Electrostatics AP Study guide

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1 Electrostatics AP Study guide 1) Know how charges interact Coulomb s Law (pg ) (Practice questions Ch. 1: Q1-4, Q6-1) (Practice problems Ch. 1: P1 P3) Positive and negative How to calculate the force on one charge from another How to calculate the distance between charged objects given the charges and the force between them How to calculate the charge of one of the objects given the other objects charge, the force and the distance ) Know how to use the shell theorem (pg. 567) (Practice questions: Ch. 1: Q5) A shell of charge of uniform charge density can be crushed down to a point charge located at the center of the shell for objects outside it and is rendered non-existent for charged objects within it 3) Know the charge of an electron and proton and know what k stands for (pg. 571, 566) 4) Know what the electric field is, how to interpret E-field lines, and how it is defined (pg , ) (Practice questions Ch. : Q1- Q6) (Practice problems Ch. : P1 P17, P39 P55) How is the electric field defined with reference to a test charge What do the electric field lines for various charge distributions look like How can one calculate the electric field at a particular Place in space from a single or multiple point charges How can one calculate the force on a charged particle at a Place in space if the E-field there is known Is E-field a scalar or vector 5) Know how to use an integral and symmetry to find the E-field at a Place in space from a continuous charge distribution (straight line, circular arc, circular ring, disc, etc.) (pg ) (Practice questions Ch. : Q7, Q8, Q11) (Practice problems Ch. : P P38) 6) Know when and how to use Gauss Law to find the electric field of a charge distribution (pg , 615-6) (Practice questions Ch. 3: Q1 Q5) (Practice problems Ch. 3: P1 P16, P P48, P53 P55) Know what flux is How to pick the correct symmetrical shape to use Gauss Law How to use symmetry to make the E-field constant so you can pull it out of the integral How to use Gauss Law when there is more than one charged distribution 7) Know the properties of charged, isolated conductors and insulators (pg ) (Practice questions Ch. 3: Q6 Q1) (Practice problems Ch. 3: P17 P1, P49 P5) E-field inside a charged conductor How charges distribute themselves on a conductor Volume Charge densities for an insulator and why volume charge densities for conductors is an oxymoron

2 8) Know about Electrical Potential Energy U e (pg. 69, ) (Practice questions Ch. 4: Q7, Q9, Q1) (Practice problems Ch. 4: P41 P 61) How is electrical potential energy defined Is Electrical potential energy a scalar or vector How is the electrical potential energy of a finite number of point particles calculated 9) Know about Electric Potential V for point individual or groups of point charges, continuous distributions of charge, and how V relates to U e (pg , , ) (Practice questions Ch. 4: Q1 Q3, Q5, Q8) (Practice problems Ch. 4: P1- P3, P1 P, P3 P33) How is electric potential defined Is electric potential a scalar or vector How is electric potential, V, related to electric potential energy, U e What is an electron volt (ev) 1) Know how electric potential and the E-field are related (pg , ) (Practice questions Ch 4: Q4, Q6) (Practice problems Ch 4: P4 P11, P34 P4) What is an equipotential surface and how do the lines of equipotential surfaces relate to E-field lines How to calculate the potential difference (ΔV) (note: ΔV = V f V i ) through an E-field given the E-field function How to calculate the E-field from the potential difference given the potential function 11) Know how potential is related to conductors especially the potential inside a conductor and the E-field inside a conductor suspended in an external E-field (pg ) (Practice problems Ch. 4: P6 P67) How does the potential change throughout a conductor What is the E-field inside a conductor (always)

3 1 r 1. Eq. 1-1 gives Coulomb s Law, F k q q, which we solve for the distance: r k q q F Nm C 6.1 C C 1 5.7N 1.39m.. (a) With a understood to mean the magnitude of acceleration, Newton s second and third laws lead to 7 c63. 1 kghc7. m s h 7 ma ma 1 1 m kg. 9. ms (b) The magnitude of the (only) force on particle 1 is q1 q 9 q 1 1 F ma k N m C. r (.3 m) Inserting the values for m 1 and a 1 (see part (a)) we obtain q = C.

4 3. The magnitude of the mutual force of attraction at r =.1 m is C C q1 q 9 F k N m C.81N. r (.1 m) 4. The fact that the spheres are identical allows us to conclude that when two spheres are in contact, they share equal charge. Therefore, when a charged sphere (q) touches an uncharged one, they will (fairly quickly) each attain half that charge (q/). We start with spheres 1 and each having charge q and experiencing a mutual repulsive force F kq / r. When the neutral sphere 3 touches sphere 1, sphere 1 s charge decreases to q/. Then sphere 3 (now carrying charge q/) is brought into contact with sphere, a total amount of q/ + q becomes shared equally between them. Therefore, the charge of sphere 3 is 3q/4 in the final situation. The repulsive force between spheres 1 and is finally ( q/ )(3 q/ 4) 3 q 3 F' F k k F r 8 r 8 F 8

5 5. The magnitude of the force of either of the charges on the other is given by 6. The unit Ampere is discussed in 1-4. Using i for current, the charge transferred is F 1 4 b qqq where r is the distance between the charges. We want the value of q that maximizes the function f(q) = q(q q). Setting the derivative df / dq equal to zero leads to Q q =, or q = Q/. Thus, q/q =.5. r g 4 6 qit.51 A 1 s.5 C.

6 7. We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb s law can be used. Let q 1 and q be the original charges. We choose the coordinate system so the force on q is positive if it is repelled by q 1. Then, the force on q is 1 qq F k qq 1 1 a r r 4 where r =.5 m. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is (q 1 + q )/. The force is now one of repulsion and is given by q q q q d id i b 1 g F k q q b 4 r 4r We solve the two force equations simultaneously for q 1 and q. The first gives the product qq 1 b g b g rf 5. m 18. N a C, 9 k Nm C The solutions are q 1 c h c h C C C. If the positive sign is used, q 1 = C, and if the negative sign is used, 6 q C. (a) Using q = ( )/q 1 with q 1 = C, we get q 1. 1 C. (b) If we instead work with the q 1 = C root, then we find q 3. 1 C. Note that since the spheres are identical, the solutions are essentially the same: one sphere originally had charge C and the other had charge C. What if we had not made the assumption, above, that q 1 + q? If the signs of the charges were reversed (so q 1 + q < ), then the forces remain the same, so a charge of C on one sphere and a charge of C on the other also satisfies the conditions of the problem. and the second gives the sum Fb q1q r. 5m k b g. 36N N m C C where we have taken the positive root (which amounts to assuming q 1 + q ). Thus, the product result provides the relation 3.1 C q q which we substitute into the sum result, producing 3. 1 q1 q C 6. 1 C. Multiplying by q 1 and rearranging, we obtain a quadratic equation q c 1. 1 C q 3. 1 C. 6 1 h 1

7 8. For ease of presentation (of the computations below) we assume Q > and q < (although the final result does not depend on this particular choice). (a) The x-component of the force experienced by q 1 = Q is 1 QQ q Q Qq Q/ q F1 x cos a a a which (upon requiring F 1x = ) leads to Q/ q, or Q/ q.83. (b) The y-component of the net force on q = q is 1 q q Q q 1 Q F y sin a a a q 9. The force experienced by q 3 is 1 q3 q1 ˆ q3 q ˆ ˆ q3 q4 ˆ F3 F31F3F34 j (cos45isin45j) i 4 a ( a) a (a) Therefore, the x-component of the resultant force on q 3 is 7 q q 1. 1 C 1 F q Nm C.17N x 4 4 a (.5 m) (b) Similarly, the y-component of the net force on q 3 is 7 q q 1. 1 C 1 F q Nm C 1.46N y 1 4 a (.5 m) which (if we demand F y = ) leads to Q/ q 1/. The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 1-1.

8 1. (a) The individual force magnitudes (acting on Q) are, by Eq. 1-1, 1 q Q q Q 1 aa/ aa/ which leads to q 1 = 9. q. Since Q is located between q 1 and q, we conclude q 1 and q are like-sign. Consequently, q 1/q = 9.. (b) Now we have 1 q Q q Q 1 a3 a/ a3 a/ which yields q 1 = 5 q. Now, Q is not located between q 1 and q, one of them must push and the other must pull. Thus, they are unlike-sign, so q 1/q = With rightwards positive, the net force on q 3 is qq qq F F F k k. L L 1 L 3 3 We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on q 3 by q 1) is negative if they are unlike charges, indicating that q 3 is being pulled toward q 1, and it is positive if they are like charges (so q 3 would be repelled from q 1). Setting the net force equal to zero L 3= L 1 and canceling k, q 3 and L 1 leads to q q q q

9 1. As a result of the first action, both sphere W and sphere A possess charge 1 qa, where q A is the initial charge of sphere A. As a result of the second action, sphere W has charge 1 q A 3e. As a result of the final action, sphere W now has charge equal to 11q A 3e 48e. Setting this final expression equal to +18e as required by the problem leads (after a couple of algebra steps) to the answer: q A = +16e. 13. (a) Eq. 1-1 gives 6.1 C 9 1.5m qq F k Nm C 1.6N. d 1 1 (b) On the right, a force diagram is shown as well as our choice of y axis (the dashed line). The y axis is meant to bisect the line between q and q 3 in order to make use of the symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges q 1 = q = q 3 = q). We see that the resultant force is along this symmetry axis, and we obtain 6.1 C 9 1.5m q Fy k d cos N m C cos3.77 N.

10 14. (a) According to the graph, when q 3 is very close to q 1 (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q 1 has the same sign as q 3. Thus, q 3 is a positive-valued charge. (b) Since the graph crosses zero and particle 3 is between the others, q 1 must have the same sign as q, which means it is also positive-valued. We note that it crosses zero at r =. m (which is a distance d =.6 m from q ). Using Coulomb s law at that point, we have 1 qq qq 3 d.6 m q q1 q19. q1, 4 r 4 d r. m or q /q 1 = (a) There is no equilibrium position for q 3 between the two fixed charges, because it is being pulled by one and pushed by the other (since q 1 and q have different signs); in this region this means the two force arrows on q 3 are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis which is nearest q and furthest from q 1 an equilibrium position for q 3 cannot be found because q 1 < q and the magnitude of force exerted by q is everywhere (in that region) stronger than that exerted by q 1 on q 3. Thus, we must look in the semi-infinite region of the axis which is nearest q 1 and furthest from q, where the net force on q 3 has magnitude qq k k qq L L L with L = 1 cm and L is assumed to be positive. We set this equal to zero, as required by the problem, and cancel k and q 3. Thus, we obtain 1 L L 1 q q L L q 3. C 3. L L q 1. C which yields (after taking the square root) L L L 1 cm 3 L 14cm L for the distance between q 3 and q 1. That is, q 3 should be placed at x 14 cm along the x-axis. (b) As stated above, y =.

11 16. Since the forces involved are proportional to q, we see that the essential difference between the two situations is F a q B + q C (when those two charges are on the same side) versus F b q B + q C (when they are on opposite sides). Setting up ratios, we have F a qb + qc F = b - q B + q C N 1 qc / qb N 1 q / q After noting that the ratio on the left hand side is very close to 7, then, after a couple of algebra steps, we are led to qc q 71 6 B C B. 17. (a) The distance between q 1 and q is r1 x x1 y y1. m.35 m.15 m.5 m.56 m. The magnitude of the force exerted by q 1 on q is Nm C 3.1 C C qq F k 35 N. 1 1 r1 (.56 m) (b) The vector F 1 is directed towards q 1 and makes an angle with the +x axis, where 1 y y cm.5 cm tan tan xx1. cm 3.5 cm (c) Let the third charge be located at (x 3, y 3), a distance r from q. We note that q 1, q and q 3 must be collinear; otherwise, an equilibrium position for any one of them would be impossible to find. Furthermore, we cannot place q 3 on the same side of q where we also find q 1, since in that region both forces (exerted on q by q 3 and q 1) would be in the same direction (since q is attracted to both of them). Thus, in terms of the angle found in part (a), we have x 3 = x r cos and y 3 = y r sin (which means y 3 > y since is negative). The magnitude of force exerted on q by q 3 is F3 k qq3 r, which must equal that of the force exerted on it by q 1 (found in part (a)). Therefore, qq qq q k k r r.645m 6.45 cm r r1 q1 Consequently, x 3 = x r cos =. cm (6.45 cm) cos( 1 ) = 8.4 cm, (d) and y 3 = y r sin = 1.5 cm (6.45 cm) sin( 1 ) =.7 cm.

12 18. (a) For the net force to be in the +x direction, the y components of the individual forces must cancel. The angle of the force exerted by the q 1 = 4 C charge on q3 C is 45, and the angle of force exerted on q 3 by Q is at where 1. cm tan cm Therefore, cancellation of y components requires qq Q q k sin 45k sin m (.3 m) (. m) from which we obtain Q = 83 C. Charge Q is pulling on q 3, so (since q 3 > ) we conclude Q = 83 C. (b) Now, we require that the x components cancel, and we note that in this case, the angle of force on q 3 exerted by Q is + (it is repulsive, and Q is positive-valued). Therefore, qq Qq k cos 45k cos m (.3 m) (. m) from which we obtain Q = 55. C 55 C. 19. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q 3 must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q 3 could not be in equilibrium. Suppose q 3 is at a distance x from q, and L x from 4.q. The force acting on it is then given by 1 qq3 4qq 3 F3 4 x L x where the positive direction is rightward. We require F 3 = and solve for x. Canceling common factors yields 1/x = 4/(L x) and taking the square root yields 1/x = /(L x). The solution is x = L/3. With L = 9. cm, we have x = 3. cm. (b) Similarly, the y coordinate of q 3 is y =. (c) The force on q is 1 qq3 4.q Fq 4 x L The signs are chosen so that a negative force value would cause q to move leftward. We require F q = and solve for q 3:. 4qx 4 q3 4 q 3 q.444 L 9 q 9 where x = L/3 is used. Note that we may easily verify that the force on 4.q also vanishes: 4qq 4 4 9q L x 1 4q 1 4q 1 4q 4q F4 q 4 L 4 L 4 9 L 4. L L

13 . (a) We note that cos(3º) = 1 3, so that the dashed line distance in the figure is r d/ 3. We net force on q 1 due to the two charges q 3 and q 4 (with q 3 = q 4 = C) on the y axis has magnitude qq 3 3 qq cos(3 ) r d This must be set equal to the magnitude of the force exerted on q 1 by q = C = 5. q 3 in order that its net force be zero: 3 3 qq qq 16 4 ( ) d D d Given d =. cm, this then leads to D = 1.9 cm. D = d =.945 d. (b) As the angle decreases, its cosine increases, resulting in a larger contribution from the charges on the y axis. To offset this, the force exerted by q must be made stronger, so that it must be brought closer to q 1 (keep in mind that Coulomb s law is inversely proportional to distance-squared). Thus, D must be decreased. 1. If is the angle between the force and the x-axis, then We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q 1 or q on q 3. Let e = C, then q 1 = q = +e and q 3 = 4.e and we have F net = F cos = cos = (e)(4e) 4 o (x + d ) x x + d. x 4e x x + d = o (x + d ) 3/. (a) To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior and as a quick way to see whether an extremum point is a maximum or a miminum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b)) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x =, which is the smallest value of the net force in the interval 5. m x. (b) The maximum is found to be at x = d/ or roughly 1 cm. (c) The value of the net force at x = is F net =. (d) The value of the net force at x = d/ is F net = N.

14 . We note that the problem is examining the force on charge A, so that the respective distances (involved in the Coulomb force expressions) between B and A, and between C and A, do not change as particle B is moved along its circular path. We focus on the endpoints ( = º and 18º) of each graph, since they represent cases where the forces (on A) due to B and C are either parallel or antiparallel (yielding maximum or minimum force magnitudes, respectively). We note, too, that since Coulomb s law is inversely proportional to r² then the (if, say, the charges were all the same) force due to C would be one-fourth as big as that due to B (since C is twice as far away from A). The charges, it turns out, are not the same, so there is also a factor of the charge ratio (the charge of C divided by the charge of B), as well as the aforementioned ¼ factor. That is, the force exerted by C is, by Coulomb s law equal to ±¼ multiplied by the force exerted by B. 3. The charge dq within a thin shell of thickness dr is dq dv Adr where A = 4r. Thus, with = b/r, we have r q z dq bz 4 rdr bcr r1 h. r1 With b = 3. C/m, r =.6 m and r 1 =.4 m, we obtain q =.38 C = C. (a) The maximum force is F and occurs when = 18º (B is to the left of A, while C is the right of A). We choose the minus sign and write F = (1 ¼) F = 4. One way to think of the minus sign choice is cos(18º) = 1. This is certainly consistent with the minimum force ratio (zero) at = º since that would also imply = 1 + ¼ = 4. (b) The ratio of maximum to minimum forces is 1.5/.75 = 5/3 in this case, which implies 5 3 = 1 + ¼ = ¼ Of course, this could also be figured as illustrated in part (a), looking at the maximum force ratio by itself and solving, or looking at the minimum force ratio (¾) at = 18º and solving for.

15 1. (a) We note that the electric field points leftward at both points. Using F q E, and orienting our x axis rightward (so î points right in the figure), we find N F ˆ C C 4 i ( N) i ˆ. We note that the symbol q is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q q. 1 which means the magnitude of the force on the proton is N and its direction ( ˆi) is leftward. (b) As the discussion in - makes clear, the field strength is proportional to the crowdedness of the field lines. It is seen that the lines are twice as crowded at A than at B, so we conclude that E A = E B. Thus, E B = N/C. The following two sketches are for the cases q 1 > q (left figure) and q 1 < q (right figure).

16 3. Since the magnitude of the electric field produced by a point charge q is given by E q /4 r, where r is the distance from the charge to the point where the field has magnitude E, the magnitude of the charge is.5m. N C 11 q 4 r E 5.61 C N m C 4. We find the charge magnitude q from E = q /4 r : 1. N C1.m 1 q4 Er C N m C

17 5. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is q E 4 R where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze, so Ze E 4 R c hb gc h c h Nm C C NC m (b) The field is normal to the surface and since the charge is positive, it points outward from the surface. 6. With x 1 = 6. cm and x = 1. cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q 1 = q =. 1 7 C, and the magnitudes and directions of the individual fields are given by: q ˆ (8.991 Nm C ).1 C 5 E1 i ˆi ( N C)i ˆ 4 ( x x1).135 m.6 m 9 7 q ˆ (8.991 Nm C )(.1 C) ˆ 5 E ˆ i i ( N C)i 4 ( x x).135 m.1 m Thus, the net electric field is E E E net 1 5 ( N C)i ˆ

18 7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q = 4. q 1 located at x = 7 cm has a greater magnitude than q 1 = C located at x 1 = cm, a point of zero field must be closer to q 1 than to q. It must be to the left of q 1. Let x be the coordinate of P, the point where the field vanishes. Then, the total electric field at P is given by 1 q q1 E. 4 ( x x) x x1 If the field is to vanish, then q q q ( x x ). ( x x ) q 1 x x1 1 x x1 Taking the square root of both sides, noting that q / q 1 = 4, we obtain x 7 cm.. x cm Choosing. for consistency, the value of x is found to be x = 3 cm. 8. (a) The individual magnitudes E 1 and E are figured from Eq. -3, where the absolute value signs for q are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on if E 1 is in the same, or opposite, direction as E. At points left of q 1 (on the x axis) the fields point in opposite directions, but there is no possibility of cancellation (zero net field) since E 1 is everywhere bigger than E in this region. In the region between the charges ( < x < L) both fields point leftward and there is no possibility of cancellation. At points to the right of q (where x > L), E 1 points leftward and E points rightward so the net field in this range is L Thus, we obtain x.7l. 1 5 E E E ˆi. net 1 Although q 1 > q there is the possibility of E net since these points are closer to q than to q 1. Thus, we look for the zero net field point in the x > L region: which leads to 1 q 1 E E x x L x L q. x q 5 1 q (b) A sketch of the field lines is shown in the figure below:

19 9. The x component of the electric field at the center of the square is given by 1 q1 q q3 q4 Ex cos 45 4 ( a/ ) ( a/ ) ( a/ ) ( a/ ) q q q q a /. Similarly, the y component of the electric field is 1 q1 q q3 q4 Ey cos 45 4 ( a/ ) ( a/ ) ( a/ ) ( a/ ) q q q q a / Nm / C (.1 C) 1 (.5 m) / N/C. 5 Thus, the electric field at the center of the square is EE ˆj (1. 1 N/C)j. ˆ y 1. We place the origin of our coordinate system at point P and orient our y axis in the direction of the q 4 = 1q charge (passing through the q 3 = +3q charge). The x axis is perpendicular to the y axis, and thus passes through the identical q 1 = q = +5q charges. The individual magnitudes E1, E, E3, and E 4 are figured from Eq. -3, where the absolute value signs for q 1, q, and q 3 are unnecessary since those charges are positive (assuming q > ). We note that the contribution from q 1 cancels that of q (that is, E1 E ), and the net field (if there is any) should be along the y axis, with magnitude equal to 1 F q q I F 1q qi Enet j j d d HG d 3 d K J HG b g KJ which is seen to be zero. A rough sketch of the field lines is shown below:

20 11. (a) The vertical components of the individual fields (due to the two charges) cancel, by symmetry. Using d = 3. m and y = 4. m, the horizontal components (both pointing to the x direction) add to give a magnitude of E q d (8.991 N m C )(3.1 C)(3. m) N/C x,net 3/ 3/ 4 ( d y ) [(3. m) (4. m) ] 1 (b) The net electric field points in the x direction, or 18 counterclockwise from the +x axis. 1. For it to be possible for the net field to vanish at some x >, the two individual fields (caused by q 1 and q ) must point in opposite directions for x >. Given their locations in the figure, we conclude they are therefore oppositely charged. Further, since the net field points more strongly leftward for the small positive x (where it is very close to q ) then we conclude that q is the negative-valued charge. Thus, q 1 is a positive-valued charge. We write each charge as a multiple of some positive number (not determined at this point). Since the problem states the absolute value of their ratio, and we have already inferred their signs, we have q 1 = 4 and q =. Using Eq. -3 for the individual fields, we find 4 E net = E 1 + E = 4 o (L + x) 4 o x for points along the positive x axis. Setting E net = at x = cm (see graph) immediately leads to L = cm. (a) If we differentiate E net with respect to x and set equal to zero (in order to find where it is maximum), we obtain (after some simplification) that location: x = L = 1.7( cm) = 34 cm. We note that the result for part (a) does not depend on the particular value of. (b) Now we are asked to set = 3e, where e = C, and evaluate E net at the value of x (converted to meters) found in part (a). The result is. 1 8 N/C.

21 13. By symmetry we see the contributions from the two charges q 1 = q = +e cancel each other, and we simply use Eq. -3 to compute magnitude of the field due to q 3 = +e. (a) The magnitude of the net electric field is 1 e 1 e 1 4e E net 4 r 4 ( a / ) 4 a (1.61 C) 6 (8.991 Nm C ) 16 N/C. (6.1 m) (b) This field points at 45., counterclockwise from the x axis. 14. The field of each charge has magnitude 19 kq e C 6 E k (8.991 N m C ) 3.61 N C. r.m (. m) The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a vector-capable calculator in polar mode) and write (starting with the proton on the left and moving around clockwise) the contributions to E net as follows: be g be 13 g be 1 g be 15 g be g. 6 This yields c h, with the N/C unit understood. (a) The result above shows that the magnitude of the net electric field is 6 E N/C. net (b) Similarly, the direction of E net is 76.4 from the x axis.

22 15. (a) The electron e c is a distance r = z =. m away. Thus, E 9 19 e (8.991 N m C )(1.61 C) 6 C N/C. 4 r (. m) (b) The horizontal components of the individual fields (due to the two e s charges) cancel, and the vertical components add to give E ez (8.991 N m C )(1.61 C)(. m) 9 19 s,net 3/ 3/ 4 ( R z ) [(. m) (. m) ] N/C. (c) Calculation similar to that shown in part (a) now leads to a stronger field N/C from the central charge. E c (d) The field due to the side charges may be obtained from calculation similar to that shown in part (b). The result is E s, net = N/C. (e) Since E c is inversely proportional to z, this is a simple result of the fact that z is now much smaller than in part (a). For the net effect due to the side charges, it is the trigonometric factor for the y component (here expressed as z/ r ) which shrinks almost linearly (as z decreases) for very small z, plus the fact that the x components cancel, which leads to the decreasing value of E s, net. 16. The net field components along the x and y axes are E net, x q q cos sin, q E net, y R R R The magnitude is the square root of the sum of the components-squared. Setting the magnitude equal to E =. 1 5 N/C, squaring and simplifying, we obtain q q qq cos E (4 R ) With R =.5 m, q 1 =. 1 6 C and q = C, we can solve this expression for cos and then take the inverse cosine to find the angle: There are two answers. (a) The positive value of angle is = q1 q1 (4 R ) E cos. qq 1 (b) The positive value of angle is = 67.8.

23 17. We make the assumption that bead is in the lower half of the circle, partly because it would be awkward for bead 1 to slide through bead if it were in the path of bead 1 (which is the upper half of the circle) and partly to eliminate a second solution to the problem (which would have opposite angle and charge for bead ). We note that the net y component of the electric field evaluated at the origin is negative (points down) for all positions of bead 1, which implies (with our assumption in the previous sentence) that bead is a negative charge. (a) When bead 1 is on the +y axis, there is no x component of the net electric field, which implies bead is on the y axis, so its angle is 9. (b) Since the downward component of the net field, when bead 1 is on the +y axis, is of largest magnitude, then bead 1 must be a positive charge (so that its field is in the same direction as that of bead, in that situation). Comparing the values of E y at and at 9 we see that the absolute values of the charges on beads 1 and must be in the ratio of 5 to 4. This checks with the 18 value from the E x graph, which further confirms our belief that bead 1 is positively charged. In fact, the 18 value from the E x graph allows us to solve for its charge (using Eq. -3):. (a) We use the usual notation for the linear charge density: = q/l. The arc length is L = r with is expressed in radians. Thus, L = (.4 m)(.698 rad) =.79 m. With q = 3( C), we obtain = C/m. (b) We consider the same charge distributed over an area A = r = (. m) and obtain = q/a = C/m². (c) Now the area is four times larger than in the previous part (A sphere = 4r ) and thus obtain an answer that is one-fourth as big: = q/a sphere = C/m². (d) Finally, we consider that same charge spread throughout a volume of V = 4 r 3 /3 and obtain the charge density = / q V = C/m 3. q 1 = 4 or²e = 4( C N m )(.6 m) ( N C ) =. 16 C. (c) Similarly, the value from the E y graph allows us to solve for the charge of bead : q = 4 or²e = 4( C N m )(.6 m) ( N C ) = C.

24 3. We use Eq. -3, assuming both charges are positive. At P, we have E Simplifying, we obtain qr q ( R) E 1 left ring right ring 3/ 3/ 4 4 R R [( R) R ] 3/ q1.56. q 5 4. Studying Sample Problem -3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by E sin 4 r along the symmetry axis, with = q/r with in radians. In this problem, each charged quarter-circle produces a field of magnitude q q E sin. r r r /4 1 1 /4 /4 4 That produced by the positive quarter-circle points at 45, and that of the negative quarter-circle points at +45. (a) The magnitude of the net field is E 1 q 1 4 q cos (8.991 Nm C )4(4.51 C).6 N/C. (5.1 m) net, x 4 r 4 r (b) By symmetry, the net field points vertically downward in the ĵ direction, or 9 counterclockwise from the +x axis.

25 5. From symmetry, we see that the net field at P is twice the field caused by the upper semicircular charge q R(and that it points downward). Adapting the steps leading to Eq. -1, we find 9 net ˆ q E j sin ˆ j. 4R 9 R (a) With R = m and q = C, E 3.8 N/C. (b) The net electric field E net the +x axis. net points in the ĵ direction, or 9counterclockwise from 6. We find the maximum by differentiating Eq. -16 and setting the result equal to zero. d dz F HG I c h K c h qz q R z 4 z R J 3 4 z R / 5 / which leads to z R/. With R =.4 cm, we have z = 1.7 cm.

26 7. (a) The linear charge density is the charge per unit length of rod. Since the charge is uniformly distributed on the rod, 15 q 4.31 C L.815 m C/m. (b) We position the x axis along the rod with the origin at the left end of the rod, as shown in the diagram. 8. We use Eq. -16, with q denoting the charge on the larger ring: qz qz 13 qq 4.19Q. 4 ( z R ) 4 [ z (3 R) ] 5 3/ 3/ Note: we set z = R in the above calculation. 3/ Let dx be an infinitesimal length of rod at x. The charge in this segment is dq dx. The charge dq may be considered to be a point charge. The electric field it produces at point P has only an x component and this component is given by de 1 x 4 dx Lax b g. The total electric field produced at P by the whole rod is the integral L L dx Ex 4 Lax 4 Lax 4 a La L q, 4 a La 4 a La upon substituting q L. With q = C, L =.815 m and a =.1 m, we 3 3 obtain N/C, or N/C. E x E x (c) The negative sign in E x indicates that the field points in the x direction, or 18 counterclockwise form the +x axis. (d) If a is much larger than L, the quantity L + a in the denominator can be approximated by a and the expression for the electric field becomes q Ex. 4 a Since a5 m L.815 m, the above approximation applies and we have 8 E x N/C, or. 8 E x N/C 15 (e) For a particle of charge q 4.31 C, the electric field at a distance a = 5 m 8 away has a magnitude N/C. E x

27 9. The smallest arc is of length L 1 = r 1 / = R/; the middle-sized arc has length L r / ( R )/ R; and, the largest arc has L3 = R)/. The charge per unit length for each arc is = q/l where each charge q is specified in the figure. Following the steps that lead to Eq. -1 in Sample Problem -3, we find E (sin 45 ) (sin 45 ) (sin 45 ) 1 3 net 4 1 r 4 r 4 3 r R which yields E net = N/C. (b) The direction is 45º, measured counterclockwise from the +x axis. Q 3. (a) It is clear from symmetry (also from Eq. -16) that the field vanishes at the center. (b) The result (E = ) for points infinitely far away can be reasoned directly from Eq (it goes as 1/z² as z ) or by recalling the starting point of its derivation (Eq. -11, which makes it clearer that the field strength decreases as 1/r² at distant points). (c) Differentiating Eq. -16 and setting equal to zero (to obtain the location where it is maximum) leads to d qz q R z R.77 3/ 5/ z R. dz 4 z R 4 z R (d) Plugging this value back into Eq. -16 with the values stated in the problem, we find E max = N/C.

28 31. First, we need a formula for the field due to the arc. We use the notation for the charge density, = Q/L. Sample Problem -3 illustrates the simplest approach to circular arc field problems. Following the steps leading to Eq. -1, we see that the general result (for arcs that subtend angle ) is E arc sin( /) sin( / ) sin( / ) 4 r 4 r. Now, the arc length is L = r if is expressed in radians. Thus, using R instead of r, we obtain ( Q/ L)sin( / ) ( Q/ R)sin( / ) Qsin( / ) Earc. 4 r 4 r 4 R The problem asks for the ratio E particle / E arc where E particle is given by Eq. -3: E Q/4 R. particle Earc Qsin( / ) / 4 R sin( / ) 3. We assume q >. Using the notation = q/l we note that the (infinitesimal) charge on an element dx of the rod contains charge dq = dx. By symmetry, we conclude that all horizontal field components (due to the dq s) cancel and we need only sum (integrate) the vertical components. Symmetry also allows us to integrate these contributions over only half the rod ( x L/) and then simply double the result. In that regard we note that sin = R/r where r x R. (a) Using Eq. -3 (with the and sin factors just discussed) the magnitude is L dq L dx y E sin 4 r 4 x R x R L / R L dx qlr x 3 x R R x R q L q 1 LR L R R L 4R With, we have E particle E arc where the integral may be evaluated by elementary means or looked up in Appendix E 1 (item #19 in the list of integrals). With q C, L.145 m and R =.6 m, we have E 1.4 N/C. (b) As noted above, the electric field E points in the +y direction, or 9counterclockwise from the +x axis.

29 33. Consider an infinitesimal section of the rod of length dx, a distance x from the left end, as shown in the following diagram. It contains charge dq = dx and is a distance r from P. The magnitude of the field it produces at P is given by 1 dx de. 4 r 34. From Eq. -6, we obtain E 1 z R C/Nm 1cm.5cm 6 z 5.31 C m 1cm N C. The x and the y components are and 1 dx dex sin 4 r 1 dx dey cos, 4 r respectively. We use as the variable of integration and substitute r = R/cos, x Rtan and dx = (R/cos ) d. The limits of integration are and / rad. Thus, and Ex Ey sind cos 4 R 4 R 4 R / d cos sin. 4 R 4 R 4 R We notice that E x = E y no matter what the value of R. Thus, E makes an angle of 45 with the rod for all values of R.

30 35. At a point on the axis of a uniformly charged disk a distance z above the center of the disk, the magnitude of the electric field is E L N M 1 z z R where R is the radius of the disk and is the surface charge density on the disk. See Eq. -6. The magnitude of the field at the center of the disk (z = ) is E c = /. We want to solve for the value of z such that E/E c = 1/. This means O Q P 36. We write Eq. -6 as E z 1 E ( z R ) max 1/ and note that this ratio is 1 (according to the graph shown in the figure) when z = 4. cm. Solving this for R we obtain R = z 3 = 6.9 cm. z 1 z 1 1. z R z R Squaring both sides, then multiplying them by z + R, we obtain z = (z /4) + (R /4). Thus, z = R /3, or z R 3. With R =.6 m, we have z =.346 m.

31 37. We use Eq. -6, noting that the disk in figure (b) is effectively equivalent to the disk in figure (a) plus a concentric smaller disk (of radius R/) with the opposite value of. That is, E (b) = E (a) R 1 o (R) + (R/) where E (a) = R 1 o (R) + R. We find the relative difference and simplify: 38. From da = r dr (which can be thought of as the differential of A = r²) and dq = da (from the definition of the surface charge density ), we have dq = Q R r dr where we have used the fact that the disk is uniformly charged to set the surface charge density equal to the total charge (Q) divided by the total area (R ). We next set r =.5 m and make the approximation dr m. Thus we get dq C. E (a) E (b) E = 1 / 4 1/ 4 1 / 17 / (a) 1/ 41 1/ or approximately 8%.

32 39. The magnitude of the force acting on the electron is F = ee, where E is the magnitude of the electric field at its location. The acceleration of the electron is given by Newton s second law: 19 4 F ee c a m m Chc. 1 N Ch ms kg 4. Eq. -8 gives using Newton s second law. F E ma q e m e a F H G I b g K J (a) With east being the i direction, we have E 1.61 C kg m s ˆi (.1 N C) ˆi 19 which means the field has a magnitude of.1 N/C (b) The result shows that the field E is directed in the x direction, or westward.

33 41. We combine Eq. -9 and Eq. -8 (in absolute values). F H G F q E q p z I KJ kep z 3 3 where we have used Eq. 1-5 for the constant k in the last step. Thus, we obtain m N m C C C m N F If the dipole is oriented such that p is in the +z direction, then F points in the z direction.. 4. (a) Vertical equilibrium of forces leads to the equality mg qemg E. e Substituting the values given in the problem, we obtain 7 mg (6.641 kg)(9.8 m/s ) E 19 e (1.61 C) N C (b) Since the force of gravity is downward, then qe must point upward. Since q > in this situation, this implies E must itself point upward..

34 43. (a) The magnitude of the force on the particle is given by F = qe, where q is the magnitude of the charge carried by the particle and E is the magnitude of the electric field at the location of the particle. Thus, 44. (a) F e = Ee = ( N/C)( C) = N. (b) F i = Eq ion = Ee = ( N/C)( C) = N. 6 F 3. 1 N 3 E NC. 9 q. 1 C The force points downward and the charge is negative, so the field points upward. (b) The magnitude of the electrostatic force on a proton is Fel ee C N C.4 1 N. (c) A proton is positively charged, so the force is in the same direction as the field, upward. (d) The magnitude of the gravitational force on the proton is Fg The force is downward. (e) The ratio of the forces is mg kg 9.8 m s N. Fel F N g N

35 45. (a) The magnitude of the force acting on the proton is F = ee, where E is the magnitude of the electric field. According to Newton s second law, the acceleration of the proton is a = F/m = ee/m, where m is the mass of the proton. Thus, c hc h C. 1 N C 1 a ms kg (b) We assume the proton starts from rest and use the kinematic equation v v ax (or else x 1 at and v = at) to show that d ib g 1 5 v ax ms. 1m ms. 46. (a) The initial direction of motion is taken to be the +x direction (this is also the direction of E ). We use vf vi a x with v f = and a F m ee m e to solve for distance x: c hc h c hc h vi mv e i x kg 5. 1 m s a ee C 1. 1 N C (b) Eq. -17 leads to x x c mh t v 5. 1 ms avg (c) Using v = ax with the new value of x, we find v i 1 K mv e v ax eex K m v v v m v 1 i e i i i e i kg5.1 m s 1.61 C 1.1 N C 8.1 m s..11. m. Thus, the fraction of the initial kinetic energy lost in the region is.11 or 11.%.

36 47. When the drop is in equilibrium, the force of gravity is balanced by the force of the electric field: mg = qe, where m is the mass of the drop, q is the charge on the drop, and E is the magnitude of the electric field. The mass of the drop is given by m = (4/3)r 3, where r is its radius and is its mass density. Thus, mg 4r g m 851kg m 9.8m s q E 3E NC and q/e = ( C)/( C) = 5, or q 5e C 48. We assume there are no forces or force-components along the x direction. We combine Eq. -8 with Newton s second law, then use Eq. 4-1 to determine time t followed by Eq. 4-3 to determine the final velocity (with g replaced by the a y of this problem); for these purposes, the velocity components given in the problem statement are re-labeled as v x and v y respectively. (a) We have aqe/ m( e/ m) E which leads to C N ˆ 13 a 1 j (.1 1 m s ) ˆ j kg C (b) Since v x = v x in this problem (that is, a x = ), we obtain x. m 7 t s 5 v 1.51 m s x d ic h v v a t 3. 1 m s 1. 1 m s 1.31 s y y y which leads to v y = m/s. Therefore, the final velocity is v ˆ 5 6 (1.5 1 m/s) i (.8 1 m/s) j. ˆ

37 49. (a) We use x = v avgt = vt/: (b) We use x 1 c x. 1 m 6 v 7. 1 ms. 8 t s at and E = F/e = ma/e: c hc h c hc h E ma 31 xm. 1 m kg NC. e et C 1.51 s h 5. Due to the fact that the electron is negatively charged, then (as a consequence of Eq. -8 and Newton s second law) the field E pointing in the +y direction (which we will call upward ) leads to a downward acceleration. This is exactly like a projectile motion problem as treated in Chapter 4 (but with g replaced with a = ee/m = m/s ). Thus, Eq. 4-1 gives x 3. m 6 t s. 6 v cos (.1 m/s)cos4. This leads (using Eq. 4-3) to vy v at m/s sin (. 1 m/s)sin4. ( m/s )( s) 5 Since the x component of velocity does not change, then the final velocity is v = ( m/s) i^ ( m/s) j^.

38 51. We take the positive direction to be to the right in the figure. The acceleration of the proton is a p = ee/m p and the acceleration of the electron is a e = ee/m e, where E is the magnitude of the electric field, m p is the mass of the proton, and m e is the mass of the electron. We take the origin to be at the initial position of the proton. Then, the coordinate of the proton at time t is x 1 apt and the coordinate of the electron is x L 1 aet. They pass each other when their coordinates are the same, or 1 1 at p L at e. This means t = L/(a p a e) and a ee m m x L L L p p e ap ae ee mp ee m mem e p kg m kg kg 5.71 m. 5. We are given = C/m and various values of z (in the notation of Eq. - 6 which specifies the field E of the charged disk). Using this with F = ee (the magnitude of Eq. -8 applied to the electron) and F = ma, we obtain a F/ m ee/ m. (a) The magnitude of the acceleration at a distance R is (b) At a distance R/1, a = (c) At a distance R/1, a = e ( ) a = = m/s. 4 m o e (11 11 ) m o = m/s. e (11 11 ) m o = m/s. (d) The field due to the disk becomes more uniform as the electron nears the center point. One way to view this is to consider the forces exerted on the electron by the charges near the edge of the disk; the net force on the electron caused by those charges will decrease due to the fact that their contributions come closer to canceling out as the electron approaches the middle of the disk.

39 53. (a) Using Eq. -8, we find c hc h c hb g b g b g F 8. 1 C 3. 1 N C i 8. 1 C 6 N C j. 4Ni. 48Nj. Therefore, the force has magnitude equal to x y F F F.4N.48N.45N. (b) The angle the force F makes with the +x axis is F 1 y 1.48N tan tan 11.3 Fx.4N 54. (a) Due to the fact that the electron is negatively charged, then (as a consequence of Eq. -8 and Newton s second law) the field E pointing in the same direction as the velocity leads to deceleration. Thus, with t = s, we find ee (1.61 C)(5 N/C) vv a t v t 4.1 m/s (1.51 s) m kg 4.71 m/s. (b) The displacement is equal to the distance since the electron does not change its direction of motion. The field is uniform, which implies the acceleration is constant. Thus, v v 5 d t5.1 m. measured counterclockwise from the +x axis. (c) With m =.1 kg, the (x, y) coordinates at t = 3. s can be found by combining Newton s second law with the kinematics equations of Chapters 4. The x coordinate is (d) Similarly, the y coordinate is Ft.4 N 3. s x x 1 x a t 18m. m.1 kg Ft y.48 N 3. s y 1 y a t 1.6m. m.1 kg

40 55. We take the charge Q 45. pc of the bee to be concentrated as a particle at the center of the sphere. The magnitude of the induced charges on the sides of the grain is q 1. pc. (a) The electrostatic force on the grain by the bee is 1. The vector area A and the electric field E are shown on the diagram below. The angle between them is = 145, so the electric flux through the area is EAEA 3 cos 18 N C 3. 1 m cos N m C. kqq kq( q) 1 1 F kq q ( d D/ ) ( D/ ) ( D/ ) ( d D/ ) where D 1. cm is the diameter of the sphere representing the honeybee, and d 4.m is the diameter of the grain. Substituting the values, we obtain 1 1 F N m C (45. 1 C)(1. 1 C) ( m) ( m) N. The negative sign implies that the force between the bee and the grain is attractive. The 1 magnitude of the force is F.56 1 N. (b) Let Q 45. pc be the magnitude of the charge on the tip of the stigma. The force on the grain due to the stigma is k Q q k Q ( q) 1 1 F k Q q ( d D) ( D) ( D) ( d D) where D 1. mm is the distance between the grain and the tip of the stigma. Substituting the values given, we have 1 1 F N m C (45. 1 C)(1. 1 C) ( m) ( m) N. The negative sign implies that the force between the grain and the stigma is attractive. 8 The magnitude of the force is F N. (c) Since F F, the grain will move to the stigma.

41 . We use zeda and note that the side length of the cube is (3. m 1. m) =. m. (a) On the top face of the cube y =. m and da da ĵ. Therefore, we have E 4i ˆ3. ˆj 4i ˆ18j ˆ. Thus the flux is EdA 4i 18j da j 18 da 18. Nm C 7 Nm C. ˆ ˆ ˆ top top top b ge j (b) On the bottom face of the cube y = and da da j. Therefore, we have c h. Thus, the flux is E 4 i3 j4i6j EdA 4i6j da j 6 da6. Nm C4 Nm C. ˆ ˆ ˆ bottom bottom bottom b g. 3. We use EA, where A A j 14. m j 6. N C ˆi 1.4 m ˆj. (a). N C ˆj 1.4 m ˆj 3.9 Nm C. (b) 3. N C ˆi 4 N C kˆ 1.4 m ˆj. (c) (d) The total flux of a uniform field through a closed surface is always zero. (c) On the left face of the cube da daî ˆ E da 4i E j da i 4 da 4. N m C 16 N m C. (e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 N m /C. Thus the net flux through the cube is. So ˆ ˆ ˆ y left left bottom (d) On the back face of the cube da daˆk EdA. Thus, =.. But since E has no z component = ( ) N m /C = 48 N m /C.

42 4. There is no flux through the sides, so we have two inward contributions to the flux, one from the top (of magnitude (34)(3.) ) and one from the bottom (of magnitude ()(3.) ). With inward flux being negative, the result is = 486 Nm /C. Gauss law then leads to q 1 9 enc (8.851 C /N m )( 486 N m C) 4.31 C. 5. We use Gauss law: q, where is the total flux through the cube surface and q is the net charge inside the cube. Thus, C 1 q C N m 5. 1 N m C.

43 6. The flux through the flat surface encircled by the rim is given by ae. Thus, the flux through the netting is 3 4 ae (.11 m) (3.1 N/C) 1.11 N m /C. 7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the 19 shape of a cube, of edge length d, with a proton of charge q C situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of flux is net = q/, and we conclude that the flux through the square is one-sixth of that. Thus, C 1 q 6 6(8.851 C N m ) N m C.

44 8. We note that only the smaller shell contributes a (non-zero) field at the designated point, since the point is inside the radius of the large sphere (and E = inside of a spherical charge), and the field points towards the x direction. Thus, with R =. m (the radius of the smaller shell), L =.1 m and x =. m, we obtain ˆ q ˆ 4R ( j) j ˆ R EE j ˆj 4 4 ( ) ( ) r Lx Lx 6 (. m) (4.1 C/m ) 4 ˆj 1.81 N/Cˆj. (8.851 C /Nm )(.1 m. m) 9. Let A be the area of one face of the cube, E u be the magnitude of the electric field at the upper face, and E l be the magnitude of the field at the lower face. Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive. The flux through the other faces is zero, so the total flux through the cube surface is AE ( E u ). The net charge inside the cube is given by Gauss law: q A E E u C 3.54 C. 1 ( ) ( C / N m )(1 m) (1 N/C 6. N/C) 6