Solution manual fundamentals of physics extended, 8th editionch22

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1 We note that the symbol q2 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 q1 = q2 The following two sketches are for the cases q1 > q2 (left figure) and q1 < q2 (right figure) G G (a) We note that the electric field points leftward at both points Using F = q0 E , and orienting our x axis rightward (so ˆi points right in the figure), we find h FGH IJ K G N × 10−19 C −40 i = −6.4 × 10−18 N i F = +16 C c which means the magnitude of the force on the proton is 6.4 × 10–18 N and its direction (−ˆi) is leftward (b) As the discussion in §22-2 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 EA = 2EB Thus, EB = 20 N/C The following diagram is an edge view of the disk and shows the field lines above it Near the disk, the lines are perpendicular to the surface and since the disk is uniformly charged, the lines are uniformly distributed over the surface Far away from the disk, the lines are like those of a single point charge (the charge on the disk) Extended back to the disk (along the dotted lines of the diagram) they intersect at the center of the disk If the disk is positively charged, the lines are directed outward from the disk If the disk is negatively charged, they are directed inward toward the disk A similar set of lines is associated with the region below the disk We find the charge magnitude |q| from E = |q|/4πε0r2: q = 4πε Er (1.00 N C )(1.00 m ) = 8.99 × 10 N ⋅ m C 2 = 1.11× 10−10 C Since the magnitude of the electric field produced by a point charge q is given by E =| q | / 4πε r , where r is the distance from the charge to the point where the field has magnitude E, the magnitude of the charge is ( 0.50 m ) ( 2.0 N C ) = 5.6 ×10−11 C q = 4πε r E = 8.99 ×109 N ⋅ m C2 With x1 = 6.00 cm and x2 = 21.00 cm, the point midway between the two charges is located at x = 13.5 cm The values of the charge are q1 = –q2 = – 2.00 × 10–7 C, and the magnitudes and directions of the individual fields are given by: G E1 = − | q1 | ˆi = −(3.196 ×105 N C)iˆ 4πε ( x − x1 ) G E2 = − q2 ˆi = −(3.196 ×105 N C)iˆ 4πε ( x − x1 ) Thus, the net electric field is G G G Enet = E1 + E2 = −(6.39 ×105 N C)iˆ 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 E= q πε 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 c hb gc h h × 10−19 C 8.99 × 109 N ⋅ m2 C 94 160 Ze E= = = 3.07 × 1021 N C −15 πε R 6.64 × 10 m c (b) The field is normal to the surface and since the charge is positive, it points outward from the surface G G (a) The individual magnitudes E1 and E2 are figured from Eq 22-3, where the absolute value signs for q2 are unnecessary since G this charge is positive Whether we add the magnitudes or subtract them depends on if E1 is in the same, or opposite, direction as G E At points left of q1 (on the –x axis) the fields point in opposite directions, but there is G G no possibility of cancellation (zero net field) since E1 is everywhere bigger than E2 in this region In the region between the charges (0 < x < L) both fields point leftward and G there is no possibility of cancellation At points to the right of q2 (where x > L), E1 points G leftward and E2 points rightward so the net field in this range is G G G Enet = | E2 | − | E1 | ˆi ( ) G Although |q1| > q2 there is the possibility of E net = since these points are closer to q2 than to q1 Thus, we look for the zero net field point in the x > L region: G G | E1 |=| E2 | Ÿ q2 | q1 | = 4πε x 4πε ( x − L )2 which leads to q2 x−L = = x | q1 | Thus, we obtain x = L ≈ 2.72 L 1− (b) A sketch of the field lines is shown in the figure below: At points between the charges, the individual electric fields are in the same direction and not cancel Since charge q2= − 4.00 q1 located at x2 = 70 cm has a greater magnitude than q1 = 2.1 ×10−8 C located at x1 = 20 cm, a point of zero field must be closer to q1 than to q2 It must be to the left of q1 Let x be the coordinate of P, the point where the field vanishes Then, the total electric field at P is given by E= | q1 | ã Đ | q2 | ă ăâ ( x x2 ) ( x x1 )2 áạ If the field is to vanish, then | q2 | | q1 | = ( x − x2 ) ( x − x1 )2 | q2 | ( x − x2 ) Ÿ = | q1 | ( x − x1 )2 Taking the square root of both sides, noting that |q2|/|q1| = 4, we obtain x − 70 = ±2.0 x − 20 Choosing –2.0 for consistency, the value of x is found to be x = −30 cm 10 We place the origin of our coordinate system at point P and orient our y axis in the direction of the q4 = –12q charge (passing through the q3 = +3q charge) The x axis is perpendicular to the y axis, and through G thus G passes G G the identical q1 = q2 = +5q charges The individual magnitudes | E1 |, | E2 |, | E3 |, and | E4 | are figured from Eq 22-3, where the absolute value signs for q1, q2, and q3 are unnecessary since those charges are positive (assuming G G q > 0) We note that the contribution from q1 cancels that of q2 (that is, | E1 | = | E2 | ), and the net field (if there is any) should be along the y axis, with magnitude equal to G E net = πε F q GH b2d g − I JK FG H IJ K 12q 3q  q3  j= − j πε 4d d d which is seen to be zero A rough sketch of the field lines is shown below: 78 Studying Sample Problem 22-4, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by G E= λ sin θ πε r θ −θ along the symmetry axis where λ = q A = q rθ with θ in radians Here A is the length of the arc, given as A = 4.0 m Therefore, θ = A r = 4.0 2.0 = 2.0 rad Thus, with q = 20 × 10–9 C, we obtain G q E = sin θ A πε r 1.0 rad −1.0 rad = 38 N C 79 (a) We combine Eq 22-28 (in absolute value) with Newton’s second law: a= (b) With v = FG H | q| E 160 × 10−19 C = 9.11 × 10−31 kg m IJ FG140 NI × 10 J = 2.46 × 10 CK KH c = 3.00 × 107 m s , we use Eq 2-11 to find 10 v − vo 3.00 × 107 × 10−10 s t= = = 122 17 a 2.46 × 10 (c) Eq 2-16 gives c c h 3.00 × 10 v − vo2 = 183 × 10−3 m ∆x = = 17 2a 2.46 × 10 h 17 ms 80 Let q1 denote the charge at y = d and q2 denote the charge at y = –d The individual G G magnitudes E1 and E2 are figured from Eq 22-3, where the absolute value signs for q are unnecessary since these charges are both positive The distance from q1 to a point on the x axis is the same as the distance from q2 to a point on the x axis: r = x + d By symmetry, the y component of the net field along the x axis is zero The x component of the net field, evaluated at points on the positive x axis, is Ex = FG IJ FG H πε K H x q +d2 IJ FG KH x x2 + d IJ K where the last factor is cosθ = x/r with θ being the angle for each individual field as measured from the x axis (a) If we simplify the above expression, and plug in x = αd, we obtain Ex = q πε d F α GG H cα + 1h 32 I JJ K (b) The graph of E = Ex versus α is shown below For the purposes of graphing, we set d = m and q = 5.56 × 10–11 C (c) From the graph, we estimate Emax occurs at about α = 0.71 More accurate computation shows that the maximum occurs at α = (d) The graph suggests that “half-height” points occur at α ≈ 0.2 and α ≈ 2.0 Further numerical exploration leads to the values: α = 0.2047 and α = 1.9864 81 (a) From Eq 22-38 (and the facts that i ⋅ i = and j ⋅ i = ), the potential energy is G G U = − p ⋅ E = − ª 3.00iˆ + 4.00jˆ (1.24 ×10−30 C ⋅ m ) ê( 4000 N C ) i ẳ ẳ 26 = 1.49 ì10 J ( ) (b) From Eq 22-34 (and the facts that i × i = and j × i = − k ), the torque is G G G τ = p × E = 3.00i + 4.00j 124 × 10−30 C ⋅ m × 4000 N C i jc e h b g  = −198 × 10−26 N ⋅ m k c h (c) The work done is G G G G G W = ∆U = ∆ − p ⋅ E = pi − p f ⋅ E d i d i = e3.00i + 4.00jj − e −4.00i + 3.00jj c124 × 10 = 3.47 × 10−26 J −30 C ⋅ m ⋅ 4000 N C i h b g 82 We consider pairs of diametrically opposed charges The net field due to just the charges in the one o’clock (–q) and seven o’clock (–7q) positions is clearly equivalent to that of a single –6q charge sitting at the seven o’clock position Similarly, the net field due to just the charges in the six o’clock (–6q) and twelve o’clock (–12q) positions is the same as that due to a single –6q charge sitting at the twelve o’clock position Continuing with this line of reasoning, we see that there are six equal-magnitude electric field vectors pointing at the seven o’clock, eight o’clock … twelve o’clock positions Thus, the resultant field of all of these points, by symmetry, is G directed toward the position midway between seven and twelve o’clock Therefore, E resultant points towards the nine-thirty position 83 (a) For point A, we have (in SI units) G ª q1 q2 º ˆ EA = « + −i 2» ¬ 4πε r1 4πε r2 ẳ ( ) (8.99 ì10 ) (1.00 ì10 = ( 5.00 ×10 ) −2 −12 C) ( −ˆi + ( ) 8.99 × 109 ) | −2.00 ×10−12 C| ( × 5.00 ×10 ) −2 ( ˆi ) = (−1.80 N C)iˆ (b) Similar considerations leads to −12 −12 G ª q1 | q2 | º ˆ ( 8.99 × 10 ) (1.00 ×10 C ) ˆ ( 8.99 ×10 ) | −2.00 × 10 C| ˆ i= i+ i EB = « + 2» −2 −2 0.500 5.00 10 0.500 5.00 10 × × × × ¬ 4πε r1 4πε r2 ¼ ( ) ( ) = (43.2 N C)iˆ (c) For point C, we have −12 −12 G ª q1 | q2 | º ˆ ( 8.99 × 10 ) (1.00 ×10 C ) ˆ ( 8.99 ×10 ) | 2.00 ì10 C| i= i i EC = ô − 2 2 » ¬ 4πε r1 r2 ẳ ( 2.00 ì 5.00 ì102 ) ( 5.00 ×10−2 ) = −(6.29 N C)iˆ (d) Although a sketch is not shown here, it would be somewhat similar to Fig 22-5 in the textbook except that there would be twice as many field lines “coming into” the negative charge (which would destroy the simple up/down symmetry seen in Fig 22-5) 84 The electric field at a point on the axis of a uniformly charged ring, a distance z from the ring center, is given by E= qz c πε z + R 2 h 3/ where q is the charge on the ring and R is the radius of the ring (see Eq 22-16) For q positive, the field points upward at points above the ring and downward at points below the ring We take the positive direction to be upward Then, the force acting on an electron on the axis is F=− eqz c πε z + R h 3/ For small amplitude oscillations z > d we have (z ± d/2)–2 ≈ z–2, so G E net ≈ FG H IJ K q q 2q + = πε z z πε z 91 (a) SupposeG the pendulum is at the angle θ with the vertical The force diagram is shown below T is the tension in the thread, mg is the magnitude of the force of gravity, and qE is the magnitude of the electric force The field points upward and the charge is positive, so the force is upward Taking the angle shown to be positive, then the torque on the sphere about the point where the thread is attached to the upper plate is τ = − ( mg − qE ) L sin θ If mg > qE then the torque is a restoring torque; it tends to pull the pendulum back to its equilibrium position If the amplitude of the oscillation is small, sinθ can be replaced by θ in radians and the torque is τ = − ( mg − qE ) Lθ The torque is proportional to the angular displacement and the pendulum moves in simple harmonic motion Its angular frequency is ω = ( mg − qE ) L I , where I is the rotational inertia of the pendulum Since I = mL2 for a simple pendulum, ω= ( mg − qE ) L mL = g − qE m L and the period is T= 2π ω = 2π L g − qE m If qE > mg the torque is not a restoring torque and the pendulum does not oscillate (b) The force of the electric field is now downward and the torque on the pendulum is τ = −(mg + qE ) Lθ if the angular displacement is small The period of oscillation is T = 2π L g + qE m ... negative-valued charge Thus, q1 is a positive-valued charge We write each charge as a multiple of some positive number ξ (not determined at this point) Since the problem states the absolute value of. .. 2πεo z3 Therefore, in the terminology of the problem, Enext = q d3/ 4πε0z5 21 Think of the quadrupole as composed of two dipoles, each with dipole moment of magnitude p = qd The moments point... charged quarter-circle produces a field of magnitude G |q| 1 2 |q| π/4 | E |= [sin θ ]− π / = r π / 4πε r 4πε π r That produced by the positive quarter-circle points at – 45°, and that of the negative
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