ELECTRIC CHARGES AND FORCES

Transcription

1 ELECTRIC CHARGES AND ORCES 5 Conceptual Questions 5.1. An insulato can be chaged. Plastic is an insulato. A plastic od can be chaged by ubbing it with wool. 5.. A conducto can be chaged. A conducto can be chaged by touching it with anothe chaged object. 5.. B and D ae both neutal because they have no effect on each othe and neutal is still attacted to eithe glass o plastic. Since ball A has been touched by plastic it is also now plastic. Since ball C is attacted to plastic (A) and neutal (B) then it must be glass (a) Like chages exet epulsive foces on each othe, so the object must also have plastic chage. Theefoe, it will attact the glass od, which has the opposite chage (i.e., glass chage). (b) You cannot pedict this because the object could be glass o neutal. Glass will epel the glass od but neutal will be attacted to the glass od Upon touching the chaged od, the metal exchanges chage with the aea of the od touched by the sphee (we ae assuming the od is an insulato). Some of the local excess chage on the od will spead ove the conducting sphee, so that both the od and the sphee will have an oveall excess chage of the same type. Thus, the sphee and the od will epel each othe Assume that the basic pemise of like chages epel, unlike chages attact still holds. Suspend an object with an excess of unknown chage fom a sting. ist, appoach a plastic-chaged od, then a glass-chaged od. An object with chage X must be attacted by both of these. Howeve, this is not sufficient because a neutal object would also be attacted by both ods. To detemine if the object is neutal o not, appoach a neutal object. If the object has chage X, it will be attacted, if the object is neutal, nothing will happen (a) The negatively chaged od will epel the negative chages on the top of the electoscope, pushing moe negative chage down onto the leaves. The leaves will sepaate moe. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist. 5-1

2 5- Chapte 5 (b) The positively chaged od will attact moe negative chages to the top of the electoscope. As they depat fom the leaves, the leaves will move close togethe The final state of each sphee and of the od is neutal. The conducting od allows the excess electons in the negatively chaged sphee to move to the positively chaged sphee and exactly neutalize the chage thee, leaving all thee conductos neutal Each sphee ends up with one unit of negative chage. Once they touch, the two sphees become essentially one conducto. The oveall net chage is 4 + =. Chage is spead unifomly ove the suface of a conducto The od will polaize the chages in the combined conducto A + B, attacting negative chages to A and leaving B with excess positive chage. The combined conducto A + B is still neutal, but A alone has net negative chage You finge becomes polaized. Positive chage is left on the tip of you finge when negative chages in the finge ae epelled by the ball. The excess positive chage in you finge is then neae to the negatively chaged ball than the negative chage in you finge, esulting in a net attactive foce that attacts the ball Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

3 Electic Chages and oces (a) The magnitude of the foce on A uaduples (inceases by a facto of 4), since the foce between the chages is popotional to the poduct of the magnitudes of the chages. Theefoe, A = 4. (b) By Newton s thid law, the foce of A on B is eual in magnitude to the foce of B on A; theefoe the foce on B also uaduples; B = A = (a) We have E() = 1000 N/C. o a point chage, E (). If the distance to the chage is doubled, E( ) 4 1 E( ) =, so = =. ( ) 4 E () 1 4 Theefoe 1 E( ) = (1000 N/C) = 50 N/C. 4 1 (b) Similaly, E () 1 E = = 4, so ( ) E = 4(1000 N/C) = 4000 N/C Since the foce on a chage in an electic field has magnitude = E, the new foce is Execises and Poblems Section 5.1 Developing a Chage Model Section 5. Chage E ( ) = = E = Model: Use the chage model. Solve: (a) In the pocess of chaging by ubbing, electons ae emoved fom one mateial and tansfeed into the othe because they ae elatively fee to move. Potons, on the othe hand, ae tightly bound in the nuclei of atoms and so ae essentially not fee to move. Thus, electons have been added to the plastic od to make it negatively chaged. 19 (b) Because each electon has a chage of C, the numbe of electons added is C = C 5.. Model: Use the chage model. Solve: (a) In the pocess of chaging by ubbing, electons ae emoved fom one mateial and tansfeed to the othe because they ae elatively fee to move. Potons, on the othe hand, ae tightly bound in the nuclei of atoms and so ae essentially not fee to move. Thus, electons have been emoved fom the glass od to make it positively chaged. (b) Because each electon has a chage of C, the numbe of electons emoved is C 10 = C whee the numeato is negative because this is the chage that is emoved, so the excess chage left behind is C. 10 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

4 5-4 Chapte Model: Use the chage model and the model of a conducto as a mateial though which electons move. Solve: (a) The chage of the glass od deceases fom +1 nc to +8.0 nc. Because it is the electons that ae tansfeed, 4.0 nc of electons has been added to the glass od. Thus, electons ae emoved fom the metal sphee and added to the glass od. 19 (b) Because each electon has a chage of C and a chage of 4.0 nc was tansfeed, numbe of electons tansfeed fom the metal sphee to the glass od is C 10 = C 5.4. Model: Use the chage model and the model of a conducto as mateial though which electons move. Solve: (a) The chage of a plastic od changes fom 15 nc to 10 nc. That is, 5 nc chage has been emoved fom the plastic. Because it is the negatively chaged electons that ae tansfeed, 5 nc has been added to the metal sphee. 19 (b) Because each electon has a chage of C and a chage of 5.0 nc was tansfeed, the numbe of electons tansfeed fom the plastic od to the metal sphee is C 10 = C 5.5. Model: Use the chage model. Solve: Each helium atom has potons and thee ae helium molecules in 1.0 mole of helium. Because 19 each poton has a chage of C, the amount of chage in 1.0 mole of oxygen is 19 5 (1.0 mol)( atoms/mol)( potons/atom)( C/poton) = C 5.6. Model: Use the chage model ml 1.0 cm 1.0 g Solve: Since the density of wate is 1.0 g/cm, the mass of 1.0 L of wate is (1.0 L) 1.0 kg. L ml = cm Each wate molecule (H0) has 10 potons (8 in the oxygen atom and one pe hydogen atom), and thus 10 electons g The numbe of moles is 100 moles. = Thee ae wate molecules in 1.0 mole of wate. Because 10 g/mole 19 one electon has a chage of C, the amount of chage in 100 mole of wate is 19 7 (100 mol)( H O/mol)(10 electon/h O)( C/electon) = C Section 5. Insulatos and Conductos 5.7. Model: Use the chage model and the model of a conducto as a mateial though which electons move. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

5 Electic Chages and oces 5-5 The chage caies in a metal electoscope ae the negative electons. As the positive od is bought nea, electons ae attacted towad it and move to the top of the electoscope. The electoscope leaves now have a net positive chage, due to the missing electons, and thus epel each othe. At this point, the electoscope as a whole is still neutal (no net chage) but has been polaized. On contact, some of the electons move to the positive od to neutalize some (but not necessaily all) of the od s positive chage. Afte contact, the electoscope does have a net positive chage. When the od is emoved, the net positive chage on the electoscope uickly coves the entie electoscope (note that no positive chages move, but the electons distibute themselves ove the suface so that thee is a net positive chage eveywhee on the suface). The net positive chage on the leaves causes them to continue to epel Model: Use the chage model. Solve: (a) No, we cannot conclude that the wall is chaged. Attactive electic foces occu between (i) two opposite chages, o (ii) a chage and a neutal object that is polaized by the chage. Rubbing the balloon does chage the balloon. Since the balloon is ubbe, its chage is negative. As the balloon is bought nea the wall, the wall becomes polaized. The positive side of the wall is close to the balloon than the negative side, so thee is a net attactive electic foce between the wall and the balloon. This causes the balloon to stick to the wall, with a nomal foce balancing the attactive electic foce and an upwad fictional foce balancing the gavitational foce on the balloon. (b) 5.9. Model: Use the chage model and the model of a conducto as a mateial though which electons move. Solve: The fist step shows two neutal metal sphees touching each othe. In the second step, the negative od epels the negative chages which will eteat as fa as possible fom the top of the left sphee. Note that the two sphees ae touching and the net chage on these two sphees is still zeo. While the od is thee on top of the left sphee, the ight sphee is moved away fom the left sphee. Because the ight sphee has an excess negative chage then, by chage consevation, the left sphee has the same magnitude of positive chage. Upon sepaation, the negative chage is tapped on the ight sphee, as shown in the thid step. As the two sphees ae moved apat fathe and the negatively chaged od is moved away fom the sphees, the chages on the two sphees edistibute unifomly ove the entie suface sphees. Thus, we ae left with two oppositely chaged sphees Model: Use the chage model and the model of a conducto as a mateial though which electons move. Solve: Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

6 5-6 Chapte 5 Chaging two neutal sphees with like chages of exactly eual magnitude can be achieved though the following six steps. (i) Bing a chaged od (say, negative) nea a neutal metal sphee. (ii) Touch the neutal sphee with the negatively chaged od, so that the od-sphee system has a net negative chage. (iii) Move the od away fom the sphee. The sphee is now negatively chaged. (iv) Bing this negatively chaged sphee close to the second neutal sphee. (v) Touch these two sphees. The excess negative chage is distibuted evenly ove the two sphees. (vi) Sepaate the sphees. The excess chage will have the same sign as the chage on the chaging od and will be evenly distibuted between the two sphees Model: Use the chage model and the model of a conducto as a mateial though which electons move. Solve: Chaging two neutal sphees with opposite chages of eual magnitude can be done though the following fou steps. (i) Touch the two neutal metal sphees togethe. (ii) Bing a chaged od (say, positive) close (but not touching) to one of the sphees (say, the left sphee). Note that the two sphees ae still touching and the net chage on the pai is zeo. The ight sphee has an excess positive chage of exactly the same magnitude as the left sphee s negative chage. (iii) Sepaate the sphees while the chaged od emains close to the left sphee, so the sepaated chage emains on the sphees. (iv) Take the chaged od away fom the two sphees. The sepaated chages edistibute unifomly ove the metal sphee sufaces. Section 5.4 Coulomb s Law 5.1. Model: Model the chaged masses as point chages. Solve: (a) The chage 1 exets a foce 1 on on to the ight, and the chage exets a foce on 1 on 1 to the left. Using Coulomb s law, K1 1 1 on on 1 1 (1.0 m) ( N m /C )(10 10 C)(10 10 C) = = = = 0.90 N (b) Applying Newton s second law on eithe 1 o gives 0.90 N 1 on = ma 1 1 a1 = = 0.90 m/s 1.0 kg Assess: Even a mico-couomb is a lot of chage. That is why 1 on (o on 1) is a measuable foce. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

7 Electic Chages and oces Model: Model the plastic sphees as point chages. Solve: (a) The chage 1 = 50.0 nc exets a foce 1 on on = 50.0 nc to the ight, and the chage exets a foce on 1 on 1 to the left. Using Coulomb s law, K1 1 on on 1 1 (.0 10 m) ( N m /C )( C)( C) = = = = N (b) The atio of the electic foce to the weight is 1 on N = =.9 mg (.0 10 kg)(9.8 m/s ) Model: Model the glass bead and the ball beaing as point chages. The ball beaing expeiences a downwad electic foce 1 on. By Newton s thid law, on 1 = 1 on. Solve: Using Coulomb s law, ( N m /C )(0 10 C) 8 1 on = = = 1 ( m) K N C Because the foce 1 on is attactive and 1 is a positive chage, the chage is a negative chage. Thus, 8 = C = 10 nc Model: The potons ae point chages. Solve: (a) The electic foce between the potons is ( N m /C )( C)( C) E = K = = 58 N 15 (.0 10 m) (b) The gavitational foce between the potons is Gmm 1 ( N m /kg )( kg)( kg) 5 G = = = N 15 (.0 10 m) (c) The atio of the electic foce to the gavitational foce is E G 58 N = = N 6 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

8 5-8 Chapte Model: Chages A, B, and C ae point chages. Please efe to igue EX5.16. Chage A expeiences an electic foce B on A due to chage B and an electic foce C on A due to chage C. The foce B on A is diected to the ight and the foce C on A is diected to the left. Solve: Coulomb s law yields: A B ( N m /C )( C)( C) 5 B on A = K = = N ( m) The net foce on A is C A ( N m /C )( C)( C) 5 C on A = K = = on A B on A C on A (.0 10 m) 5 5 = + = ( N) i + ( N)( i) = 0.0 N N Model: Chages A, B, and C ae point chages. Please efe to igue EX5.17. Solve: The foce on B fom chage A is diected downwad since two negative chages epel. Coulomb s law gives the magnitude of the foce as A B ( Nm /C )( C)(.0 10 C) 5 A on B = K = = N (.0 10 m ) 5 So A on B = ( j) N The foce on B fom chage C is diected downwads since the two opposite chages attact. Coulomb s law gives the magnitude of the foce as C B ( Nm /C )(.0 10 C)(.0 10 C) 4 C on B = K = =.6 10 N ( m) 4 So C on B = (.6 10 j) N The net electic foce on chage A is 5 4 A = B on A + C on A = ( j) N (.6 10 j) N 4 = ( j ) N Model: Objects A and B ae point chages. Because thee ae only two chages A and B, the foce on chage A is due to chage B only, and the foce on B is due to chage A only. Solve: Coulomb s law gives the magnitude of the foces between the chage: ( N m /C )( C)( C) 4 A on B = B on A = = N (.0 10 m) Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

9 Electic Chages and oces 5-9 Because the chage on object A is positive and on object B is negative, B on A is upwad and A on B is downwad. Thus, 4 4 =+ (7. 10 N) j and = (7. 10 N) j B on A Assess: By Newton s thid law, the two foces have eual magnitudes but opposite diections because they fom an action-eaction pai, just as we found. A on B Model: Assume the plastic bead, the poton, and the electon ae point chages. Solve: Coulomb s law gives ( N m /C )(15 10 C)( C) 1 bead on electon = bead on poton = = N ( m) (a) Becaue the bead is much moe massive than both the electon and the poton, we can ignoe any acceleation of the bead. Newton s second law is = ma, so 1 bead on poton N 14 apoton = = = m/s m poton kg Because opposite chages attact, 14 a poton = (1. 10 m/s, towad bead) (b) Similaly, 1 bead on electon N 17 aelecton = = = m/s m electon kg 17 Thus a electon = (.4 10 m/s, away fom bead). Assess: Although the foce on the poton has the same magnitude as the foce on the electon, the electon has a much geate acceleation because it has a much smalle mass. Section 5.5 The ield Model 5.0. Model: Potons and electons poduce electic fields. Solve: (a) The electic field of the poton is C E = K = ( N m /C ) ( N/C, away fom poton) = ( m) (b) The electon caies the opposite chage, so the electic field of the electon is E = ( N/C)( ) = ( N/C, towad electon) 5.1. Model: Model the poton and the electon as point chages. Solve: (a) The foce that an electic field E exets on a chage is = E. A poton has = e. Thus, 17 = e(400i j) N/C = (6.4i j) 10 N 19 poton whee we used e = C. 17 (b) The chage on an electon is = e. Thus, ( = = i j) 10 N electon poton Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

10 5-10 Chapte 5 (c) om Newton s second law, a 17 poton x + y N 10 poton = = = = m/s mpoton m poton kg (d) The electon expeiences a foce of the same magnitude but it has a diffeent mass. Thus, N 1 aelecton = = m/s kg The foces may be the same, but the electon has a much lage acceleation due to its much smalle mass. Assess: The two foces in pats (a) and (b) ae eual in magnitude but opposite in diection. 5.. Model: The electic field is due to a chage and extends to all points in space. Solve: The magnitude of the electic field at a distance fom a chage is 9 10 E = K 1.0 N/C = ( N m /C ) = C = 0.11 nc (1.0 m) 5.. Model: The electic field is that of a negative chage on the plastic bead. Model the small bead as a point chage. Solve: The electic field is C 4 E = K = ( N m /C ) = N/C ( m) whee is the unit vecto fom the chage to the point at which we calculate the field. Assess: The diection of the electic field is towad the bead, which it should be since the bead is negative Model: Teat the chage on the object is a point chage. Solve: The electic field at a distance fom a point chage is E = K. Because the electic field points away fom the object, E = (70,000 N/C). Thus, K = 70,000 N/C (70,000 N/C)( m) 8 = = C = 1 nc 9 ( N m /C ) Assess: Since the field points away fom it, we know that the chage must be positive Model: A field is the agent that exets an electic foce on a chage. Solve: Newton s second law on the plastic ball is Σ ( net ) y = on G. To balance the gavitational foce with the electic foce, mg ( kg)(9.8 N/kg) 6 on = G E = mg E = = =. 10 N/C C Because on must be upwad and the chage is negative, the electic field at the location of the plastic ball must be 6 pointing downwad. Thus E = (. 10 N/C, downwad). Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

11 Electic Chages and oces 5-11 Assess: = E means the sign of the chage detemines the diection of o E. o positive, E and ae pointing in the same diection. But E and point in opposite diections when is negative Model: The electic field is that of a positive point chage located at the oigin. The positions (5.0 cm, 0.0 cm), (5.0 cm, 5.0 cm), and (5.0 cm, 5.0 cm) ae denoted by A, B, and C, espectively. Solve: (a) The electic field fo a positive chage is E = K,away fom 9 9 Using K = N m /C and = 1 10 C, 108 N m /C E =, away fom The electic fields at points A, B, and C ae 108 N m /C 4 EA = i = i N/C ( m) E 108 N m /C B ( ) ( )N/C ( m) ( m) i j = + = i + + j E 108 N m /C C ( ) ( ) N/C ( m) ( m) i j = = i + j (b) The thee vectos ae shown in the diagam. Assess: The vectos E A, E B, and E C ae pointing away fom the positive chage Model: The electic field is that of a negative point chage located at (x, y) = (1.0 cm, 0 cm). Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

12 5-1 Chapte 5 Solve: The electic field fo a positive chage is 9 9 E = K Using K = N m /C and = 1 10 C, 108 N m /C E = The electic fields at points A, B, and C (see figue above) ae We need to take components to find E B : 108 N m /C 4 EA = i = i N/C ( m) 108 N m /C 4 EC = i =.0 10 i N/C ( m) 108 N m /C m m EB = i j ( m) + ( m) ( m) ( m) ( m) ( m) = ( i j) N/C Assess: Since the field points towad the negative chage, it should have a positive x-component and a negative y-component, as we have found Model: Use the chage model. Solve: The numbe of moles in the penny is M.1 g n = = = mol A 6.5 g/mol The numbe of coppe atoms in the penny is 1 N = nn A = ( mol)( mol ) = Since each coppe atom has 9 electons and 9 potons, the total positive chage in the coppe penny is Similaly, the total negative chage is 19 5 ( )( C) = C C. Assess: Total positive and negative chages ae eual in magnitude Model: The beads ae point chages. Solve: The beads ae oppositely chaged so ae attacted to one anothe. The foce on each is the same by Newton s thid law. Coulomb s law give the foce as ( C)( C) 4 = K = ( Nm /C ) = N (.0 10 m) The beads acceleate at diffeent ates because thei masses ae diffeent. By Newton s second law, the acceleation of the plastic bead is 4 (7. 10 N) aplastic = = = 0.6 m/s towad glass bead. m (.0 10 kg) Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

13 Electic Chages and oces 5-1 o the glass bead, the acceleation is a 4 glass (7. 10 N) = = 0.18 m/s towad plastic bead. ( kg) 5.0. Model: The 15 Xe nucleus and the poton will be teated as point chages. That is, all the chage on the Xe nucleus is assumed to be at its cente. Solve: (a) The magnitude of the foce between the nucleus and the poton is given by Coulomb s law: K nucleus poton ( N m /C )( C)( C) nucleus on poton = = = N 15 ( m) (b) Applying Newton s second law to the poton, N on poton = mpotonapoton apoton = =.0 10 m/s kg Model: Teat the two chaged sphees as point chages. Solve: The electic foce on one chaged sphee due to the othe chaged sphee is eual to the sphee s mass times its acceleation. Because the sphees ae identical and eually chaged, m1 = m = mand 1 = =. We have K1 K on 1 = 1 on = = = ma ( kg)(150 m/s )(.0 10 m) 15 = = = C 9 ma K N m /C 8 = C = 8 nc 5.. Model: Objects A and B ae point chages. Solve: (a) It is given that A on B = 0.45 N. By Newton s thid law, B on A = A on B = 0.45 N. Coulomb s law gives 1 KA K ( B A)( A) B on A = A on B = 0.45 N = = (0.45 N) (0.45 N)(10 10 m) 6 = = = C K ( N m /C ) A 9 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

14 5-14 Chapte 5 (b) Newton s second law is B on A = m A a A. Hence, B on A A on B 0.45 N aa = 4.5 m/s m = m = kg = A B 5.. Model: The chages ae point chages. Please efe to igue P5.. Solve: The electic foce on chage 1 is the vecto sum of the foces on 1 and on 1, whee 1 is the 1.0 nc chage, is the.0 nc chage, and is the othe.0 nc chage. We have K1 on 1 =, away fom ( N m /C )( C)(.0 10 C) =, away fom ( m) 4 4 = ( N, away fom ) = ( N)[cos(60 ) i+ sin(60 ) j] on on 1 = on 1 + on 1 = ( N)sin(60 ) j = (.1 10 j) N 4 K =, away fom = ( N, away fom ) = ( N)[ cos(60 ) i+ sin(60 ) j] The foce on the 1.0 nc chage is.1 10 N diected upwad Model: The chages ae point chages. Please efe to igue P5.4. Solve: The electic foce on chage 1 is the vecto sum of the foces on 1 and on 1, whee 1 is the 1.0 nc chage, is the.0 nc chage, and is the.0 nc chages. We have K1 on 1 =, away fom ( N m /C )( C)(.0 10 C) =, away fom ( m) 4 4 = ( N, away fom ) = ( N)[cos(60 ) i+ sin(60 ) j] on 1 on 1 on 1 K1 4 4 on 1 =, towad = ( N, towad ) = ( N)[cos(60 ) i sin(60 ) j] 4 4 = + = ( N) cos(60 ) i = in So, the foce on the 1.0 nc chage is N and it is diected to the ight Model: The chages ae point chages. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

15 Electic Chages and oces 5-15 Solve: The electic foce on chage 1 is the vecto sum of the foces on 1 and on 1. We have K1 on 1 =, away fom ( N m /C )(10 10 C)( C) =, away fom ( m) = ( N, away fom ) = j N K1 on 1 =, towad ( N m /C )(10 10 C)(15 10 C) =, towad (.0 10 m) ( m) + = ( N, towad ) = ( N)( cosθi+ sin θ j) om the geomety of the figue, cm θ = tan = cm This means cos θ = and sin θ = Theefoe, on 1 ( = i j) N on 1 = on 1 + on 1 = ( i j) N The magnitude and diection of the esultant foce vecto ae on 1 = ( N) +( N) = N N 1 tanφ = =.180 φ = tan (.180) = 7 below the x axis, N o on 1 points 5 counteclockwise fom the +x-axis Model: The chages ae point chages. Solve: The point chages ae 1 = 10 nc, =+ 8.0 nc, and = 10 nc. The electic foce on chage 1 is the vecto sum of the foces on 1 and on 1. We have K = 1 on 1, towad ( N m /C )(10 10 C)( C) =, towad ( m) = (7. 10 N, towad ) = (7. 10 N) j Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

16 5-16 Chapte 5 K = ( N m /C )(10 10 C)(10 10 C) =, towad (.0 10 m) = =( ) 1 on 1, towad ( N, towad ) N i on 1 = on 1 + on 1 = + ( i j) N The magnitude and diection of the esultant foce vecto ae on 1 = ( N) + (7. 10 N) = N N 1 tanφ = φ = tan (7.) = 8 above the x-axis, N o on 1 points 98 counteclockwise fom the +x-axis Model: The chages ae point chages. Solve: The electic foce on chage 1 is the vecto sum of the foces on 1 and on 1. We have K = 1 on 1, away fom ( N m /C )(5 10 C)(10 10 C) =, away fom ( m) (.0 10 m) + = ( N, away fom ) = ( N)( cosθisin θ j) 4 4 om the geomety of the figue, 4.0 cm 4 tanθ = θ = 5.1 on 1 = ( N)( 0.60i 0.80 j).0 cm Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

17 Electic Chages and oces ( N m /C )(5 10 C)(5 10 C) 4 4 on 1 = = =, towad (.5 10 N, towad ).5 10 i N (.0 10 m) 4 4 = + = ( i j) N on 1 on 1 on 1 The magnitude and diection of the esultant foce vecto ae on 1 = ( N) + ( N) =.0 10 N N φ = tan 45 4 = clockwise fom the +x-axis N 5.8. Model: The chages ae point chages. Solve: The chages ae 1 =+ 5.0 nc, =+ 10 nc, and = 10 nc. The electic foce on 1 is the vecto sum of the foces on 1 and on 1. We have K = 1 on 1, away fom ( N m /C )(5 10 C)(10 10 C) =, away fom ( m) = ( N, away fom ) = ( N) j 1 on 1, towad 4 4 K = ( N m /C )( C)(10 10 C) =, towad ( m) (.0 10 m) + 4 = ( N)(cosθi+ sin θ j) Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

18 5-18 Chapte 5 om the geomety of the figue, 4.0 cm 1 4 tanθ = θ = tan = cm 4 4 = ( N)[cos(5.1 ) i + sin (5.1 ) j] = (1.80 i j) 10 N on 1 Theefoe, 4 4 on 1 = on 1 + on 1 = ( i j) N The magnitude and diection of the esultant foce vecto ae on 1 = ( N) + ( N) = N N tanφ = φ = clockwise fom the +x-axis. N 5.9. Model: The chages ae point chages. Please efe to igue P5.9. Solve: Placing the 1.0 nc chage at the oigin and calling it 1, the chage is in the fist uadant, the chage is in the fouth uadant, the 4 chage is in the thid uadant, and the 5 chage is in the second uadant. The electic foce on 1 is the vecto sum of the electic foces fom the othe fou chages,, 4, and 5. The magnitude of these fou foces is the same because all fou chages ae eual in magnitude and ae euidistant fom 1. So, ( N m /C )(.0 10 C)( C) 4 on 1 = on 1 = 4 on 1 = 5 on 1 = =.6 10 N ( m) + ( m) Thus, on 1 = (.6 10 N, away fom ) + (.6 10 N, away fom ) + (.6 10 N, towad 4) + (.6 towad 5 ). In component fom, 4 10 N, on 1 = on 1{ [ cos(45 ) i sin (45 ) j] + [cos(45 ) i+ sin (45 ) j] + [ cos(45 ) i sin (45 ) j] + [ cos(45 ) i+ sin (45 ) j] } 4 = (.6 10 N) [ 4cos(45 ) i]= in Assess: By symmety, we see that the vetical foces must cancel, and that the hoizontal foce must be in the negative diection, which agees with the calculation Model: The chages ae point chages. Please efe to igue P5.40. Solve: Placing the 1.0 nc chage at the oigin and calling it 1, the 6.0 nc is, the chage is in the fist uadant, and the 4 chage is in the second uadant. The net electic foce on 1 is the vecto sum of the electic foces fom the thee chages,, and 4. We have K = ( N m /C )( C)(.0 10 C) =, away fom ( m) = ( N, away fom ) = ( N)[ cos(45 ) i sin (45 ) j] 1 on 1, away fom 5 5 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

19 Electic Chages and oces 5-19 K = ( N m /C )( C)( C) =, towad ( m) 5 5 = ( N, away fom ) = j N K, away fom ( N)[cos(45 ) i sin(45 ) j] 1 on 1, towad on 1 = 4 = Summing these foces vectoially gives = + + = [( N) ( N)sin (45 )] j = j N on 1 on 1 on 1 4 on 1 Assess: By symmety, we see that the hoizontal foces must cancel, and that the vetical foce must be upwad, which agees with the calculation Model: The chages ae point chages. Please efe to igue P5.41. Solve: Place the 1.0 nc chage at the oigin and call it 1 ; the 6.0 nc is, the chage is in the fist uadant, and the 4 chage is in the second uadant. The net electic foce on 1 is the vecto sum of the electic foces fom the othe thee chages,, and 4. We have K = ( N m /C )( C)(.0 10 C) =, towad ( m) ( N, towad ) ( N)[cos(45 ) i sin (45 ) j] 1 on 1, towad 5 5 = = + K1 5 5 on 1 =, towad = (.1 = 6 10 N, towad ) j N K on 1 =, away fom 4 = ( N)[cos(45 ) i sin (45 ) j] Adding these components togethe vecto-wise gives 5 5 on 1 = on 1 + on on 1 = ( N) cos(45 ) i + ( N) j 5 5 = ( i j) N 5.4. Model: The chaged paticles ae point chages. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

20 5-0 Chapte 5 Solve: (a) The mathematical poblem is to find the position fo which the foces 1 on p and on p ae eual in magnitude and opposite in diection. If the poton is at position x, it is a distance x fom 1 and d x fom, whee d = 1.0 cm. The magnitudes of the foces ae K K K 1p x ( d x) Euating the two foces and using d = 1.0 cm, 1 p 1 p p 1 on p = = on p = K1 p K p.0 nc 4.0 nc = = 1+ x x= x x + x 1= 0 x ( d x) x ( d x) The solutions to the euation ae x = cm and.41 cm. Both ae points whee the magnitudes of the two foces ae eual, but cm is a point whee the magnitudes ae eual and the diections ae the same. The solution we want is that the poton should be placed at x =.4 cm. (b) Yes, the net foce on the electon located at x =.4 cm will also be zeo. This is because the solution in pat (a) does not depend specifically on the type of the chage that expeiences zeo foce fom the othe two chages. uthemoe, if on p is zeo fo a poton, the electic field at that point must be zeo. Thus, thee will be no foce on any chaged paticle at that point Model: The chaged paticles ae point chages. Solve: The two.0 nc chages exet an upwad foce on the 1.0 nc chage. Since the net foce on the 1 nc chage is zeo, the unknown chage must exet a downwad foce of eual magnitude. This implies that is a positive chage. The foce of chage on chage 1 is K1 on 1 =, away fom ( N m /C )( C)(.0 10 C) = (cosθ i+ sin θ j) (0.00 m) + (0.00 m) 1 om the figue, θ = tan (/) =.69. Thus 5 5 ( = i j) N on 1 om symmety, on 1 is the same except the x-component is evesed. When we add on 1 and on 1, x-components cancel and the y-components add to give 5 on 1+ on 1 = j N the Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

21 Electic Chages and oces 5-1 on 1 must have the same magnitude, pointing in the ĵ diection, so A positive chage 5 5 K1 ( N)(0.00 m) on = N = = = 0.68 nc ( N m /C )( C) = 0.68 nc will cause the net foce on the 1.0 nc chage to be zeo Model: The chaged paticles ae point chages. Please efe to igue P5.44. Solve: The chage is in static euilibium, so the net electic field at the location of is zeo. We have 9 1 ( C) E net = E + E 1 5 nc = K ( ± i) + K ( i) = 0N/C (0.0 m) (0.10 m) We have used the ± sign to indicate that a positive chage on 1 leads to an electic field along + î and a negative chage on 1 leads to an electic field along. î Because the above euation can only be satisfied if we use i, we infe that the chage 1 is a negative chage. Thus, C ( i) + ( i) = 0 N/C 1 = 0 nc 1= 0 nc (0.0 m) (0.10 m) Model: The chaged paticles ae point chages. Solve: The foce on is the vecto sum of the foce fom Q and +Q. We have Q + KQ Q on + = K, towad Q ( cosθi sin θ j) = a + y a + y K + Q + KQ + Q on + =, away fom + Q ( cosθi sin θ j) = + a y + a + y KQ net = ( cos θ ) i + 0 j N a + y om the figue we see that cos θ = a a + y. Thus ( ) KQa = ( a + y ) net x / Assess: Note that ( net ) x = net because the y-components of the two foces cancel each othe out Model: The chaged paticles ae point chages. Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

22 5- Chapte 5 Solve: (a) The foce on is the vecto sum of the foce fom Q and Q. We have K + Q + KQ + Q on + =, away fom + Q ( i) = ( a x) ( ax) K Q + KQ Q on + =, towad Q ( i) = ( a x) + ( a+ x) ( ) 1 1 KQ( a + x ) = KQ + = ( a x) ( a+ x) ( a x ) net x To aive at the final expession we used ( a x) ( a+ x) = [( a x)( a+ x)] = ( a x ). (b) Thee ae two cases when x > a. o x > a, K + Q + KQ + Q on + =, away fom + Q ( i) = + ( x a) ( xa) K Q + KQ Q on + =, towad Q ( i) = ( x a) + ( x+ a) ( ) 1 1 4KQax = KQ = ( x a) ( x+ a) ( x a ) net x o x < a (that is, fo negative values of x), KQ KQ + Q on + = ( i) Q on ( i) ( x a) + = + ( a+ x) ( ) 4KQax = ( x a ) net x That is, the net foce is to the ight when x > a and to the ight when x < a. We can combine these two cases into a single euation fo x > a: Hee, the foce is always to the ight when x > a. ( ) 4KQa x = ( x a ) net x Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

23 Electic Chages and oces Model: The chages ae point chages. Solve: We will denote the chages Q, 4Q and Q by 1,, and, espectively. K Q KQ 1 on =, towad Q ( i) = L L K 4Q 4 on =, away fom 4 KQ [ cos (45 ) sin (45 ) ] KQ Q = + i + j on = = 1 on ( L) L L K Q KQ on =, towad Q ( j) = L L The net electic foce on the chage + is the vecto sum of the electic foces fom the othe thee chages. The net foce is KQ KQ i j KQ KQ KQ net = ( i) + ( j) i(1 ) j(1 ) = L L L L L KQ KQ KQ = (1 ) + (1 ) = ( ) L L L net Model: The chages ae point chages. Solve: Label the chages,, and Q with numbes 1,, and, espectively. The positive chage exets an attactive foce on the chage at the ogin that is given by 1 1 on = K j = K j L L The chage Q exets an attactive foce on at the oigin that is given by Q α on = K i = K i ( L) 4L Because the net foce is at 45 to the x-axis (o y-axis), the the î and ĵ components must be of eual magnitude. Setting them eual an solving fo α gives α K K = α = 4 L 4L Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

24 5-4 Chapte Model: The chages ae point chages. We must fist identify the egion of space whee the thid chage is located. You can see fom the figue that the foces can t possibly add to zeo if is above o below the axis o outside the chages. Howeve, at some point on the x-axis between the two chages the foces fom the two chages will be oppositely diected. Solve: The mathematical poblem is to find the position fo which the foces 1 on and on ae eual in magnitude. If is the distance x fom 1, it is the distance L x fom. The magnitudes of the foces ae K 1 K K K(4 ) 1 on = = on = = 1 x ( L x) Euating the two foces gives K K(4 ) L = ( L x) = 4 x x= and L x ( L x) The solution x = L is not allowed as you can see fom the figue. To find the magnitude of the chage, we apply the euilibium condition to chage 1 : 1 1 on 1 on L 1 ( L ) K K 4 = = = = 9 We ae now able to check the static euilibium condition fo the chage 4 (o ): 4 1 K 9 1 on = on = = = ( ) ( L ) K L L x L L The sign of the thid chage must be negative. A positive sign on will not have a net foce of zeo eithe on the chage o the chage 4. In summay, a chage of 4 9 placed x = 1 L fom the chage will cause the -chage system to be in static euilibium Model: Use the chage model and assume the coppe sphees ae point objects with point chages. Solve: (a) The mass of the coppe sphee is 4π 4π 5 M = ρv = = (890 kg/m ) ( m) = kg = g The numbe of moles in the sphee is M g 4 n = = = mol A 6.5 g/mol The numbe of coppe atoms in the sphee is N = nn A = ( mol)( mol ) = Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

25 Electic Chages and oces The numbe of electons in the coppe sphee is thus = The total positive o negative 19 chage in the sphee is ( )( C) = 164 C. Hence, the sphees will have a net chage of C = C. The foce between two such sphees is 9 6 K1 ( N m /C )( C) = = =.4 10 N ( m) ( m) (b) This is a foce that is easily detectable. Since we don t obseve such foces, any diffeence between the poton 9 chage and the electon chage must be smalle than 1 pat in Model: The electon and the poton ae point chages. Solve: The electic Coulomb foce between the electon and the poton povides the centipetal acceleation fo the electon s cicula motion. Thus, Ke ()() e mv = = mω 9 19 Ke ( N m /C )( C) 16 1 ev 15 f = = = ( ad/s) ev/s 1 11 = m ( kg)(5. 10 ) π ad 5.5. Model: Model the chaged balls as point chages and ignoe fiction. Solve: Combining Coulomb s law and Newton s second law fo the balls at an abitay distance d apat gives thei acceleation: = K = ma a= K d md Thus, if we plot acceleation as a function of invese distance suaed, the magnitude of the chage can be calculated fom the slope s: K s= = m sm K The fit gives a slope of 4 s = N m /kg, so the magnitude of the chage is 4 ( N m /kg)(.0 10 kg) = = 8.1 nc 9 ( N m /C ) Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

26 5-6 Chapte Model: Model the bee as a point chage. Solve: (a) The foce on the bee due to gavity is G = mg, and the electic foce on the bee is e = E. The atio of the electic foce to the bee s weight is e G 1 E (100 N/C)( 10 C) = = =. 10 mg ( kg)(9.8 m/s ) (b) o the bee to be suspended by the electic foce, this foce must have the same magnitude as the foce due to gavity and be diected upwad. Euating the magnitudes of the foces and solving fo E give mg ( kg)(9.8 m/s ) 7 E = mg E = = = N/C 1 ( 10 C) The electic field must be diected upwad so that the foce on a positive chage is upwad Model: Model the metal plate and youself as point chages. Solve: At the beginning, both you and the metal plates ae neutal. As the electons ae pumped fom the metal plate into you, the plate gains as much positive chage as you gain negative chage. When enough chage diffeence builds up between you and the metal plate, the gavitational foce on you will be countebalanced by the upwad electical foce. You will begin to hang suspended in the ai when K plate on you = G = mg (.0 m) mg(.0 m) (60 kg)(9.8 N/kg)(.0 m) 4 = = = C K Nm /C Dividing this chage by the chage on an electon yields electons Model: The chaged plastic beads ae point chages and the sping is an ideal sping that obeys Hooke s law. Solve: Let be the chage on each plastic bead. The epulsive foce between the beads pushes the beads apat. The sping is stetched until the estoing sping foce on eithe bead is eual to the epulsive Coulomb foce: kδx = kδx = K K The sping constant k is obtained by noting that the weight of a 1.0 g mass stetches the sping 1.0 cm. Thus ( kg)(9.8 N/kg) mg = k( m) k = = 0.98 N/m m 6 (0.98 N/m)( m m)( m) = = nc N m /C Model: Take the cente of mass of the dipole to be midway between the two chages. s Solve: The toue about the cente of mass due to each chage is ( ) τ = and each is in the same diection, so the total toue is τ = s e. The electic foce e = E, so the toue may be witten as τ = Es = pe, whee we have used p s in the last step Solve: (a) Kinetic enegy is K = 1 mv, so the velocity suaed is v = K/ m. om kinematics, a paticle moving though distance Δx with acceleation a, stating fom est, finishes with v = aδ x. To gain K = 10 J of kinetic enegy in Δx =.0 μm euies an acceleation of 18 v K/ m K.0 10 J a = = = = = m/s m/s Δx Δx mδ x 1 6 ( kg)(.0 10 m) e 18 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

27 Electic Chages and oces 5-7 (b) The foce that poduces this acceleation is (c) The electic field euied is = ma= ( kg)( m/s ) = N N N N/C E = = = e C (d) The foce on an electon due to a chage is = K e/. To have beakdown, the foce on the electon must be 1 at least N. The minimum chage that could cause a beakdown will be the chage that causes exactly a 1 foce of N: 1 Ke 1 (0.010 m) ( N) 8 = = N = = = C = 69 nc Ke 9 19 ( N m /C )( C) Model: The chaged sphees behave as point chages. Each sphee is in static euilibium and the sting makes an angle θ with the vetical. The thee foces acting on each sphee ae the electic foce, the gavitational foce on the sphee, and the tension foce. Solve: In static euilibium, Newton s fist law gives net = T + + = 0. In component fom, Dividing the two euations gives ( ) = T + ( ) + ( ) = 0N ( ) = T + ( ) + ( ) = 0N net x x G x e x net y y G e y K Tsinθ + 0N+ = 0N Tcosθ mg + 0N= 0N d K K Tsinθ = = Tcosθ =+ mg d (Lsin θ ) 9 9 K ( N m /C )( C) 4 θ θ = = = sin tan Lmg 4(1.0 m) ( kg)(9.8 N/kg) o small-angles, tanθ sin θ. With this appoximation we obtain sinθ = ad,, so θ = Model: The chaged sphees behave as point chages. G e Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

28 5-8 Chapte 5 Each sphee is in static euilibium when the sting makes an angle of 0 with the vetical. The thee foces acting on each sphee ae the electic foce, the gavitational foce on the sphee, and the tension foce. Solve: In the static euilibium, Newton s fist law gives net = T + ( ) + = 0. In component fom, we have ( ) = T + ( ) + ( ) = 0N ( ) = T + ( ) + ( ) = 0N net x x G x e x net y y G e y K Tsinθ + 0N+ = 0N Tcosθ mg + 0N= 0N d K + K Tsinθ = = Tcosθ =+ mg d (Lsin θ ) Dividing the two euations and solving fo gives 4sin θ tanθlmg 4(sin 0 tan 0 )(1.0 m) (.0 10 kg)(9.8 N/kg) = = = 0.75 μc K N m /C G e Model: The electic field is that of a positive point chage located at the oigin. Please efe to igue P5.60. Place the +10 nc chage at the oigin. Solve: The electic field is 9 9 ( N m /C )(10 10 C) 90 N m /C E = K, away fom, away fom, away fom = = The electic field at each of the thee points is 90.0 N m /C 5 E1 =, away fom = j N/C (.0 10 m) 90.0 N m /C 4 E =, away fom = (.6 10 N/C)(cosθi + sin θ j) ( m) = (.6 10 N/C) i + j = (.9 10 i j) N/C ( ) N m /C 4 E = = ( m), away fom i N/C Model: The electic field is that of a negative point chage. Please efe to igue P5.61. Place point 1 at the oigin. Solve: The electic field is 9 9 ( N m /C )(.0 10 C) E = K, towad, towad = 18.0 N m /C =, towad Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

29 Electic Chages and oces 5-9 The electic fields at the two points ae 18.0 N m /C E1 =, towad ( m) 5 = ( N/C, 60 counte clockwise fom the + x-axis o 60 noth of east) 18.0 N m /C E =, towad ( m) 5 = ( N/C, 60 clockwise fom the x-axis o 60 noth of west) 5.6. Model: The electic field is that of a positive point chage located at the oigin. Please efe to igue P5.6. Place the 5.0 nc chage at the oigin. Solve: The electic field is 9 9 (9 10 N m /C )( C) E = K, away fom, away fom = 45 N m /C =, away fom The electic field at the thee points is 45 N m /C 4 E1 =, away fom = ( N/C)(cosθi+ sin θ j) (.0 10 m) ( m) = ( N/C) ( i + j= i + j) N/C N m /C 5 E =, away fom = i N/C ( m) 45Nm /C 4 4 E =, away fom ( i j) N/C = (.0 10 m) + ( m) 5.6. Model: The electic field is that of a negative chage at (x, y) = (.0 cm, 1.0 cm). Solve: (a) The electic field of a negative chage points towad the chage, so we can oughly locate whee the field has a paticula value by inspecting the signs of E x and E y. At point a, the electic field has no y-component and the x-component points to the left, so its location must be to the ight of the chage along a hoizontal line. Using the euation fo the field of a point chage, a 9 9 K K ( N m /C )( C) Ex = E = a = = = 0.00 m =.0 cm E 5,000 N/C Thus, point a is at the position ( xa, y a) = (4.0 cm, 1.0 cm). x Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

30 5-0 Chapte 5 (b) Point b is above and to the left of the chage. The magnitude of the field at this point is Using the euation fo the field of a point chage, x y E = E + E = (161,000 N/C) + (80,500 N/C) = 180,000 N/C 9 9 K K ( N m /C )(10 10 C) E = b = = =.6 cm b E 180,000 N/C This gives the total distance but not the hoizontal and vetical components. Howeve, we can detemine the angle θ because E b points staight towad the negative chage. Thus, E 1 y 1 80, θ = tan = tan = tan = 6.57 Ex 161,000 The hoizontal and vetical distances ae then dx = b cosθ =.00 cm and dy = b sinθ = 1.00 cm. Thus, point b is at the position ( xb, y b) = (0.0 cm,.0 cm). (c) Point c, which is below and to the left of the chage, is calculated by following a simila pocedue. We fist find that E = 6,000 N/C. om this we find that the total distance c = 5.00 cm. The angle φ is E 1 y 1 1,600 φ = tan = tan = 6.87 Ex 8,000 which gives the distances dx = c cosφ = 4.00 cm and dy = c sinφ =.00 cm. Thus point c is at position ( x c, y c ) = (.0 cm,.0 cm) Model: The electic field is that of a positive chage at (x, y) = (1.0 cm,.0 cm). Solve: (a) The electic field of a positive chage points staight away fom the chage, so we can oughly locate the points of inteest based simply on whethe the signs of E x and E y ae positive o negative. o point a, the electic field has no y-component and the x-component points to the left, so point a must be to the left of the chage along a hoizontal line. Using the field of a point chage, a 9 9 K K ( N m /C )( C) Ex = E = a = = = 0.00 m =.0 cm E 5,000 N/C x Thus, ( xa, y a) = ( 1.0 cm,.0 cm). (b) Point b is above and to the ight of the chage. The magnitude of the field at this point is x y E = E + E = (161,000 N/C) + (80,500 N/C) = 180,000 N/C Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.

31 Electic Chages and oces 5-1 Using the field of a point chage, b 9 9 K K ( N m /C )( C) E = b = = =.6 cm E 180,000 N/C This gives the total distance but not the hoizontal and vetical components. Howeve, we can detemine the angle θ because E points staight away fom the positive chage. Thus, b E 1 y 1 80,500 N/C 1 1 θ = tan = tan = tan = 6.57 Ex 161,000 N/C The hoizontal and vetical distances ae then dx = b cos θ = (.6 cm)cos6.57 =.00 cm and dy = b sinθ = 1.00 cm. Thus, point b is at position ( xb, y b) = (.0 cm,.0 cm). (c) To calculate point c, which is below and to the ight of the chage, a simila pocedue is followed. We fist find E = 6,000 N/C fom which we find the total distance c = 5.00 cm. The angle φ is E 1 y 1 8,800 φ = tan = tan = 5.1 Ex 1,600 which gives distances dx = c cosφ =.00 cm and dy = c sinφ = 4.00 cm. Thus, point c is at position ( x c, y c ) = (4.0 cm,.0 cm) Model: The electic field is that of thee point chages. Solve: (a) In the figue, the distances ae 1= = (1.0 cm) + (.0 cm) =.16 cm and the angle is θ = tan (1.0/.0) = Using the euation fo the field of a point chage, 9 9 K1 1 E 1 (0.016 m) ( N m /C )( C) E = = = = 9.0 kn/c We now use the angle θ to find the components of the field vectos: E1= E 1cosθi E1sin θ j = (8540i 840 j) N/C = (8.5i.8 j) kn/c E = Ecosθi + Esin θ j = (8540i j) N/C = (8.5 i +.8 j) kn/c E is easie since it has only an x-component. Its magnitude is E = = = 10,000 N/C E = E i= 10ikN/C 9 9 K ( N m /C )( C) (0.000 m) (b) The electic field is defined in tems of an electic foce acting on chage E : = /. Since foces obey a pinciple of supeposition ( net = ) it follows that the electic field due to seveal chages also obeys a pinciple of supeposition. (c) The net electic field at a point.0 cm to the ight of is E net = E1 + E + E = 7i kn/c. The y-components of E 1 and E cancel, giving a net field pointing along the x-axis. 1 Copyight 01 Peason Education, Inc. All ights eseved. This mateial is potected unde all copyight laws as they cuently exist.