The mass of a water molecule is the mass of two hydrogens and one oxygen: 1u + 1u + 16u = 18u, where. = = 0.05 C/kg m. 65 kg 0.

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1 EECTRIC CHARGE, FORCE, AND FIED 0 EXERCISES Section 0.1 Electic Chage 1. INTERPRET We ll estimate the chage you body would cay if the electon chage slightly diffeed fom the poton chage. DEVEOP Since the human body is about 60% wate, let s assume that the numbe of potons/electons pe kilogam in you body is the same as that of a wate molecule. Wate is hydogen atoms (with one poton and one electon each) and one oygen atom (with 8 potons and 8 electons). If the chages on the electons and potons ae diffeent by one pat in a billion, then the net chage on a wate molecule would be Δ q 10 q q e C C poton electon The mass of a wate molecule is the mass of two hydogens and one oygen: 1u + 1u + 16u 18u, whee 7 1 u kg. Theefoe, the net chage to mass atio fo a wate molecule would be: 7 Δ q C 0.05 C/kg 7 m kg HO EVAUATE We ll assume you mass is 65 kg. If we appoimate this mass as pue wate, then the total chage on you body would be appoimately Δq Δq m m HO 65 kg 0.05 C/kg C ASSESS This is a huge amount of chage. Imagine half of the chage was in you head/chest and the othe half was in you legs, about a mete away. Then the magnitude of the foce of epulsion between the uppe and lowe pats of you body would be This would ip you apat! F 9 1 ( N m /C )( C) kq q N 1 1 ( 1 m) 14. INTERPRET This poblem deals with quantity of chage in a typical lightning flash. We want to epess the quantity in tems of the elementay chage e. 19 DEVEOP Since the magnitude of elementay chage e is e C, the numbe N of electons involved in the lightening flash is given by N Q/e. EVAUATE Substituting the values given in the poblem statement, we find N 5 C 0 Qe / C ASSESS Since 1 coulomb is about elementay chages, ou esult has the ight ode of magnitude. 15. INTERPRET This poblem asks us to find the combination of u and d quaks needed to make a poton, which has a positive unit chage e, and a neuton, which has zeo chage. DEVEOP The u quak has chage e/ and the d quak has chage e/, so we can combine these so the chages sum to unity (fo the poton) o zeo (fo the neuton). 0-1 Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

2 0- Chapte 0 EVAUATE (a) Two u quaks and a d quak make a total chage of (/) 1/ 1, so the poton can be constucted fom the quak combination uud. (b) Two d quaks and a u quak make a total chage of (1/) + / 0, so the neuton can be constucted fom the quak combination udd. ASSESS Because of a phenomenon called colo confinement, quaks can only be found in hadons, of which potons and neutons ae the most stable eamples. 16. INTERPRET This poblem deals with the quantity of net chage Eath caies. We want to epess the quantity in tems of the elementay chage e. 19 DEVEOP Since the chage caied by an electon is q e C, the numbe N of electons caied by Eath is given by N Q/q. EVAUATE Substituting the values given in the poblem statement, we find 5 Q 5 10 C 4 N e C ASSESS Since 1 coulomb is about elementay chages, ou esult has the ight ode of magnitude. 17. INTERPRET This poblem deals with the numbe of electons a honeybee has to lose to acquie a net positive chage of +180 pc. We epess the quantity in tems of the elementay chage e. 19 DEVEOP Since the chage caied by an electon is q e C, the numbe N of electons the honeybee needs to lose to have a net chage of +Q is given by N +Q/e. EVAUATE Substituting the values given in the poblem statement, we find 1 Q C 9 N e C ASSESS Being positively chaged, honeybees ae attacted to the negatively chaged spide webs. Section 0. Coulomb s aw 18. INTERPRET The electon and poton each cay one unit of electic chage, but the sign of the chage is opposite fo the two paticles. Given the distance between them, we ae to find the foce (magnitude and diection) between these paticles. DEVEOP Coulomb s law (Equation 0.1) gives the foce between two paticles. Fo this poblem, 19 q1 q e C and m. Because the chages have opposite signs, the foce will be attactive. EVAUATE The foce between these paticles is attactive and has the magnitude 9 19 ke ( N m /C )( C) 8 F N m ASSESS Because the electon is much less massive than the poton, the electon does most of the acceleating in this two-paticle system. 19. INTERPRET We want to know how fa a poton must be fom an electon to eet an electical attaction equal to the electon s weight on Eath. DEVEOP The foce between a poton and electon has a magnitude given by Coulomb s law (Equation 0.1): 19 F ke /, whee 1 e C. We ll solve this fo the distance,, at which F me g. EVAUATE The distance at which the electical foce is equal to the gavitational foce is e 9 19 ( N m /C )( C) 1 ( kg)( 9.8 m/s ) ke 5.1 m mg 10 ASSESS On the molecula scale, potons and electons ae oughly 10 m apat. Since the Coulomb attaction scales as 1/, electic foces will clealy ovewhelm any gavity effects. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

3 Electic Chage, Foce, and Field 0-0. INTERPRET The poblem is to calculate the chage on tiny Styofoam pieces. DEVEOP Since the chages ae equal, the epulsion foce at the given distance is F kq EVAUATE Solving fo the chage in Coulomb s law gives ( 0.01 N)( m) F q nc k 9 ( N m /C ) / (Equation 0.1). ASSESS This seems easonable fo small chaged objects. 1. INTERPRET This poblem involves finding the unit vecto associated with the electical foce one chage eets on anothe, given the coodinates of the positions of both chages. DEVEOP A unit vecto fom the position of chage q at q (1 m, 0), to any othe point (, y) is ( q ) ( 1 m, y) n q 1 m + y EVAUATE (a) When the othe chage is at position (1 m, 1 m), the unit vecto is (0,1 m) n (0,1) j 0 + (1 m) 0, 0, (b) When (c) Finally, when ( 1m,0) ( ) + n 1,0 i 1m 0 m, m, the unit vecto is ( 1 m, m) ( 1, ) n 0.16i j 1 m + m 10 The sign of q doesn t affect this unit vecto, but the signs of both chages do detemine whethe the foce eeted by q is epulsive o attactive; that is, in the diection of + n o n. ASSESS The unit vecto always points away fom the chage q located at (1 m, 0).. INTERPRET This poblem equies us to find the foce acting on a poton due to an electon, given the positions of the two paticles. DEVEOP Use the esult of the pevious poblem to find the diection of the foce acting on the poton. The elevant quantities ae q ( 0,0) fo the poton and ( 0.41nm, 0.6 nm), so the unit vecto indicating the diection of the foce is ( q ) ( 0.41 m, 0.6 nm) n 0.75i j q 0.41 m nm Because the chages ae opposite in sign, the foce will act in the positive n diection. The magnitude of the foce may be found using Coulomb s law (Equation 0.1). EVAUATE The magnitude of the foce is 9 19 ke (9 10 N m /C )( C) 10 Fp N 18 ( ) 10 m so the complete (i.e., vecto) foce is ( i + 5.1j) 10 N ASSESS This foce is two odes of magnitude less than the foce between the electon and poton in a hydogen 1 atom. The diection of the foce is θ tan ( ) 41 above the ais. Section 0. The Electic Field. INTERPRET This poblem is about calculating the electic field stength due to a souce, when the foce epeienced by the electon is known. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

4 0-4 Chapte 0 DEVEOP Equation 0.a shows that the electic field stength (magnitude of the field) at a point is equal to the foce pe unit chage that would be epeienced by a chage at that point: E F / q. EVAUATE With q, e we find the field stength to be 9 F N 9 E N/C e C ASSESS Since the chage of electon is negative, the electic field will point in the opposite diection as the foce. 4. INTERPRET This poblem asks us to find the magnitude of the foce on an electic chage that is placed in an electic field of the given stength. DEVEOP The electic field is the foce pe unit chage, so we multiply the magnitude of the chage by that of the electic field to find the foce (see Equation 0.b). 6 4 EVAUATE The magnitude of the foce is F qe (.0 10 C)( 100 N/C).0 10 N. ASSESS The diection of this foce will be in the same diection as that of the electic field because the chage is positive. 5. INTERPRET This poblem involves calculating the electic field needed to poduce the given foce on the given chage, and then find the foce epeienced by a second chage is the same electic field. DEVEOP Equation 0.a shows that the electic field stength (magnitude of the field) at a point is equal to the foce pe unit chage that would be epeienced by a chage at that point: E F / q. The equation allows us to calculate E give that F 150 mn and q 68 nc. Fo pat (b), the foce epeienced by anothe chage q in the same field is given by Equation 0.b: F q E. EVAUATE (a) With q 68 nc, we find the field stength to be F E e 68 nc (b) The foce epeienced by a chage q 5 μc in the same field is 150 mn N/C 6 F qe 5 μc.1 10 N/C 77 N ASSESS The foce a test chage paticle epeiences is popotional to the magnitude of the test chage. In ou poblem, since q 5 μc> q 68 nc, we find F > F. 6. INTERPRET We want the foce on an ion inside a cell with an intenal electic field. DEVEOP A singly-chaged ion has a chage of q, e so the magnitude of the foce will be F e E. EVAUATE The foce on the ion in the cell is ASSESS 19 6 F e E C N/C 1. pn Although a pico-newton is vey small, it is a significant foce at the molecula scale. 7. INTERPRET This poblem is simila to Poblem 0.5, in that we ae given the foce eeted on a given chage by an unknown electic field, and we ae to find the foce eeted by this field on anothe chage (a poton, in this case). DEVEOP Apply Equation 0.b E F q to find the electic field, with q 1 μc. The foce on the poton (q p C) is given by Equation 0.a, Fp qpe. Inseting the esult of Equation 0.b gives F Fp qpe qp q EVAUATE Inseting the given quantities into the epession above fo the foce on the poton gives 19 10i N F ( C) 1.6 p i pn C ASSESS The foce on the poton acts in the opposite diection compaed to the foce on the oiginal chage, because the two chages have opposite signs. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

5 Electic Chage, Foce, and Field INTERPRET Fo this poblem, we ae to calculate the electic field stength due to a positive point chage the poton. DEVEOP The electic field stength at a distance fom a point souce chage q is given by Equation 0.: kq E The poton in a hydogen atom behaves like a point chage, so we can apply this equation to find the electic field of the poton. The chage of the poton is q +e C. EVAUATE At a distance of one Boh adius (a nm) fom the poton, the electic field stength is ASSESS ( N m /C )( C ) 11 ( m) ke E a N/C The field stength at the position of the electon is enomous because of the close poimity. Section 0.4 Fields of Chage Distibutions 9. INTERPRET Fo this poblem, we ae to find the electic field at seveal locations in the vicinity of two given point chages. We can apply the pinciple of supeposition to solve this poblem. DEVEOP Take the oigin of the -y coodinate system to be at the midpoint between the two chages, as indicated in the figue below, with unit vectos î pointing to the ight, and ĵ pointing up. We use Equation 0.4 to find the electic field at the given points. et q μc and q.0 μc. et ± (.5 cm) ± j denote the positions of the chages and denote that of the field point (i.e., the point at which we ae calculating the electic field). A unit vecto fom chage i to the field point is ( ± ) ± [whee the plus (minus) sign coesponds to the positive (negative) chage)]. Thus, the spatial factos in Coulomb s law ae i i i i ( ± ) ±. By the pinciple of supeposition (Equation 0.4), the total electic field at any point is q 1 1 q E k + 1 EVAUATE (a) Fo the point at 5.0 cm to the left of P, we have ( 5.0 cm) i 1 ( 5.0 cm) (.5 cm) (.5 cm) + i i i 5.0 cm i.5 cm i 7.5 cm i so the electic field is q 1 1 q qi 1 qi E k + k 1 1 C ( 0.05 m) ( m) o 6 MN/C, pointing to the left N m.0 10 C i.0 10 C i MN/C i Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

6 0-6 Chapte 0 (b) Fo the point at 5.0 cm diectly above P, we have ( 5.0 cm) j 1 (.5 cm) i+ ( 5.0 cm) j ( 5.0 cm) + (.5 cm) (.5 cm) i+ ( 5.0 cm) j / ( 5.0 cm) + (.5 cm) so the electic field is / 6 9 N m.0 10 C ( 0.05) i j ( 0.05) i j E / / C m ( 0.05) + ( 0.050) ( 0.05) + ( 0.050) 6 9 N m.0 10 C / ( i ) (5. MN/C)i C ( 0.05) + ( 0.050) m o 5. MN/C to the ight. (c) Fo 0, we have 1 i 1 i (.5 cm) i (.5 cm) (.5 cm) (.5 cm) i (.5 cm) (.5 cm) + + so the electic field is E MN/C C m (0.05) (0.05) o 58 MN/C to the ight. ASSESS 6 9 N m.0 10 C i i i The electic field fo pat (b) is much weake because the fields fom the two chages lagely cancel. 0. INTERPRET Given the magnitude of the dipole moment, we ae asked to calculate the distance between the pai of opposite chages that make up the dipole. DEVEOP As shown in Equation 0.5, the electic dipole moment p is the poduct of the chage q and the sepaation d between the two chages making up the dipole: p qd EVAUATE Using the equation above, the distance sepaating the chages of a dipole is 0 p C m d 19 9 pm 0.09 nm q C ASSESS The distance d has the same ode of magnitude as the Boh adius ( a nm). 1. INTERPRET We ae given a long wie with a unifom chage density and ae asked to find the electic field stength 8 cm fom the wie. We can assume that the wie length is much, much geate than 8 cm. DEVEOP Fo a vey long wie ( >> 8 cm), Eample 0.7 shows that the magnitude of the electic field falls off like 1/. Theefoe, the electic field is simply scaled by the atio of the distances, o 1 E E1 Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

7 Electic Chage, Foce, and Field 0-7 EVAUATE Inseting the given quantities into the epession above gives E ( 1.9 kn/c) 1.1kN/C 8 ASSESS The electic field gets weake the fathe we ae fom the wie, as epected.. INTERPRET In this poblem we ae asked to find the line chage density, given the field stength at a distance fom a long wie. We can assume that the wie is much, much longe than the distance involved (45 cm), so that the esult of Eample 0.7 applies. DEVEOP If the electic field points adially towad the long wie ( >> 45 cm), the chage on the wie must be negative. The magnitude of the field is given by the esult of Eample 0.7, kλ E EVAUATE Using the equation above, we find the line chage density to be E ( 60 kn/c)( 0.45 m) λ 6.5 μc/m 9 k Nm/C ASSESS The electic field stength due to a line chage density deceases as 1/. Compae this to the 1/ dependence of the electic field of a point chage.. INTERPRET We will use Coulomb s law and the definition of electic field to find the electic field at a point on the ais of a chaged ing. DEVEOP Fom Eample 0.6, which is done fo a geneal distance, we see that E kq ( + a ) / We want to know the field E at position a. EVAUATE Inseting a into the epession above gives kq kqa kq E ( + a ) ( a ) / / 8a ASSESS The units ae kq/(distance), which ae coect fo an electic field. Section 0.5 Matte in Electic Fields 4. INTERPRET Fo this poblem, we ae to find the foce geneated by 10 elementay chages in the given electic field, and calculate the mass that can be suspended by this foce in the Eath s gavitational field. DEVEOP By Newton s second law, the foce due to gavity and the foce due to the electic field must cancel each othe fo the oil dop to have no acceleation. This condition gives F,o g Fe mg qe The equation can be used to compute the mass m. EVAUATE Using the equation above, the mass is 19 7 ( C)(.0 10 N/C) ( 9.8 m/s ) qe m g kg ASSESS Because this mass is so small, the size of such a dop may be bette appeciated in tems of its adius, 1/ R ( m/4 πρoil). Millikan used oil of density ρ oil g/cm, so R 9.46 μm fo this dop. 5. INTERPRET This poblem involves kinematics, Newton s second law, and Coulomb s law. We can use these concepts to find the electic field stength necessay to acceleate an electon fom est to c/10 within 5.0 cm. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

8 0-8 Chapte 0 DEVEOP If the electic filed is constant in space, the foce applied to the chage will be constant, so we can use Equation.11, which applied fo constant acceleation, to find the necessay acceleation fo the electon. This gives 0 v v0 + a( 0) whee v c/10, cm, and v 0 0 because the electon stats fom est. Thus the acceleation is v c a cm ( ) Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may 0 The foce needed to povide this acceleation is given by Newton s second law which, combined with Coulomb s law, gives mc F Eq ma 10 m which we can solve fo E. EVAUATE Inseting q C fo the electon s chage (since we ae only concened with the stength of the electic field, not its diection), we find ASSESS mc E q 1 8 ( kg)(.0 10 m/s) 10 m C 10 m 19 ( ) N/C Because the electon has as negative chage, it would move opposite to the diection of this electic field. 6. INTERPRET This poblem is about the motion of a poton, which has chage +e, in an electic field that points to the left. The poton entes the field egion with the given velocity, and we ae to find how fa it tavels in the field befoe it eveses diection. We ae also to descibe its subsequent motion. To addess this poblem, we will use Newton s second law and Coulomb s law, and some kinematics fom Chapte. DEVEOP Choose the ais to the ight, in the diection of the poton s initial velocity, so that the electic field is oiented to the left. If only the Coulomb foce (Equation 0.b) acts on the poton, the acceleation can be found fom Newton s second law (fo constant mass: Equation 4.): ( ee F qe qe i) ma a m whee a is the magnitude of the poton s acceleation along the ais. The negative sign means that the poton deceleates as it entes the electic field. Because the acceleation is constant, we can apply Equation.11, v v0 + a( 0). When the poton eveses diection, its velocity v 0 momentaily, so we inset this into Equation.11 to find the distance taveled 0. 5 EVAUATE (a) Using Equation.11, with v m/s, we find the maimum penetation into the field egion to be 7 5 ( kg)(.8 10 m/s) ( )( ) v mv 1.46 cm 1. cm C N/C a ee 19 (b) The poton subsequently moves to the left, with the same constant acceleation in the field egion, until it eits with v f v. 0 ASSESS The deceleation of the poton inceases with the field stength E. Note that it is unealistic to have an electic field that begins instantaneously in space o time, as we will find in late chaptes. 7. INTERPRET This poblem involves an electostatic analyze like that in Eample 0.8, so we will use the esults of that eample. We ae to find the coefficient E 0 in the epession fo the electic field stength in the analyze that will pemit potons to eit the analyze. DEVEOP Fom the analysis of Eample 0.8, we know that the coefficient E 0 of the analyze is elated to the paticle mass m, its velocity v, and it chage q by

9 Electic Chage, Foce, and Field 0-9 PROBEMS mv E0 qb 7 19 Given that m kg fo a poton, q C, and v and b ae given, we can find E 0. EVAUATE Inseting the given quantities in the epession fo E 0 gives 7 ( kg)( m/s) mv E0 980 N/C 19 eb ( C)( m) to two significant figues. ASSESS Note that the poton eits the analyze with the same speed with which it enteed that analyze, because the foce is always pependicula to the poton s tajectoy (i.e., it s a centipetal foce). Thus, the foce does no wok on the poton, but the poton s velocity has changed diection. 8. INTERPRET The poblem asks fo an estimate of the faction of electons emoved fom a -g ping pong ball due to ubbing. DEVEOP Suppose that half the ball s mass is potons (the othe half is neutons). The numbe of potons is then N Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may p,0 1 g m Fo the ping pong ball to be initially neutal, it must have the same numbe of electons N e,0. Theefoe, N e,0 N p,0. The numbe of electons emoved is 6 1μC 1 10 C 1 1 Δ N e C C 1.6 The faction emoved is then ΔN e /N e,0. EVAUATE Inseting the numbes, we find the faction of the ping pong ball s electons that ae emoved is 1 1 Δ Ne Ne,0 ( 1 g) mp ( 1 g ) ( kg) ASSESS The faction is about a hunded billionth. Thus, although we emoved some 10 tillion electons, it is only a vey small faction of the total numbe of electons. 9. INTERPRET This poblem involves Coulomb s law, which we can use to elate the foce epeienced by the two paticles to thei chages. DEVEOP Coulomb s law (Equation 0.1) gives the foce between chaged paticles 1 and as kq1q F We ae given that q 1 q, and the 1.5 cm 0.15 m, so we can solve fo the magnitude of the lage chage q 1. EVAUATE Substituting fo q in Coulomb s law and solving fo q 1 gives q1 F qq 1 q1 k F 14N q1 ± ± ( 0.15 m) ± μc 9 k N m /C ASSESS Because the foce is epulsive, the chages must have the same sign. Howeve, fom the infomation given in the poblem statement, we cannot tell whethe the sign is positive o negative. 40. INTERPRET In solving this poblem we follow Poblem Solving Stategy 0.1. The souce chages ae the poton and the electon, and the chage on which the foces act is the helium nucleus (with chage +e). Using the pinciple of supeposition, we can sum the individual foces to find the total foce. p

10 0-10 Chapte 0 DEVEOP Make a sketch of the situation that shows the chages and thei positions (see figue below). The unit vecto fom the poton s position towad the oigin is î. Using Equation 0.1, the Coulomb foce eeted by the poton on the helium nucleus is 9 kq ( N m /C )( ) pq e e He FP,He ( i) ( i) ( nn) i 9 p,he m whee we have etained an eta significant figue in the esult because this is an intemediay esult. Similaly, the unit vecto fom the electon s position to the oigin is ĵ, so the foce it eets on the helium nucleus is kq ( eqhe ),He ( k e e F ) 0.68 nn e i j j m e,he EVAUATE To two significant figues, the net Coulomb foce on the helium nucleus is the sum of these: F F + F 0.18 nn i nn j net p,he e,he ASSESS In situations whee thee ae moe than one souce chage, we apply the supeposition pinciple and add the electic foces vecto-wise. Because the electon is close to the He nucleus than the poton in this poblem ( e,he < p,he ), we epect Fp,He < Fe,He. The diection of the foce is θ atan( ) 106 with espect to the positive ais. 41. INTERPRET This involves finding the electic foce on one chage fom anothe chage when both ae located in the -y plane. DEVEOP Denote the positions of the chages by 1 15i j cm fo q μc, and 4.4i + 11 j cm fo q. μc. The vecto fom q 1 to q is, fom which we can find the chage sepaation,, that goes 1 into Coulomb s law, F kqq /, fo the foce q eets on q. The diection of this foce can be designated with the unit vecto, /. EVAUATE Fist, we find the vecto displacement: 4.4i+ 11 j cm 15i+ 5.0 j cm 10.6i+ 6.0 j cm The magnitude of this vecto is F cm 1. cm. So the electic foce has a magnitude of 9 ( N m /C )( 9.5 μc)(. μc) ( 0.1 m) kq q 18.4 N 1 1 Using the unit vecto, we can wite the foce in component fom: 10.6i+ 6.0 j cm F 1 F1 ( 18.4 N) 16i 9.0 j N 1. cm ASSESS Without calculating the eact diections, we can see that points to the left of the positive y-ais, wheeas F 1 points to the ight of the negative y-ais. This seems to agee with the fact that the foce between oppositely chaged paticles is attactive, pointing in the opposite diection as the vecto that sepaates thei positions. 4. INTERPRET Coulomb s law applies hee. Since moe than one souce chage is involved, we make use of the supeposition pinciple to find the net foce on the test chage. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

11 Electic Chage, Foce, and Field 0-11 DEVEOP Make a sketch of the situation (see figue below). By symmety, the test chage must be placed on the ais; if not, it will epeience a nonzeo foce in the ĵ diection. If we place a chage to the left of the oigin, it will epeience a nonzeo foce because the chage q is both lage (in magnitude) and close than the chage q, so it will always geneate a geate foce fo all < 0. If we place a positive (negative) test chage between the two chages, the net foce will be to the ight (left); that is, nonzeo in both cases. Thus, the test chage Q must be placed on the ais to the ight of the q chage; that is, at > a. Using the pinciple of supeposition (Equation 0.4) with Coulomb s law (Equation 0.1), the net Coulomb foce on the test chage is Set F 0 to solve fo. ( ) ( ) ( a) kq q kq q F + a, o EVAUATE The condition F 0 implies that Only the solution ASSESS At ( 6) a 6a+ a 0. Thus, a± 9a a ± 6 a + 6 a 5.45a is to the ight of a. + the foces acting on Q fom q and q eactly cancel each othe. Notice that ou esult is independent of the sign and magnitude of the thid chage Q. 4. INTERPRET This poblem involves Coulomb s law and the pinciple of supeposition, which we can use to find the condition such that the thee given chages all epeience zeo net foce. DEVEOP By symmety, the negative chage placed at the midpoint between the two positive chages epeiences zeo net foce. On the othe hand, the Coulomb foce on +Q is kq( q) kq F + a a EVAUATE Setting F 0, we obtain Q 4 q. ASSESS To veify that we have the coect positioning, we calculate the net foce on the left-hand chage. Coulomb s law and the supeposition pinciple give 4q 16 k i + k i 0 a q a The epession fo the foce on the ight-hand chage is the same, ecept that the sign of the unit vectos is evesed. The foce on the cental chage is 4q ( 4q k i ) + k ( i) 0 a a Thus, all thee chages epeience zeo net foce. The situation is depicted below. Note that the equilibium is unstable, since if q is displaced slightly towad one chage, the net foce on it will be in the diection of that chage. 44. INTERPRET Moe than one souce chage is involved in this poblem. Theefoe, we use Coulomb s law and apply the supeposition pinciple to find the net foce on q. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

12 0-1 Chapte 0 1 m j, m i, figue below). The unit vecto pointing fom q 1 towad q is ( 1) 1 DEVEOP We denote the positions of the chages by 1 1 i j and ( m) + ( m) (see Similaly, the unit vecto pointing fom q towad q is ( ) The vecto fom of Coulomb s law and the supeposition pinciple give the net electic foce on q as kq1q( 1) kqq( ) F F1 + F + 1 EVAUATE Substituting the values given in the poblem statement, we find the foce acting on q to be kq1q( 1) kqq( ) F F1 + F + 1 ( 6 ) ( C.0i +1.0j 4 10 C ) j N m /C C + ( 1.6i0. j) N m 8.0 m 1 o F F+ F y 1.7 N at an angle of θ tan ( Fy/ F) 11 to the ais. ASSESS The foce between q 1 and q is epulsive ( qq 1 > 0 ) while the foce between q and q is attactive ( qq< 0 ). The two foces add vectoially to give the net foce on q. 45. INTERPRET This poblem is simila to the peceding one, only the magnitude of the chages has changed. Theefoe, we can use the same stategy to solve this poblem. DEVEOP The position of the chages ae again denoted by 1 1 m j, ( m ) i, and ( m) i+ ( m) j The unit vecto pointing fom q towad q 1 is ( 1) ( 1m) j( m) i( m) j 1 1 j m 5 5 Similaly, the unit vecto pointing fom q towad q 1 is ( 1) ( 1m) j( m) i 1 1 i + j m 5 5 The vecto fom of Coulomb s law and the supeposition pinciple give the net electic foce on q as ( ) ( q ) 1 q q i + j q i j 1 F1 kq1 + kq 1 + / / m 5 m We ae told that the foce on q 1 is in the î diection, so the ĵ component of F 1 must be zeo. This gives q q 0 EVAUATE (a) Fom the equation above, we find that q q, o q 0 μc. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

13 (b) Inseting the esult fom pat (a) into the epession fo F 1 gives 9 / F N m /C 5 0 μc 4.0i i N. 1 Electic Chage, Foce, and Field 0-1 ASSESS Because the chages q and q ae positioned symmetically above and below q 1, the esult that q q is epected. 46. INTERPRET We ae given the data of the electic foces on the DNA segments in an electophoesis appaatus. We would like to detemine the electic field stength of the appaatus based on the data given. DEVEOP The electic chage q caied by the DNA segments is diectly popotional to the numbe of base pais, with each base pai caying a chage of q e. Since F q E, plotting F vesus the numbe of base pais give a staight line, and the slope is equal to ee, whee E is the field stength. EVAUATE (a) A plot of the electic foce vesus the numbe of base pais is given below. The electic field is 4 16 ( ) pn N N/C E ( C) ( C) ASSESS Electophoesis is a widely used application of electic fields fo sepaating molecules by size and molecula weight. It s especially useful in biochemisty and molecula biology fo distinguishing lage molecules like poteins and DNA fagments. 47. INTERPRET We e asked to calculate the electic field fom a point chage at seveal locations. DEVEOP The electic field fom a point chage is given by Equation 0.: E kq /. The chage is at the oigin, so the vecto,, has the same coodinate values as the points in -y plane that we e asked to conside. Since the unit vecto is /, we can wite the electic field in a moe compact fom: E kq /. EVAUATE (a) At 50 cm and y 0 cm, the electic field is (b) At (c) At ( 9 Nm )( 65 μc) C ( 50 cm) kq E ( 50 i cm). i MN/C 50 cm and y 50 cm, the electic field is ( 9 Nm )( 65 μc) C E 50i+ 50 j cm 0.87i j MN/C 0.8i+ 0.8 j MN/C ( cm) 5 cm and y 75 cm, the electic field is ( 9 Nm )( 65 μc) C E ( 5i 75 j cm) 0.0i0.89 j MN/C cm ASSESS The magnitude of the electic field in pats (a), (b), (c) is. MN/C, 1. MN/C, and 0.9 MN/C, espectively. This shows that the field deceases as one moves futhe away fom the point chage, as we would epect. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

14 0-14 Chapte INTERPRET This poblem involves Coulomb s law and, since moe than one souce chage is involved, the supeposition pinciple. We will use these to find the point whee the field stength vanishes. DEVEOP We fist note that the field can be zeo only along the line joining the chages (the ais). To the left o ight of both chages, the fields due to each both have y components in the same diection, and so cannot sum to zeo. Between the two chages, a distance > 0 fom the 1 μc chage, the electic field is kq 1 kq E i + i 10 cm EVAUATE With q μc and q.0 μc, the field vanishes when 1.0 μc.0 μc ( 10 cm ) ASSESS Since E 0 at 10 cm 4.1 cm cm a chage placed at that point does not epeience any foce. 49. INTERPRET This poblem involves two souce chages, a poton at 0 and an ion at 5.0 nm. We can apply the pinciple of supeposition to find the electic field at 5 nm. DEVEOP The field at the point 5 nm due to the poton is ke E p i 5nm The field at the same point due to the ion is E i kq ( ) i i 10 nm The total electic field is the sum of these two, and is zeo at 5.0 nm, so we can solve fo q i. EVAUATE Solving fo the ion s chage gives ke kqi i + i 0 5nm 10nm ( 10 nm) ( 5nm) q e 4e i ASSESS Note that the field due to the ion was defined fo a positive chage, so the final chage is negative, as indicated. 50. INTERPRET Coulomb s law applies hee. The two souce chages q ae positioned on the ais, and we use the supeposition pinciple to find the field stength at a point on the y ais. DEVEOP As in Eample 0., we apply the symmety agument to show that the components of the electic field due to both chages cancel, and the net electic field points in the +y diection. EVAUATE (a) Since electic field is the foce pe unit chage, fom Eample 0., we obtain E kq kqy cosθ a y ( + ) net, y / (b) The magnitude of the field, a positive function, is zeo fo y 0 and y, hence it has a maimum in between. Setting the spatial deivative of the electic field equal to zeo, we find ( / 5/ ) ( a y y a y ) ( y) o a + y y 0. Thus, the field stength maima ae at y ± a. ASSESS By symmety, we epect the diections of the electic field at y ± a to be opposite. Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

15 Electic Chage, Foce, and Field INTERPRET This poblem is that same as Eample 0.5, ecept that it is otated by 90. Thee ae two souce paticles: a poton at (0, 0.60 nm) and an electon at (0, 0.60 nm), so the system is dipole. DEVEOP We can use the esult of Eample 0.5, with y eplaced by, and by y (o, equivalently, j by i, and i by j). The electic field on the ais is then E kqaja + / 19 whee q e C and a 0.60 nm (see Figue 0.1 otated 90 clockwise). The constant 9 19 kq N m /C C.88 GN/C nm. EVAUATE (a) Midway between the two chages (at 0), the electic field is kqj j E(0,0) (8.0 GN/C) a (.88 nm GN/C) j ( 0.60 nm) (b) fo nm, / E.0 nm,0 (.88 nm GN/C) 0.60 nm nm j 190 MN/C j (c) Fo 0 nm, / E ( 0 nm,0) (.88 nm GN/C)( 0.60 nm) ( 0) ( j) ( kn/c) j( 0 kn/c ) j. to two significant figues. ASSESS Fo pat (c), because a, we can apply Equation 0.6a, which gives ( )( ) ( 0 nm) kp.88 nm GN/C 0 nm kqa E 0 nm,0 j j 16.0 kn/c which diffes by only 0.1% fom the moe pecise esult of pat (c). 5. INTERPRET We find the electic field on the ais of a dipole, and show that Equation 0.6b is coect. To do this we will use the equation fo electic field. q DEVEOP The spacing between the + and chages is a. We will use E k fo each chage to find the total field at a point >> a. EVAUATE + q q E k i + k i kq[( a) ( + a) ] i ( a) ( + a) kq a a E i a a Fo >> a, 1± 1, so kq a a kq a k( qa) E i[1 1 ] 4 i. + kp But p qd qa, so E i. ASSESS We have shown what was equied. 5. INTERPRET This poblem involves finding the net chage o an unknown chage distibution. We ae given the behavio of the electic field as a function of distance fom the chage, fo distances much, much geate than the size of the chage distibution. DEVEOP Taking the hint, we suppose that the field stength vaies with an invese powe of the distance, n E. Unde this hypothesis, 96/87.7 (1.44/.16) n, o n ln(96/87.7)/ln(1.44/.16).0. EVAUATE A dipole field falls off like chage is zeo. fo >> a, so the chage distibution must be a dipole, whose net Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

16 0-16 Chapte 0 ASSESS Note that this esult is only valid if the dipole sepaation a 1 m. Because this is nomally the case, the esult is appeas valid. 54. INTERPRET Coulomb s law and the pinciple of supeposition applies hee. Thee ae thee souce chages, and we ae to find the field stength at a point on the y ais above the uppe-most chage. DEVEOP Make a sketch of the situation (see figue below). The electic field on the y ais (fo y > a/) due to the two chages on the ais follows fom Eample 0.. The only diffeence is that in this poblem, the chages on the ais ae sepaated by a instead of a. Thus, kqy E 1 j / y + a /4 On the othe hand, using Equation 0., we find the electic field due to the chage on the y ais fo y > a/ is kq E j y a/ Using the pinciple of supeposition, the total field fo y > a/ is the sum of E 1 and E. EVAUATE (a) Fo y > a/, the total field is simply the sum of both tems: y 1 E E 1+ E kq + j / ( y + a /4 ) ( y / a ) (b) Fo y >> a, the electic field may be appoimated as y 1 kq E kq + j j ( y ) / y y which is like that due to a point chage of magnitude q. ASSESS At lage distances much, much geate than a, the chage distibution looks like a point chage located at the oigin with chage q. 55. INTERPRET This poblem involves Coulomb s law, which geneates the given foces between two chaged metal sphees. The sphees initially epeience an attactive foce, but when the chage on the sphees is equilibated, the foce becomes epulsive. DEVEOP The chages initially attact, so q 1 q, and, by Coulomb s law (Equation 0.1), we have kq1q.5 N 1 m When the sphees ae bought togethe, they shae the total chage equally, each acquiing 1 ( q + q ). The 1 magnitude of thei epulsion is.5 N k ( q + q ) 1 Because the foces have the same magnitude, we can equation them to find the oiginal chages q 1 and q. 4 m Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

17 1 EVAUATE Equating these two foces, we find a quadatic equation Electic Chage, Foce, and Field 0-17 q1 q 4 qq 1, with solutions q1 ( ± 8 ) q. Both solutions ae possible, but since 1 9 a elabeling of the chages. Since qq 1.5 N m / N m /C 1.67 μc q 1 ± + 8 ( 16.7 μc) ± 40 μc and q ( 40. μc) ( + 8) 6.9 μc, + o q + 6qq + q 0, , they meely epesent, the solutions ae o the same values with q 1 and q intechanged. ASSESS The esults ae epoted to two significant figues, which is the pecision to which the data is known. 56. INTERPRET Two foces ae involved in this poblem: the Coulomb foce and the sping foce. The sping is stetched due to the Coulomb epulsion between the chages, and we ae to find by how much the sping stetches. DEVEOP The Coulomb foce is given by Equation 0.1 and the sping foce is given by Equation 4.9, F s k, whee is the displacement fom equilibium of the sping. We assume that the Coulomb epulsion F e is the only foce stetching the sping. When balanced with the sping foce, F F, s o kq ( + ) 0 s e k 0 whee 0 is the equilibium length. This cubic equation can be solved by iteation o by Newton s method. EVAUATE Substituting the values given in the poblem statement gives 0+ kq k s 9 ( N m /C )( 8 μc ) 0.56 m m 145 N/m Newton s method yields 18.0 cm, to two significant figues. ASSESS Ou esult makes sense because the amount stetched is seen to decease with inceasing sping constant k s and incease with the magnitude of the chage q. 57. INTERPRET We will calculate the electic field magnitude a distance fom eithe end of a unifomly chaged od of chage Q and length. kdq DEVEOP We will use the integal fom fo electic field, Equation 0.7: E. Because the chage is distibuted unifomly along the length, the diffeential chage element is dq ( Q / ) d, whee is the location of this chage along the od, see figue below. We ae only inteested in the electic field along the -ais at points with >, in which case the distance to dq is just. EVAUATE The integal fo the electic field stength is kdq kq E d 0 0 ( ) We can change vaiables: u, d du, so the integal becomes kq du kq 1 1 kq E u ( ) ( ) ASSESS Fo, the electic field educes to E kq /, which is what we d epect since the od will appea as a point chage fom a geat distance. 58. INTERPRET The electon undegoes cicula motion whee the centipetal foce (Chapte ) is povided by the Coulomb foce in the fom of an electic field aound a long cuent-caying wie. DEVEOP The electic field of the wie is adial and falls off like 1/ ( E kλ /, see Eample 0.7). Fo an attactive foce (negative electon encicling a positively chaged wie), this is the same dependence as the Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

18 0-18 Chapte 0 centipetal acceleation (see Equation.9, a c v /). Fo cicula motion aound the wie, the Coulomb foce povides the electon s centipetal acceleation. Thus, applying Newton s second law gives F ma c v ee m keλ v m The equation can be used to deduce the tangential speed v of the electon. EVAUATE Substituting the values given, we find the speed to be ( )( )( ) keλ N m /C C.5 10 C/m 6 v.8 10 m/s 1 m kg ASSESS The speed of the electon is independent of, the adial distance fom the long wie. This is because both the electic field and the centipetal acceleation fall off as 1/, so the -dependence cancels out. 59. INTERPRET The chage obiting the wie undegoes cicula motion, and the centipetal foce (Chapte ) is povided by the Coulomb foce. We ae asked to find the line chage density, given the paticle s obital speed. This poblem is simila to Poblem DEVEOP The electic field of the wie is adial and falls off like 1/ ( E kλ /, see Eample 0.7). Fo an attactive foce (positive chage encicling a negatively chaged wie), this is the same dependence as the centipetal acceleation (Equation.9, a c v /). Fo cicula motion aound the wie, the Coulomb foce povides the centipetal acceleation. Thus, Newton s second law gives F qe mac v qe kqλ ac m m The equation can be used to deduce the line chage density λ give the speed. EVAUATE The above equation gives 9 mv ( kg)( 80 m/s) λ 14 μc/m 9 9 kq N m /C.1 10 C ASSESS Fo the foce to be attactive, the line chage density must be negative. 60. INTERPRET Fo this poblem, we ae to find the toque on the given dipole placed at 0 with espect to a linea electic field (simila to what is shown in Figue 0.0). We ae also to find the wok done by the electic field in oienting the dipole so that it is antipaallel to the field. DEVEOP The toque on the dipole is given by Equation 0.9, τ P E pesinθ The wok done in oienting the dipole antipaallel to the field is the change in the potential enegy fom the initial to the final states, whee the potential enegy of a dipole in an electic field is given by Equation Thus, W Δ U p E p E EVAUATE (a) Evaluating the epession fo toque gives (b) Evaluating the epession fo wok gives τ pesinθ 1.5 nc m 4.0 MN/C sin 0.0 mn m W pe(cos0 cos180 ) ( 1.5 nc m)( 4.0 MN/C)( 1.866) 11 mj ASSESS When the dipole is oiented antipaallel to the field, the toque is also zeo, but this is an unstable equilibium because the slightest petubation will allow the dipole to evese its oientation in the electic field. f i Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

19 Electic Chage, Foce, and Field INTERPRET You want to check if a patent fo an isotope sepaato will wok. Since diffeent isotopes have the same chage but diffeent mass, the device can wok if it disciminates between objects with diffeent chage-tomass atios. DEVEOP You can assume the acceleating field, E, 1 is constant and that the plates in the figue ae sepaated by a distance. Theefoe, if an atom stipped of its electons stats at est at the bottom plate, it will be acceleated by the field to a final speed of v a qe1 m So, it is tue that isotopes of diffeent chage-to-mass ations ( q/ m) will leave the fist half of the device with diffeent speeds. Fo eample, an isotope with a elatively lage chage-to-mass atio will attain a highe speed in the acceleating field than anothe isotope with a lowe chage-to-mass atio. But the question is: can the second half of the device select just one of these speeds so that only one type of isotope emeges? EVAUATE The second half of the device is an electostatic analyze, as descibed in Eample 0.8. It has a cuved field, E E0( b/ ), which points towad the cente of cuvatue. The paametes E 0 and b ae constants with units of electic field and distance, espectively. It was shown in the tet that paticles enteing the device fom below will only emege fom the hoizontal outlet if thei speed satisfies: qe0b v m If you equate this speed with the speed fom the acceleating field, you find that the chage-to-mass atio cancels out. This means thee s no discimination between isotopes. If one type of isotope can emege, then they all can. The device doesn t wok. ASSESS The poblem with this device is that both the acceleating field and the cuving field depend on the chage-to-mass atio in the same way. One way to get aound this is to acceleate the isotopes by heating them to high tempeatue. The speeds in this case won t depend on the chage. Anothe way is to use a magnetic field to cuve the path of the isotopes. In this case, the chage-to-mass atio doesn t cancel out, as we ll see in Eample INTERPRET The poblem asks us to find the electic field nea a stand of DNA. DEVEOP We e looking fo the electic field at a distance, y 5 nm, fom a stand of DNA that has length, 5.0 μm. Since the point we e consideing is not nea eithe end of the stand and since y/ 0.005<< 1, the situation is appoimately the same as fo an infinite wie. In Eample 0.7, it was shown that the electic field at a distance y fom an infinite wie has magnitude E kλ / y, whee λ is the chage pe unit length. EVAUATE We e told that the DNA caies + e of chage fo evey nm of length, so λ e /nm. Using the above fomula: Nm ( C ) Nm ( C ) 9 ( ) kλ C/nm C/m E 0.1 GN/C y 5 nm 5 10 m ASSESS Since the chage pe unit length is positive, the electic field will point outwad fom the DNA stand, as in Figue The actual field will deviate fom this pictue nea eithe end. But we often can get away with teating a wie o a sheet as infinite, as long as we only conside points that ae shot distances away. 6. INTERPRET This poblem is about an electic dipole aligned with an etenal electic field. We ae to find the dipole moment given the electic field and the amount of wok needed to evese the dipole s oientation. DEVEOP Using Equation 0.10, the enegy equied to evese the oientation of such a dipole is EVAUATE Δ U pe cosθf cosθi pe cos 180 cos 0 pe Using the equation above, the electic dipole moment is 7 Δ U.1 10 J 0 p C m E N/C Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

20 0-0 Chapte 0 ASSESS An electic dipole tends to align itself in the diection of the etenal electic field. Thus, enegy is equied to change its oientation. 64. INTERPRET This poblem involves 4 souce chages, aanged as two dipoles. The dipoles ae oiented paallel and on the same line. We ae to find the epession fo the foce between the two dipoles, so Coulomb s law and the pinciple of supeposition will apply. DEVEOP Make a sketch of the situation (see figue below). The ight-hand dipole has chages +q at + a/, q at a/, each of which epeiences a foce fom both chages of the left-hand dipole, which ae +q at a/ and q at a/. The Coulomb foce on a chage in the ight-hand dipole due to a chage in the left-hand dipole is kqrq( R ) i R (see solution to Poblem 15), so the total foce on the ight-hand dipole is kq a a F kq i + i + a a a EVAUATE (a) In the limit a<< kq a ( ) 6 6 kq a i kp i F i whee p qa is the dipole moment of both dipoles. (b) The foce on the ight-hand dipole is in the negative diection, indicating an attactive foce. ASSESS By Newton s thid law, each dipole epeiences a foce of equal magnitude but in the opposite diection. 65. INTERPRET This poblem is about the inteaction between a dipole and the electic field due to a souce chage. DEVEOP With the ais in the diection fom Q to p and the y ais paallel to the dipole in Figue 0.0, we have p ( qa) j and ( / ) E kq i. In the limit >> a, the toque on the dipole is given by Equation 0.9, τ p E, whee E is the field fom the point chage Q, at the position of the dipole. EVAUATE (a) Using Equation 0.9, we find the toque to be kq kqqa τ p E ( qaj) i k The diection is into the page, o clockwise, to align p with E. (b) The Coulomb foce obeys Newton s thid law. The field of the dipole at the position of Q is (Eample 0.5 adapted to new aes) kqa E dipole j Thus, the foce on Q due to the dipole is kqqa F on Q QEdipole j The foce on the dipole due to Q is the opposite of this: kqqa F on dipole Fon Q j The magnitude of F on dipole is kqqa/. (c) The diection of F is in +, j o paallel to the dipole moment. on dipole Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may

21 Electic Chage, Foce, and Field 0-1 ASSESS The net foce F on dipole will cause the dipole to move in the + j diection. In addition, thee is a toque that tends to align p with E. So, the motion of the dipole involves both tanslation and otation. 66. INTERPRET This poblem gives the position of two souce chages, and we ae to find a position at which the electic field is zeo. DEVEOP The electon s field is diected towad the electon (a negative chage) and the ion s field is diected away fom the ion (a positive chage). Theefoe, the fields can cancel only at points on the negative ais ( < 0) since the diections ae opposite and the smalle chage is close. The field fom a point chage is ( q ) E q kq i whee q e and q 0 fo the electon, and q 5e and q 10 nm fo the ion. Setting the total field to zeo gives k( e) 5e( 10 nm) 0 i + i 10 nm nm 4 10 nm 10 nm + which we can solve fo. EVAUATE The negative solution to this quadatic is 10 nm ( 10 nm) 4( 10 nm + ) (.5 nm)( 1+ 5) 8.1 nm 4 ASSESS Thus, if we label the distance between the two chages as a, the field goes to zeo not fa fom a fom the smalle chage. 67. INTERPRET You e asked to estimate the chage distibution on two diffeent molecules given thei dipole moments. DEVEOP You given the dipole moment, p, fo H O and CO in debyes. A debye (D) is a unit with dimensions of 0 chage times distance. You can look in an outside efeence and see that 1 D.4 10 C m. You can model these dipole molecules as two opposite chages, ± q, sepaated by a distance d. Since the atomic sepaation 10 fo many molecules is about an Angstom ( 1 Å 10 m ), we can estimate how the chage is distibuted on each molecule: q p/ d. EVAUATE et s fist convet the dipole moments to SI units: C m 0 pho ( 1.85 D) C m 1 D C m 1 pco ( 0.1 D) C m 1 D Assuming the atoms in the molecules ae sepaated by about 1 Angstom, the amount of chage on each atom is 0 p C m 1 e qho 0.4 e d 10 m C 1 p C m 1 e qco 0.0 e d 10 m C We ve witten the esult in tems elementay chage. ASSESS In the case of wate, the oygen atom patially steals the electons fom the hydogen atoms. That esults in a negative factional chage ( q 0.4e) on the oygen atom, and a coespondingly positive factional chage on the hydogen atoms. A simila situation occus with cabon monoide, but this time the cabon atom is q Copyight 016 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently eist. No potion of this mateial may