is added in series to this combinan'on.
ff; and Rs are parallel, ﬂy, is added in series to this combinaﬁon. > Sumry 0 of 11 items compléte (0% avg. score
0 of 118 points ‘ 19 of19 10/31/2008 11:21 AM ...
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 Spring '08
 Smith,D
 Physics

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