© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.1

CHAPTER 16: Electric Charge and Electric FieldAnswers to Questions1. A plastic ruler is suspended by a thread and then rubbed with a cloth. As discussed in section 16-1,the ruler is negatively charged. Bring the charged comb close to the ruler. If the ruler is repelled bythe comb, then the comb is negatively charged. If the ruler is attracted by the comb, then the comb ispositively charged.2. The clothing gets charged by frictional contact in the tumbling motion of the dryer. The air insidethe dryer is dry, and so the clothes can sustain a relatively large static charge. That charged objectwill then polarize your clothing, and be attracted to you electrostatically.3. Water is a polar molecule – it has a positive region and a negative region. Thus it is easily attractedto some other charged object, like an ion or electron in the air.4. The positively charged rod slightly polarizes the molecules in the paper. The negative charges in thepaper are slightly attracted to the part of the paper closest to the rod, while the positive charges in thepaper are slightly repelled from the part of the paper closest to the rod. Since the opposite chargesare now closer together and the like charges are now farther apart, there is a net attraction betweenthe rod and the paper.5. The plastic ruler has gained some electrons from the cloth and thus has a net negative charge. Thischarge polarizes the charge on the piece of paper, drawing positives slightly closer and repellingnegatives slightly further away. This polarization results in a net attractive force on the piece ofpaper. A small amount of charge is able to create enough electric force to be stronger than gravity,and so the paper can be lifted.On a humid day this is more difficult because the water molecules in the air are polar. Those polarwater molecules are able to attract some fraction of the free charges away from the plastic ruler.Thus the ruler has a smaller charge, the paper is less polarized, and there is not enough electric forceto pick up the paper.6. The net charge on the conductor is the unbalanced charge, or excess charge after neutrality has beenestablished. The net charge is the sum of all of the positive and negative charges in the conductor. Ifa neutral conductor has extra electrons added to it, then the net charge is negative. If a neutralconductor has electrons removed from it, then the net charge is positive. If a neutral conductor hasthe same amount of positive and negative charge, then the net charge is zero.Free charges in a conductor refer to those electrons (usually 1 or 2 per atom) that are so looselyattracted to the nucleus that they are “free” to be moved around in the conductor by an externalelectric force. Neutral conductors have these free electrons.7. For each atom in a conductor, only a small number of its electrons are free to move. For example,every atom of copper has 29 electrons, but only 1 or 2 from each atom are free to move easily. Also,not even all of the free electrons move. As electrons move toward a region, causing an excess ofnegative charge, that region then exerts a large repulsive force on other electrons, preventing themfrom all gathering in one place.Chapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.28. The force of gravity pulling down on the leaves, tending to return them to the vertical position.9. The magnitude of the constant in Newton’s law is very small, while the magnitude of the constant inCoulomb’s law is quite large. Newton’s law says the gravitational force is proportional to theproduct of the two masses, while Coulomb’s law says the electrical force is proportional to theproduct of the two charges. Newton’s law only produces attractive forces, since there is only onekind of gravitational mass. Coulomb’s law produces both attractive and repulsive forces, since thereare two kinds of electrical charge.

10. For the gravitational force, we don’t notice it because the force is very weak, due to the very smallvalue of G, the gravitational constant, and the small value of ordinary masses. For the electric force,we don’t notice it because ordinary objects are electrically neutral to a very high degree. We noticeour weight (the force of gravity) due to the huge mass of the Earth, making a significant gravityforce. We notice the electric force when objects have a static charge (like static cling from theclothes dryer), creating a detectable electric force.11. The electric force is conservative. You can “store” energy in it, and get the energy back. Forexample, moving a positive charge close to another stationary positive charge takes work (similar tolifting an object in the Earth’s gravitational field), but if the positive charge is then released, it willgain kinetic energy and move away from the “stored energy” location (like dropping an object in theEarth’s gravitational field). Another argument is that the mathematical form of Coulomb’s law isidentical to that of Newton’s law of universal gravitation. We know that gravity is conservative, andso we would assume that the electric force is also conservative. There are other indications as well.If you move a charge around in an electric field, eventually returning to the starting position, the network done will be 0 J. The work done in moving a charge around in an electric field is pathindependent – all that matters is the starting and ending locations. All of these are indications of aconservative force.12. The charged plastic ruler has a negative charge residing on its surface. That charge polarizes thecharge in the neutral paper, producing a net attractive force. When the piece of paper then touchesthe ruler, the paper can get charged by contact with the ruler, gaining a net negative charge. Then,since like charges repel, the paper is repelled by the comb.13. The test charge creates its own electric field, and so the measured electric field is the sum of theoriginal electric field plus the field of the test charge. By making the test charge small, the field thatit causes is small, and so the actual measured electric field is not much different than the originalfield to measure.14. A negative test charge could be used. For purposes of defining directions, the electric field mightthen be defined as the OPPOSITE of the force on the test charge, divided by the test charge.Equation (16-3) might be changed to E F q , q 0 .15.Giancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.316. The electric field is strongest to the right of the positive charge, because the individual fields fromthe positive charge and negative charge both are in the same direction (to the right) at that point, sothey add to make a stronger field. The electric field is weakest to the left of the positive charge,because the individual fields from the positive charge and negative charge are in opposite directionsat that point, and so they partially cancel each other. Another indication is the spacing of the fieldlines. The field lines are closer to each other to the right of the positive charge, and further apart tothe left of the positive charge.17. At point A, the net force on a positive test charge would be down and to the left, parallel to thenearby electric field lines. At point B, the net force on a positive test charge would be up and to theright, parallel to the nearby electric field lines. At point C, the net force on a positive test chargewould be 0. In order of decreasing field strength, the points would be ordered A, B, C.18. Electric field lines show the direction of the force on a test charge placed at a given location. Theelectric force has a unique direction at each point. If two field lines cross, it would indicate that theelectric force is pointing in two directions at once, which is not possible.19. From rule 1: A test charge would be either attracted directly towards or repelled directly away froma point charge, depending on the sign of the point charge. So the field lines must bedirected either radially towards or radially away from the point charge.From rule 2: The magnitude of the field due to the point charge only depends on the distance fromthe point charge. Thus the density of the field lines must be the same at any locationaround the point charge, for a given distance from the point charge.From rule 3: If the point charge is positive, the field lines will originate from the location of the

point charge. If the point charge is negative, the field lines will end at the location ofthe point charge.Based on rules 1 and 2, the lines are radial and their density is constant for a given distance. This isequivalent to saying that the lines must be symmetrically spaced around the point charge.20. If the two charges are of opposite sign, then E 0 at a point closer to the weaker charge, and on theopposite side of the weaker charge from the stronger charge. The fields due to the two charges are ofopposite direction at such a point. If the distance between the two charges is l, then the point atwhich E 0 is 2.41 l away from the weaker charge, and 3.41 l away from the stronger charge.If the two charges are the same sign, then E 0 at a point between the two charges, closer to theweaker charge. The point is 41% of the distance from the weaker charge to the stronger charge.21. We assume that there are no other forces (like gravity) acting on the test charge. The direction of theelectric field line gives the direction of the force on the test charge. The acceleration is alwaysparallel to the force by Newton’s 2nd law, and so the acceleration lies along the field line. If theparticle is at rest initially and then released, the initial velocity will also point along the field line,and the particle will start to move along the field line. However, once the particle has a velocity, itwill not follow the field line unless the line is straight. The field line gives the direction of theacceleration, or the direction of the change in velocity.Chapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.422. Since the line of charge is infinitely long, it has no preferred left or right direction. Thus bysymmetry, the lines must point radially out away from the center of the line. A side view and an endview are shown for a line of positive charge. As seen from the end view, the field is not uniform.As you move further away from the line of charge, the field lines get further apart, indicating that thefield gets weaker as you move away from the line of charge.23. Just because the electric flux through a closed surface is zero, the field need not be zero on thesurface. For example, consider a closed surface near an isolated point charge, and the surface doesnot enclose the charge. There will be electric field lines passing through the surface, but the totalelectric flux through the surface will be zero since the surface does not enclose any charge. Thesame number of field lines will enter the volume enclosed by the surface as leave the volumeenclosed by the surface.On the contrary, if E 0 at all points on the surface, then there are no electric field lines passingthrough the surface, and so the flux through the surface is zero.24. The electric flux depends only on the charge enclosed by the gaussian surface, not on the shape ofthe surface. E will be the same for the cube as for the sphere.

Solutions to Problems1. Use Coulomb’s law to calculate the magnitude of the force.6 21 2 9 2 22 2 2

3.60 10 C8.988 10 N m C 13.47N 13N9.3 10 mQQF kr2. Use the charge per electron to find the number of electrons.6 1419

1 electron30.0 10 C 1.87 10 electrons1.602 10 C3. Use Coulomb’s law to calculate the magnitude of the force.19 191 2 9 2 2 32 12 2

1.602 10 C 26 1.602 10 C8.988 10 N m C 2.7 10 N1.5 10 mQQF kr4. Use Coulomb’s law to calculate the magnitude of the force.19 21 2 9 2 22 15 2

1.602 10 C8.988 10 N m C 9.2N5.0 10 mQQF krGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.55. Use Coulomb’s law to calculate the magnitude of the force.6 31 2 9 2 2 32 1 2

25 10 C 3.0 10 C8.988 10 N m C 5.5 10 N3.5 10 mQQF kr6. Since the magnitude of the force is inversely proportional to the square of the separation distance,2

1Fr, if the distance is multiplied by a factor of 1/8, the force will be multiplied by a factor of 64.2

0 F 64F 64 3.2 10 N 2.0 N7. Since the magnitude of the force is inversely proportional to the square of the separation distance,2

1Fr, if the force is tripled, the distance has been reduced by a factor of 3 .0 8.45 cm4.88 cm3 3rr8. Use the charge per electron and the mass per electron.6 14 14193114 16

1 electron42 10 C 2.622 10 2.6 10 electrons1.602 10 C9.11 10 kg2.622 10 e 2.4 10 kg1 e9. Convert the kg of H2O to moles, then to atoms, then to electrons. Oxygen has 8 electrons per atom,

and hydrogen has 1 electron per atom.23 1922 2 27

1mole H O 6.02 10 molec. 10 e 1.602 10 C1.0 kg H O 1.0 kg H O1.8 10 kg 1 mole 1 molec. e5.4 10 C10. Take the ratio of the electric force divided by the gravitational force.1 2 9 2 2 19 22E 1 2 3911 2 2 31 27G 1 2 1 22

8.988 10 N m C 1.602 10 C2.3 106.67 10 N m kg 9.11 10 kg 1.67 10 kgQQkF r kQQF m m Gm m GrThe electric force is about 2.3 1039 times stronger than the gravitational force for the given scenario.11. (a) Let one of the charges be q , and then the other charge is T Q q . The force between thecharges is2T 2 2E 2 2 T 2 TT T

q Q q k k q qF k qQ q Qr r r Q Q. If we letT

qxQ, then2 2E 2 T

kF Q x xr, where 0 x 1. A graph of 2 f x x x between the limits of 0 and 1Chapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.6shows that the maximum occurs at x 0.5, or T q 0.5Q . Both charges are half of the total,and the actual maximized force is 2

E 2 T 0.25kF Qr.(b) If one of the charges has all of the charge, and the other has no charge, then the force betweenthem will be 0, which is the minimum possible force.12. Let the right be the positive direction on the line of charges. Use the fact that like charges repel andunlike charges attract to determine the direction of the forces. In the following expressions,

9 2 2 k 8.988 10 N m C .275 2 2248 2 2285 2 2

75 C 48 C 75 C 85 C147.2N 1.5 10 N0.35m 0.70m75 C 48 C 48 C 85 C563.5N 5.6 10 N0.35m 0.35m85 C 75 C 85 C 48 C416.3N 4.2 10 N0.70m 0.35mF k kF k kF k k13. The forces on each charge lie along a line connecting the charges. Let thevariable d represent the length of a side of the triangle, and let the variable Qrepresent the charge at each corner. Since the triangle is equilateral, each angle is60o.2 2 2o o12 2 12 2 12 22 2 2o o13 2 13 2 13 22 2o1 12 13 1 12 13 2 2

cos60 , sin 60cos60 , sin 600 2 sin 60 3x yx yx x x y y y

Q Q QF k F k F kd d dQ Q QF k F k F kd d dQ QF F F F F F k kd d6 2 22 2 9 2 21 1 1 2 2

11.0 10 C3 3 8.988 10 N m C 83.7 N0.150m x y

QF F F kdThe direction of 1 F is in the y-direction . Also notice that it lies along the bisector of the oppositeside of the triangle. Thus the force on the lower left charge is of magnitude 83.7 N , and will pointo 30 below the x axis . Finally, the force on the lower right charge is of magnitude 83.7 N , andwill point o 30 below the x axis .14. Determine the force on the upper right charge, and then use thesymmetry of the configuration to determine the force on the other threecharges. The force at the upper right corner of the square is the vectorsum of the forces due to the other three charges. Let the variable d

represent the 0.100 m length of a side of the square, and let the variableQ represent the 6.00 mC charge at each corner.41 F 1 Q2 Q 3 Qd4 Q43 F 42 F1 Q2 Q 3 Q12 Fd13 FddGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.72 241 2 41 2 412 2 2 2o42 2 42 2 2 42 22 243 2 43 43 2

, 02 2cos45 ,2 2 4 40 ,x yx yx y

Q QF k F k Fd dQ Q Q QF k F k k F kd d d dQ QF k F F kd dAdd the x and y components together to find the total force, noting that 4x 4 y F F .2 2 24 41 42 43 2 2 2 4

2 20 14 4 x x x x y

Q Q QF F F F k k k Fd d d2 22 24 4 4 2 2

2 11 2 24 2 x y

Q QF F F k kd d3 29 2 2 72

6.00 10 C 18.988 10 N m C 2 6.19 10 N0.100m 21 4 o4

tan 45 yx

FFabove the x-direction.For each charge, the net force will be the magnitude determined above, and will lie along the linefrom the center of the square out towards the charge.15. Determine the force on the upper right charge, and then the symmetry of the configuration says thatthe force on the lower left charge is the opposite of the force on the upper right charge. Likewise,determine the force on the lower right charge, and then the symmetry of the configuration says thatthe force on the upper left charge is the opposite of the force on the lower right charge.The force at the upper right corner of the square is the vector sum of theforces due to the other three charges. Let the variable d represent the0.100 m length of a side of the square, and let the variable Q representthe 6.00 mC charge at each corner.2 241 2 41 2 412 2 2 2o42 2 42 2 2 42 22 243 2 43 43 2

, 02 2cos45 ,2 2 4 40 ,x yx yx y

Q QF k F k Fd dQ Q Q QF k F k k F kd d d dQ QF k F F kd dAdd the x and y components together to find the total force, noting that 4x 4 y F F .2 2 2 24 41 42 43 2 2 2 2 4

2 20 1 0.646454 4 x x x x y

Q Q Q QF F F F k k k k Fd d d d2 22 2

4 4 4 2 2 0.64645 2 0.9142 x y

Q QF F F k kd d41 F1 Q2 Q 3 Q

d4 Q43 F42 FChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.83 29 2 2 72

6.00 10 C8.988 10 N m C 0.9142 2.96 10 N0.100m1 4 o4

tan 225 yx

FFfrom the x-direction, or exactly towards the center of the square.For each charge, the net force will be the magnitude of 7 2.96 10 N and each net force will liealong the line from the charge inwards towards the center of the square.16. Take the lower left hand corner of the square to be the origin of coordinates. Each charge will have ahorizontal force on it due to one charge, a vertical force on it due to one charge, and a diagonal forceon it due to one charge. Find the components of each force, add the components, find the magnitudeof the net force, and the direction of the net force. At the conclusion of the problem is a diagramshowing the net force on each of the two charges.(a)2 2o2 2 2 2 2

2 2 42 : cos 45 2 2 2 4.82842 Qx

Q Q Q Q Q kQQ F k k kl l l l2 2o2 2 2 2 222 2 1 2 1 o2 2 2 2 22

2 3 2 4sin 45 6 2 2 8.828428.828410.1 tan tan 614.8284QyyQ Qx Qy Qx

Q Q Q Q Q kQF k k kl l l lkQ FF F Fl F(b)2 2o3 2 2 2 2

3 4 3 33 : cos 45 12 2 13.06072 4 Qx

Q Q Q Q Q kQQ F k k kl l l l2 2o3 2 2 2 222 2 1 3 1 o3 3 3 2 33

3 2 3 3sin 45 6 2 7.06072 47.060714.8 tan tan 33213.0607QyyQ Qx Qy Qx

Q Q Q Q Q kQF k k kl l l lkQ FF F Fl F17. The forces on each charge lie along a line connecting the charges. Letthe variable d represent the length of a side of the triangle. Since thetriangle is equilateral, each angle is 60o. First calculate the magnitudeof each individual force.Q 2Q4Q 3Ql3Q F2Q F1 Q2 Q 3 Q12 Fd13 Fdd32 F31 F 21 F23 FGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.96 61 2 9 2 212 2 2216 61 3 9 2 213 2 231

4.0 10 C 8.0 10 C8.988 10 N m C1.20m0.1997N

4.0 10 C 6.0 10 C8.988 10 N m C1.20m0.1498NQQF kdFQQF kdF6 62 3 9 2 223 2 2 32

8.0 10 C 6.0 10 C8.988 10 N m C 0.2996N1.20mQ QF k FdNow calculate the net force on each charge and the direction of that net force, using components.o o 21 12 13o o 11 12 1312 2 1 1 1 o1 1 1 1 21

0.1997N cos 60 0.1498N cos 60 2.495 10 N0.1997N sin 60 0.1498N sin 60 3.027 10 N3.027 10 N0.30N tan tan 2652.495 10 Nx x xy y yyx yx

F F FF F FFF F FFo 12 21 23o 12 21 2312 2 1 2 1 o2 2 2 2 12

0.1997N cos60 0.2996N 1.998 10 N0.1997N sin 60 0 1.729 10 N1.729 10 N0.26N tan tan 1391.998 10 Nx x xy y yyx yx

F F FF F FFF F FF

o 13 31 32o 13 31 3212 2 1 3 1 o3 3 3 3 13

0.1498N cos60 0.2996N 2.247 10 N0.1498N sin 60 1.297 10 N1.297 10 N0.26N tan tan 302.247 10 N0x x xy y yyx yx

F F FF F FFF F FF18. Since the force is repulsive, both charges must be the same sign. Since the total charge is positive,both charges must be positive. Let the total charge be Q. Then if one charge is of magnitude q, thenthe other charge must be of magnitude Q q . Write a Coulomb’s law expression for one of thecharges.22222 6 6 229 2 24 6 6 4

04 4 22.8N 1.10m 560 10 C 560 10 C8.988 10 N m C2 25.54 10 C , 5.54 10 C 5.54 10 C , 5.54 10 Cq Q q FrF k q Qqr kFrQ Qq kQ qChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.1019. The negative charges will repel each other, and so the third chargemust put an opposite force on each of the original charges.Consideration of the various possible configurations leads to theconclusion that the third charge must be positive and must be betweenthe other two charges. See the diagram for the definition of variables.For each negative charge, equate the magnitudes of the two forces on the charge. Also note that0 x l .2 20 0 0 02 2 2 20 02 22 20 02 2 0 2 0 2 0

3 3 3

left: right:30.3663 13 33 0.4023 1Q Q Q Q Q Qk k k kx l l x lQ Q Q Q lk k x lx l xQ Q Q xk k Q Q Q Qx l lThus the charge should be of magnitude 0 0.40Q , and a distance 0 0 0.37l from Q towards 3Q .20. Assume that the negative charge is d = 18.5 cm to the right of thepositive charge, on the x-axis. To experience no net force, thethird charge Q must be closer to the smaller magnitude charge(the negative charge). The third charge cannot be between thecharges, because it would experience a force from each charge inthe same direction, and so the net force could not be zero. And the third charge must be on the linejoining the other two charges, so that the two forces on the third charge are along the same line. Seethe diagram. Equate the magnitudes of the two forces on the third charge, and solve for x > 0.1 2 21 2 2 21 2626 61 2

3.5 10 C18.5cm 116cm4.7 10 C 3.5 10 CQ Q Q Q Qk k x dd x x Q QQx dQ QF F21. (a) If the force is repulsive, both charges must be positive since the total charge is positive. Call thetotal charge Q.21 2 1 1 21 2 2 2 1 12 22 2126 6 29 2 26 6

04 42 212.0N 1.06m90.0 10 C 90.0 10 C 48.988 10 N m C2

69.9 10 C , 22.1 10 CkQQ kQ Q Q FdQ Q Q F Q QQd d kFd FdQ Q Q QQ k k+ x4.7 C –3.5 C–1 Q 2 QQdx0 Q Q0 3Qll – xGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.11(b) If the force is attractive, then the charges are of opposite sign. The value used for F must thenbe negative. Other than that, the solution method is the same as for part (a).21 2 1 1 21 2 2 2 1 12 22 2126 6 29 2 26 6

04 42 212.0N 1.06m90.0 10 C 90.0 10 C 48.988 10 N m C2104.4 10 C , 14.4 10 CkQQ kQ Q Q FdQ Q Q F Q QQd d kFd FdQ Q Q QQ k k22. The spheres can be treated as point charges since they are spherical, and so Coulomb’s law may beused to relate the amount of charge to the force of attraction. Each sphere will have a magnitude Qof charge, since that amount was removed from one sphere and added to the other, being initiallyuncharged.2 21 22 2 9 2 27 1219

1.7 10 N0.12 m8.988 10 N m C

1 electron1.650 10 C 1.0 10 electrons1.602 10 CQQ Q FF k k Q rr r k23. Use Eq. 16–3 to calculate the force.19 16 q 1.602 10 C 2360N C east 3.78 10 N westqFE F E24. Use Eq. 16–3 to calculate the electric field.14519

3.75 10 Nsouth2.34 10 N C southq 1.602 10 CFE25. Use Eq. 16–3 to calculate the electric field.56

8.4Ndown9.5 10 N C upq 8.8 10 CFE26. Use Eq. 16–4a to calculate the electric field due to a point charge.69 2 2 62 1 2

33.0 10 C8.988 10 N m C 7.42 10 N C up2.00 10 mQE krNote that the electric field points away from the positive charge.Chapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.1227. Assuming the electric force is the only force on the electron, then Newton’s 2nd law may be used tofind the acceleration.1914 2net 31

1.602 10 C750N C 1.32 10 m s9.11 10 kgqm q a EmF a ESince the charge is negative, the direction of the acceleration is opposite to the field .28. The electric field due to the negative charge will pointtoward the negative charge, and the electric field dueto the positive charge will point away from the

positive charge. Thus both fields point in the samedirection, towards the negative charge, and so can beadded.1 2 1 21 2 2 2 2 2 2 1 21 29 2 26 6 72 2

4/ 2 / 24 8.988 10 N m C8.0 10 C 7.0 10 C 8.4 10 N C8.0 10 mQ Q Q Q kE E E k k k k Q Qr r d d dThe direction is towards the negative charge .29.30. Assuming the electric force is the only force on the electron, then Newton’s 2nd law may be used tofind the electric field strength.27 6 219

1.67 10 kg 1 10 9.80m s0.102N C 0.1N C1.602 10 C net

maF ma qE Eq31. Since the electron accelerates from rest towards the north, the net force on it must be to the north.Assuming the electric force is the only force on the electron, then Newton’s 2nd law may be used tofind the electric field.312 10net 19

9.11 10 kg115m s north 6.54 10 N C south1.602 10 Cmm qqF a E E a32. The field due to the negative charge will point towardsthe negative charge, and the field due to the positivecharge will point towards the negative charge. Thusthe magnitudes of the two fields can be added togetherto find the charges.1 Q 0 2 Q 01 Ed 2 2 EQ E Qd 2Q EQGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.132 1

10net 2 2 9 2 2

8 745N C 1.60 10 m2 2 2.65 10 C/ 2 8 8 8.988 10 N m C Q

Q kQ EdE E k Qd d k33. The field at the center due to the two negative charges on oppositecorners (lower right and upper left in the diagram) will canceleach other, and so only the positive charge and the oppositenegative charge need to be considered. The field due to thenegative charge will point directly toward it, and the field due tothe positive charge will point directly away from it. Accordingly,the two fields are in the same direction and can be addedalgebraically.1 2 1 21 2 2 2 269 2 2 6 o2

2 2 247.0 27.0 10 C8.988 10 N m C 4.70 10 N C at 450.525m 2Q Q Q QE E E k k kd d d34. The field at the upper right corner of the square is the vector sum ofthe fields due to the other three charges. Let the variable d representthe 1.0 m length of a side of the square, and let the variable Qrepresent the charge at each of the three occupied corners.1 2 1 2 1o2 2 2 2 2 2 23 2 3 1 2

, 02 2cos45 ,2 2 4 40 ,x yx yx y

Q QE k E k Ed dQ Q Q QE k E k k E kd d d dQ QE k E E kd dAdd the x and y components together to find the total electric field, noting that x y E E .1 2 3 2 2 2

2 20 14 4 x x x x y

Q Q QE E E E k k k Ed d d2 2

2 2

2 11 2 24 2 x y

Q QE E E k kd d69 2 2 42

2.25 10 C 18.988 10 N m C 2 3.87 10 N C1.00m 21 o tan 45 yx

EEfrom the x-direction.1 E3 E2 E1 Q2 Q 3 Qd2 Q1 Q 45.0 Cd2 Q1 E2 E2 Q 27.0 CChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.1435. Choose the rightward direction to be positive. Then the field due to +Q will be positive, and thefield due to –Q will be negative.2 2 2 2 2 2 2

Q Q 1 1 4kQxaE k k kQx a x a x a x a x aThe negative sign means the field points to the left .36. For the net field to be zero at point P, the magnitudes of the fields created by 1 Q and 2 Q must beequal. Also, the distance x will be taken as positive to the left of 1 Q . That is the only region wherethe total field due to the two charges can be zero. Let the variable d represent the 12 cm distance,and note that 1

1 2 2 Q Q .1 21 2 2 211 2 212 1 2 2 2

12cm29cm2 1 2 1Q Qk kx x dQ Q d

x d dQ Q Q QE E37. (a) The field due to the charge at A will point straight downward, and thefield due to the charge at B will point along the line from A to theorigin, 30o below the negative x axis.A 2 A A 2oB 2 B 2 2oB 2 2A B 2 A B 22 2 2 2 2 22 24 4 4 2

0 ,3cos302sin 3023 32 23 9 12 34 4 4tax xxyx x x y y yx y

Q QE k E E kl lQ Q QE k E k kl l lQ QE k kl lQ QE E E k E E E kl lk Q k Q k Q kQE E El l l l21 1 1 1 o2

32 3 n tan tan tan 3 2403 32yx

QE k lE Qkl(b) Now reverse the direction of A EA 2 A A 2o oB 2 B 2 2 B 2 2A B 2 A B 2

0 ,3cos30 sin 302 232 2,x xx yx x x y y y

Q QE k E E kl lQ Q Q Q QE k E k k E k kl l l l lQ QE E E k E E E kl lA EQlllQB EABGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.152 2 2 2 2 22 24 4 4 221 1 1 o2

3 44 4 42 1 tan tan tan 3303 32x yyx

k Q k Q k Q kQE E El l l lQE k lE Qkl38. In each case, find the vector sum of the field caused by the charge on the left left E and the fieldcaused by the charge on the right right EPoint A: From the symmetry of the geometry, incalculating the electric field at point A only the verticalcomponents of the fields need to be considered. Thehorizontal components will cancel each other.1 o2 2

5.0tan 26.610.0d 5.0cm 10.0cm 0.1118m69 2 2 o 6 oA 2 2 A

7.0 10 C2 sin 2 8.988 10 N m C sin 26.6 4.5 10 N C 900.1118mkQEdPoint B: Now the point is not symmetrically placed, andso horizontal and vertical components of each individualfield need to be calculated to find the resultant electricfield.1 o 1 o2 22 2left leftleftright

5.0 5.0tan 45 tan 18.45.0 15.05.0cm 5.0cm 0.0707m5.0cm 15.0cm 0.1581mddleft right 2 left 2 rightleft righto o9 2 2 6 62 2left right 2 left 2 rightleft right

cos coscos45 cos18.48.988 10 N m C 7.0 10 C 6.51 10 N C0.0707m 0.1581msin sin8.988x x xy y y

Q QE k kd dQ QE k kd dE EE Eo o9 2 2 6 62 22 2 7 1 oB

sin45 sin18.410 N m C 7.0 10 C 9.69 10 N C0.0707m 0.1581m1.2 10 N C tan 56 yx y Bx

EE E E

Edleft EQ Qright EdAright Eleft Eleft dQ left right Qright dChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.16The results are consistent with Figure 16-31b. In the figure, the field at Point A points straight up,matching the calculations. The field at Point B should be to the right and vertical, matching thecalculations. Finally, the field lines are closer together at Point B than at Point A, indicating that thefield is stronger there, matching the calculations.39. Both charges must be of the same sign so that the electric fields created by the two charges opposeeach other, and so can add to zero. The magnitudes of the two electric fields must be equal.1 2 2 11 2 2 2 12

9 193 2 3 4 4Q Q Q QE E k k Ql l Q40. From the diagram, we see that the x components of the two fields will cancel each other at the pointP. Thus the net electric field will be in the negativey-direction, and will be twice the y-component ofeither electric field vector.net 2 22 2 2 2 1/ 22 2 3/ 2

2 sin 2 sin22in the negative directionkQE Ex akQ ax a x akQayx a41. We assume that gravity can be ignored, which is proven in part (b).(a) The electron will accelerate to the right. The magnitude of the acceleration can be found fromsetting the net force equal to the electric force on the electron. The acceleration is constant, soconstant acceleration relationships can be used.net2 2019 42 631

2 2 21.602 10 C 1.45 10 N C2 1.10 10 m 7.49 10 m s9.11 10 kgq EF ma q E amq Ev v a x v a x xm(b) The value of the acceleration caused by the electric field is compared to g.19 415 23115 2142

1.602 10 C 1.45 10 N C2.55 10 m s9.11 10 kg2.55 10 m s2.60 109.80m sq EamagThe acceleration due to gravity can be ignored compared to the acceleration caused by theelectric field.42. (a) The electron will experience a force in the opposite direction to the electric field. Since theelectron is to be brought to rest, the electric field must be in the same direction as the initialvelocity of the electron, and so is to the right .xQ EQQaaQ EGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.17(b) Since the field is uniform, the electron will experience a constant force, and therefore have aconstant acceleration. Use constant acceleration relationships to find the field strength.2 2 20 02 2 31 6 2 20 0 219 2

2 29.11 10 kg 3.0 10 m s6.4 10 N C2 2 2 1.602 10 C 4.0 10 mqE qEF qE ma a v v a x v xm mm v v mvE

q x q x43. Use Gauss’s law to determine the enclosed charge.encl 3 2 12 2 2 8

E encl E 1.45 10 N m C 8.85 10 C N m 1.28 10 C oo

QQ44. (a)2 2 1 2 2

E E A E r 5.8 10 N C 1.8 10 m 59N m C(b)o 2 2 o 1 2 2

E E A Ecos45 r 5.8 10 N C cos45 1.8 10 m 42N m C(c) o 2

E E A Ecos90 r 045. (a) Use Gauss’s law to determine the electric flux.6encl 5 2E 12 2 2

1.0 10 C1.1 10 N m C8.85 10 C N m o

Q(b) Since there is no charge enclosed by surface A2, E 0 .46. (a) Assuming that there is no charge contained within the cube, then the net flux through the cube is0 . All of the field lines that enter the cube also leave the cube.(b) There are four faces that have no flux through them,because none of the field lines pass through those faces.In the diagram shown, the left face has a positive fluxand the right face has the opposite amount of negativeflux.2 3 2left3 2right other

6.50 10 N C6.50 10 N C 0EA El ll47. Equation 16-10 applies.12 2 2 2 900

8.85 10 C N m 130N C 1.0m 1.15 10 CQ AE Q EA48. The electric field can be calculated by Eq. 16-4a, and that can be solved for the charge.2 2 2 2112 9 2 2

2.75 10 N C 3.50 10 m3.75 10 C8.988 10 N m CQ ErE k Qr kThis corresponds to about 2 108 electrons. Since the field points toward the ball, the charge mustbe negative .lChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the

publisher.1849. See Example 16-11 for a detailed discussion related to this problem.(a) Inside a solid metal sphere the electric field is 0 .(b) Inside a solid metal sphere the electric field is 0 .(c) Outside a solid metal sphere the electric field is the same as if all the charge were concentratedat the center as a point charge.69 2 2 32 2

3.50 10 C8.988 10 N m C 3.27 10 N C3.10mQE kr(d) Same reasoning as in part (c).69 2 2 22 2

3.50 10 C8.988 10 N m C 8.74 10 N C6.00mQE kr(e) The answers would be no different for a thin metal shell.50. See Figure 16-33 in the text for additional insight into this problem.(a) Inside the shell, the field is that of the point charge, 2

QE kr.(b) There is no field inside the conducting material: E 0 .(c) Outside the shell, the field is that of the point charge, 2

QE kr.(d) The shell does not affect the field due to Q alone, except in the shell material, where the field is0. The charge Q does affect the shell – it polarizes it. There will be an induced charge of –Quniformly distributed over the inside surface of the shell, and an induced charge of +Quniformly distributed over the outside surface of the shell.51. (a) The net force between the thymine and adenine is due to the following forces.O – H attraction:2OH 2 2 o o

0.4 0.2 0.081.80A 1.80Ae e keF kO – N repulsion:2ON 2 2 o o

0.4 0.2 0.082.80A 2.80Ae e keF kN – N repulsion:2

NN 2 2 o o

0.2 0.2 0.043.00A 3.00Ae e keF kH – N attraction:2HN 2 2 o o

0.2 0.2 0.042.00A 2.00Ae e keF k2 2A-T OH ON NN HN 2 2 2 2 10 2

0.08 0.08 0.04 0.04 11.80 2.80 3.00 2.00 1.0 10 mkeF F F F FdGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.199 2 2 19 210 1010 2

8.988 10 N m C 1.602 10 C.02004 4.623 10 N 4.6 10 N1.0 10 m(b) The net force between the cytosine and guanine is due to the following forces.O – H attraction:2OH 2 2 o o

0.4 0.2 0.081.90A 1.90Ae e keF k (2 of these)O – N repulsion:2ON 2 2 o o

0.4 0.2 0.082.90A 2.90Ae e keF k (2 of these)H – N attraction:2HN 2 2 o o

0.2 0.2 0.042.00A 2.00Ae e keF kN – N repulsion:2NN 2 2 o o

0.2 0.2 0.043.00A 3.00Ae e keF k2 2C-G OH ON NN HN 2 2 2 2 10 29 2 2 19 210 1010 2

0.08 0.08 0.04 0.04 1

2 2 21.90 2.90 3.00 2.00 1.0 10 m8.988 10 N m C 1.602 10 C.03085 7.116 10 N 7.1 10 N1.0 10 m2keF F F F Fd(c) For 5 10 pairs of molecules, we assume that half are A-T pairs and half are C-G pairs. Weaverage the above results and multiply by 5 10 .5 5 10 10net A-T C-G5 51

210 10 4.623 10 N 7.116 10 N5.850 10 N 6 10 NF F F52. Set the magnitude of the electric force equal to the magnitude of the force of gravity and solve forthe distance.2E G 29 2 21931 2

8.988 10 N m C1.602 10 C 5.08m9.11 10 kg 9.80m seF F k mgrkr emg53. Calculate the total charge on all electrons in 3.0 g of copper, and then compare the 38 C to thatvalue.23 195

1 mole 6.02 10 atoms 29 e 1.602 10 CTotal electron charge 3.0g63.5g mole atoms 1 e1.32 10 CChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.206105

38 10 CFraction lost 2.9 101.32 10 C54. Since the gravity force is downward, the electric force must be upward. Since the charge is positive,the electric field must also be upward. Equate the magnitudes of the two forces and solve for theelectric field.27 27E G 19

1.67 10 kg 9.80m s1.02 10 N C, up1.602 10 C

mgF F qE mg Eq55. Use Eq. 16-4a to calculate the magnitude of the electric charge on the Earth.6 2 252 9 2 2

150 N C 6.38 10 m6.8 10 C8.988 10 N m CQ ErE k Qr kSince the electric field is pointing towards the Earth’s center, the charge must be negative .56. (a) From problem 55, we know that the electric field is pointed towards the Earth’s center. Thus anelectron in such a field would experience an upwards force of magnitude E F eE . The forceof gravity on the electron will be negligible compared to the electric force.E1913 2 13 231

1.602 10 C 150N C2.638 10 m s 2.6 10 m s , up9.11 10 kgF eE maeEam(b) A proton in the field would experience a downwards force of magnitude E F eE . The force ofgravity on the proton will be negligible compared to the electric force.E1910 2 10 227

1.602 10 C 150N C1.439 10 m s 1.4 10 m s , down1.67 10 kgF eE maeEam(c) For the electron:13 2122

2.638 10 m s2.7 109.80m sagFor the proton:10 292

1.439 10 m s1.5 109.80m sag57. For the droplet to remain stationary, the magnitude of the electric force on the droplet must be thesame as the weight of the droplet. The mass of the droplet is found from its volume times the density

of water. Let n be the number of excess electrons on the water droplet.4 3E 35 3 3 3 2 36 719

4 4 1.8 10 m 1.00 10 kg m 9.80m s9.96 10 1.0 10 electrons3 3 1.602 10 C 150N CF q E mg neE r gr gneE58. There are four forces to calculate. Call the rightward direction the positive direction. The value of kis 8.988 109 N m2 C2 and the value of e is 1.602 10 19C .Giancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.21net CH CN OH ON 9 2 2 2 2 210 10

0.40 0.20 1 1 1 11 10 m 0.30 0.40 0.18 0.282.445 10 N 2.4 10 Nk e eF F F F F59. The electric force must be a radial force in order for the electron to move in a circular orbit.2 2E radial 2orbit orbit19 2 29 2 2 10orbit 2 2 31 6

1.602 10 C8.988 10 N m C 2.1 10 m9.11 10 kg 1.1 10 m sQ mvF F kr rQr kmv60. Set the Coulomb electrical force equal to the Newtonian gravitational force on one of the bodies (theMoon).2Moon EarthE G 2 2orbit orbit11 2 2 22 24Moon Earth 139 2 2

6.67 10 N m kg 7.35 10 kg 5.98 10 kg5.71 10 C8.988 10 N m CQ M MF F k Gr rGM MQk61. (a) The electron will experience a force in the opposite direction to the electric field. Thus theacceleration is in the opposite direction to the initial velocity. The force is constant, and so

constant acceleration equations apply. To find the stopping distance, set the final velocity to 0.2 2031 6 2 2 2 20 019 3

29.11 10 kg 21.5 10 m s0.115m2 2 2 1.602 10 C 11.4 10 N CeEF eE ma a v v a xmv v mvxa eE(b) To return to the starting point, the velocity will reverse. Use that to find the time to return.031 60 0 0 0 819 3

2 2 9.11 10 kg 21.5 10 m s2.14 10 s1.602 10 C 11.4 10 N Cv v atv v v v mvta a qE62. Because of the inverse square nature of the electric field,any location where the field is zero must be closer to theweaker charge 2 Q . Also, in between the two charges,the fields due to the two charges are parallel to each other and cannot cancel. Thus the only placeswhere the field can be zero are closer to the weaker charge, but not between them. In the diagram,this means that x must be positive.2 1 2 2

2 2 2 1 0Q QE k k Q l d Qll l d1 Q2 Qd lChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.2262 25 61 2 1

5.0 10 C 1.6mfrom ,2.0m2.5 10 C 5.0 10 C 2.6mfromQ Ql dQ Q Q63. The sphere will oscillate sinusoidally about the equilibrium point, with an amplitude of 5.0 cm. Theangular frequency of the sphere is given by k m 126N m 0.800kg 12.5rad s . Thedistance of the sphere from the table is given by r 0.150 0.050cos 12.5t m. Use this distanceand the charge to give the electric field value at the tabletop. That electric field will point upwards at

all times, towards the negative sphere.9 2 2 6 42 2 2 272

8.988 10 N m C 3.00 10 C 2.70 10N C0.150 0.050cos 12.5 m 0.150 0.050cos 12.51.08 10N C, upwards3.00 cos 12.5QE kr t tt64. The wires form two sides of an equilateral triangle, and so the two charges areseparated by a distance d 78 cm and are directly horizontal from each other. Thusthe electric force on each charge is horizontal. From the free-body diagram for one ofthe spheres, write the net force in both the horizontal and vertical directions and solvefor the electric force. Then write the electric force by Coulomb’s law, and equate thetwo expressions for the electric force to find the charge.T TT E E T2E 23 2 o1 6 69 2 2

cos 0cossin 0 sin sin tancos2 tantan 224 10 kg 9.80m s tan302 7.8 10 m 6.064 10 C 6.1 10 C8.988 10 N m Cyx

mgF F mg FmgF F F F F mgQ mgF k mg Q dd k65. The electric field at the surface of the pea is given by Equation (16-4a). Solve that equation for thecharge.6 3 2 292 9 2 2

3 10 N C 3.75 10 m5 10 C8.988 10 N m CQ ErE k Qr kThis corresponds to about 3 billion electrons.66. There will be a rightward force on 1 Q due to 2 Q , given by Coulomb’s law. There will be a leftwardforce on 1 Q due to the electric field created by the parallel plates. Let right be the positive direction.1 22 1

QQF k Q ExmgE FT FGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.236 69 2 2 6 42

6.7 10 C 1.8 10 C8.988 10 N m C 6.7 10 C 7.3 10 N C0.34m0.45N, right67. Since the electric field exerts a force on the charge in thesame direction as the electric field, the charge ispositive. Use the free-body diagram to write theequilibrium equations for both the horizontal and verticaldirections, and use those equations to find the magnitudeof the charge.1 oE T E TT T3 2 o74

43cos 38.655sin 0 sincos 0 tancostan 1.0 10 kg 9.80m s tan 38.66.5 10 C1.2 10 N Cxy

F F F F F QEmgF F mg F QE mgmgQE68. The weight of the mass is only about 2 N. Since the tension in the string is more thanthat, there must be a downward electric force on the positive charge, which means thatthe electric field must be pointed down . Use the free-body diagram to write anexpression for the magnitude of the electric field.T E E T2T 77

05.67N 0.210kg 9.80m s1.06 10 N C3.40 10 CF F mg F F QE F mgF mgE

Q69. To find the number of electrons, convert the mass to moles, the moles to atoms, and then multiply bythe number of electrons in an atom to find the total electrons. Then convert to charge.23 1928

1 mole Al 6.02 10 atoms 13 electrons 1.602 10 C15 kg Al 15 kg Al2.7 10 kg 1 mole 1 molecule electron7.0 10 CThe net charge of the bar is 0C , since there are equal numbers of protons and electrons.70. (a) The force of sphere B on sphere A is given by Coulomb’s law.2

AB 2 , away from BkQFR(b) The result of touching sphere B to uncharged sphere C is that the charge on B is shared betweenthe two spheres, and so the charge on B is reduced to Q 2 . Again use Coulomb’s law.2AB 2 2

2, away from B2QQ kQF kR RmgE FT FL 55cmE FT Fmg43cmChapter 16 Electric Charge and Electric Field© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.24(c) The result of touching sphere A to sphere C is that the charge on the two spheres is shared, andso the charge on A is reduced to 3Q 4 . Again use Coulomb’s law.2AB 2 2

3 4 2 3, away from B8Q Q kQF kR R71. On the x-axis, the electric field can only be zero at a location closer to the smaller magnitude charge.Thus the field will never be zero to the left of the midpoint between the two charges. Also, inbetween the two charges, the field due to both charges will point to the left, and so the total fieldcannot be zero. Thus the only place on the x-axis where the field can be zero is to the right of thenegative charge, and so x must be positive. Calculate the field at point P and set it equal to zero.2 22 2

2

0 2 2.412 1Q Q dE k k x x d x dx x dThe field cannot be zero at any points off the x-axis. For any point off the x-axis, the electric fieldsdue to the two charges will not be along the same line, and so they can never combine to give 0.72. The electric field will put a force of magnitude E F QE oneach charge. The distance of each charge from the pivot point isL 2 , and so the torque caused by each force is E 2QELF r .Both torques will tend to make the rod rotate counterclockwisein the diagram, and so the net torque is net 22QELQEL .73. A negative charge must be placed at the center of the square. LetQ 8.0 C be the charge at each corner, let -q be the magnitude ofnegative charge in the center, and let d 9.2cm be the side length ofthe square. By the symmetry of the problem, if we make the net forceon one of the corner charges be zero, the net force on each othercorner charge will also be zero.2 241 2 41 2 412 2 2 2o42 2 42 2 2 42 22 243 2 43 43 2o4 2 4 2 2 4

, 02 2cos45 ,2 2 4 40 ,2 2cos 452x yx yx yq qx qy

Q QF k F k Fd dQ Q Q QF k F k k F kd d d dQ QF k F F kd dqQ qQ qQF k F k k Fd d dThe net force in each direction should be zero.2 262 2 2

2 2 1 10 0 7.66 10 C

4 2 4 x

Q Q qQF k k k q Qd d dSo the charge to be placed is 6 q 7.66 10 C .41 F 1 Qd

2 Q 3 Qq4q F4 Q43 F 42 FQQEE FE FGiancoli Physics: Principles with Applications, 6th Edition© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as theycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.25This is an unstable equilibrium . If the center charge were slightly displaced, say towards the right,then it would be closer to the right charges than the left, and would be attracted more to the right.Likewise the positive charges on the right side of the square would be closer to it and would beattracted more to it, moving from their corner positions. The system would not have a tendency toreturn to the symmetric shape, but rather would have a tendency to move away from it if disturbed.