HW1solutions - Physics 213 HW #1 – Solutions q3=+5 q1=?...

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Unformatted text preview: Physics 213 HW #1 – Solutions q3=+5 q1=? Spring 2009 21.13. We solve this problem by applying Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions. The force F2 that q2 exerts on qq q3 has magnitude F2 k 2 2 3 and is in the +x direction. F1 must be in the r2 q2=-3 0 2 4 x direction, so q1 must be positive. F1 F2 gives k 2 2 q1 q3 qq k 2 2 3 . r12 r2 r 2.00 cm q1 q2 1 (3.00 nC) 0.750 nC . r2 4.00 cm The result for the magnitude of q1 doesn’t depend on the magnitude of q3 . 21.33. [Electron in an Electric Field] Use F qE to relate the force on the particle to its charge and the electric field between the plates. The force is constant and produces constant acceleration. The motion is similar to projectile motion, so use constant acceleration equations for the horizontal and vertical components of the motion. For an electron q e. F qE and q < 0 F and E are in opposite directions, so F is upward. Free-body diagram for the electron: F y ma y eE ma We need to find the acceleration of the electron: Just misses upper plate x x0 2.00 cm when y y0 0.500 cm. x-component: v0 x v0 1.60 106 m/s, ax 0, x x0 0.0200 m, t ? x x0 v0 xt 1 axt 2 2 t x x0 0.0200 m 1.25 108 s v0 x 1.60 106 m/s In this same time t the electron travels 0.0050 m vertically. y-component: t 1.25 108 s, v0 y 0, y y0 0.0050 m, ay ? y y0 v0 yt 1 a yt 2 2 ay Now find E 2( y y0 ) 2(0.0050 m) 6.40 1013 m/s 2 2 t (1.25 108 s)2 ma (9.109 1031 kg)(6.40 1013 m/s2 ) 364 N/C e 1.602 1019 C Note: The acceleration produced by the electric field is >> g, so it is OK to neglect the gravitational force on the electron. (d) (b) a eE (1.602 1019 C)(364 N/C) 3.49 1010 m/s2 in the same direction as the electric field (downward). Still << g. (d) mp 1.673 1027 kg This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the proton won’t hit the plates. The proton has the same initial speed, so it takes the same time t 1.25 108 s to travel horizontally the length of the plates. Displacement: y y0 v0 yt 1 ayt 2 1 (3.49 1010 m/s2 )(1.25 108 s)2 2.73 106 m. = 2.73 106 m, downward. 2 2 (c) The displacements are in opposite directions because the electron has negative charge and the proton has positive charge. Th e electron and proton have the same magnitude of charge, so the force due the electric field has the same magnitude for both charges. But the proton has a mass larger by a factor of 1836 so its acceleration and its vertical displacement are smaller by this fa ctor. qq for each pair of charges and find the vector sum of the r2 forces that q1 and q2 exert on q3 . 21.72. Apply F k The three charges and the forces on q3 are shown to the right. (a) F1 k q1q3 (5.00 109 C)(6.00 109 C) (8.99 109 N m2 /C2 ) 1.079 104 N r12 (0.0500 m)2 q2q3 (2.00 109 C)(6.00 109 C) (8.99 109 N m2 /C2 ) 1.20 104 N . 2 r2 (0.0300 m)2 36.9 . F1x F1 cos 8.63 105 N . F1 y F1 sin 6.48 105 N . F2 k F2 x 0 , F2 y F2 1.20 104 N . Fx F1x F2 x 8.63 105 N . Fy F1 y F2 y 6.48 105 N (1.20 104 N) 5.52 105 N . (b) F Fx2 Fy2 1.02 104 N . tan Fy Fx 0.640 . 32.6 , below the x axis. 21.106. (a) [Triangle] (b) The first diagram is the only one in which the electric field must point in the negative y-direction, so q1 3.00 C, and q2 0 . (c) Calculate the electric field at P due to each charge and add vectors to get the net field. Just from the geometry of the triangle, cos1 5 12 12 5 , sin 1 , cos 2 and sin 2 . 13 13 13 13 k q1 k q2 5 12 Ex 0 q2 7.2 C , so q2 7.2 C . (0.050 m)2 13 (0.120 m) 2 13 Ey k q1 12 kq2 5 1.17 107 N/C . (negative because it’s downward) (0.050 m)2 13 (0.120 m)2 13 21.23. Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other three and then add these forces as vectors. (a) The charges are placed as shown to the right. q1 q2 q3 q4 Q Consider forces on q4 . The free-body diagram is given in the figure below. Take the y-axis to be parallel to the diagonal between q2 and q4 and let y be in the direction away from q2 . Then F2 is in the y -direction. -2- | F3 || F1 | | F2 | 1 1 Q2 4 0 L2 Q2 4 0 2 L2 F1x F1 sin 45 F1 / 2 F1 y F1 cos45 F1 / 2 F3 x F3 sin 45 F3 / 2 F3 y F3 cos45 F3 / 2 F2 x 0, F2 y F2 (b) Fxtotal F1x F2 x F3x 0 Fytotal F1 y F2 y F3 y (2/ 2) F total kQ2 kQ 2 kQ 2 2 2 (1 2 2) L2 2L 2L kQ 2 (1 2 2). Same for all four charges. 2 L2 (c) The charge q will be at a distance L / 2 from each of the corner charges. 2 It will exert a force along the diagonal, so the total force along the diagonal will be F Qk qQk (1 2 2) 2 . This will be zero when 2L2 ( L / 2) q 1 (1 2 2)Q 0.957Q . 4 Note that q is independent of L. The answer is reasonable because q is proportional in magnitude and opposite in sign to the charges whose repulsive forces it is counteracting. #1. [Movable Charge] (a) Consider the two limits y<<D and y>>D. If y<<D, the –Q charge will dominate, so Ey<0, that is, the E vector points downward. On the other hand, if y>>D, then the two far-away charges on the x-axis will look like a single +Q charge, so Ey>0, that is, the E vector points upward. Therefore, there must be an intermediary value of y where E y=0, that is, (b) From the above figure, E is horizontal. r D2 y 2 and sin y / r , 2Qqk sin kQq 2y 1 2 kQq 2 2 . so E y 2 2 3/2 r y y (D y ) One can easily check that this formula has the correct limits given in r y +2Q D -Q 1.3D . Since D part (a). Setting Ey=0, we find y D(2 1) is the only length scale in the problem, y had to be D times some numerical constant. 2/3 1/2 -3- #Q21.10 [Metal Objects] Here’s one way to do it. (i) Put the objects together. (ii) Bring a charged object near them. This polarizes the touching objects, inducing negative charge on the left, and positive charge on the right. (iii) Now separate the objects. (iv) And remove the external charge. By charge conservation, the charges on the two metal objects must be equal and opposite. There are other ways to get equal and opposite charge on two metal objects. Can you come up with another? 21.43. [Positive Pair] We find the total electric field by calculating the field due to each charge and taking the vector sum of these two fields. Only the x component of each field is nonzero on the x-axis. The electric field points away from positive charges. (a) Halfway between the two charges, E 0. q q ax (b) For | x | a , Ex k . 4kq 2 2 (a x) (a x) 2 ( x a 2 )2 q q x2 a2 For x a , Ex k . 2kq 2 2 (a x) 2 ( x a 2 )2 (a x) q q x2 a2 2kq 2 For x a , Ex k . 2 2 (a x) ( x a 2 )2 (a x) The magnitude of the field approaches infinity at the location of either one of the point charges. (c) By symmetry, the net electric field is in the +y-direction for y > 0 and in the -y-direction for y < 0. ER r L a y a +y Ey EL r R +x q ER E L E Ey 2E cos kq r2 where with r y2 a2 so cos y / r , Ey 2kqy ( y a 2 )3/2 2 q Is this answer physically reasonable? First, let’s check the units . The units of the right hand side are ( Nm2C 2 )(C )(m) N (m2 )3/2 C , -4- the appropriate units for an electric field. Now let’s check that the formula has reasonable limits. As should. When y=0, y , EY 0 , as it EY 0 , which must be the case because of symmetry. (d) Graph of EY vs. y: For y>0, EY > 0 electric field is upward, or positive. For y<0, EY < 0 electric field is downward, or negative. -a 0 a x x2 a 2 x 2 , so, using the second equation in part (b), we have 2kq 2kq Ex 2 . Similarly, when y=0, x<<a, the third equation in part (b) gives Ex 2 . x x Similarly, we can find the electric field at x=0, | y | a by taking the limit of the equation in part (c) to get 2kq E y 2 . In all cases, the lowest order term is the same as the electric field due to a point with charge 2q at the origin. y (e) Let’s consider the case y=0, x>>a. In this limit, This makes sense because at very large distances, the charges are so close together (relative to your distance away from the m) that this system should resemble that of a single charge with magnitude 2q. The electric field decreases with increasing dis tance, as expected. However, the electric field from a positive pair does not drop off as quickly with distance as it would fr om a dipole, since in a dipole the opposite charges tend to cancel each other out, whereas in a pair of two positive charges (or two negative ch arges), they add. #2 . [Electric Dipole-Dipole Force] First, let’s assign labels to the charges (1, 2, 3, and 4 ) and the dipoles (L and R): L 1 2 3 R 4 Now let FL be the force on dipole L, and FR be the force on dipole R. FL will be the sum of the forces on 1 and 2. We’ll call the force on charge a by charge b Fab. By Newton’s 3rd law, Fba = -Fab. (a) FL i 1 F1i i 2 F2i F13 F14 F23 F24 . Now, F13 is repulsive, and F14 is attractive, but F13 is stronger (because 1 is closer to 3), so charge 1 is repelled from dipole R. F23 is attractive, and F24 is repulsive, but F23 is stronger, so charge 2 is attracted to dipole R. Because 2 is closer to R than 1 is, this attractive force is stronger, so dipole L is attracted to dipole R. Similarly, dipole R is attracted to dipole L. (b) Using the above expression for FL and applying Coulomb’s law: FL kQ 2 kQ 2 kQ 2 kQ 2 2 2 2 kQ 2 (2 x 2 ( x D)2 ( x D) 2 ) . 2 d13 d14 d 23 d 24 By a similar calculation (or Newton’s 3rd law), FR FL . (c) Combining over a common denominator and expanding the numerator, FL kQ 2 (6 D 2 x 2 2 D 4 ) 6 D 2 x 2 kQ 2 6D 2 kQ 2 x 2 ( x D) 2 ( x D) 2 x6 x4 . Therefore, n=4. (d) If one dipole is flipped, all of the forces reverse direction, so the dipoles repel. The magnitude of the force remains unchanged. If both are flipped, nothing changes. (e) The electric dipole-dipole force decreases with distance x much more rapidly (1 /x ) than the Coulomb charge-charge force 2 3 (1/x ) or even the charge-dipole force discussed in lecture (1/x ). Furthermore, the direction of the dipole-dipole force depends on the angular orientations of the dipoles. Because the orientations of the molecules are usually random, the attractive and repulsive forces from individual molecules tend to cancel. Random orientations of electric dipoles in real objects combined with the rather short ranged nature of the dipole-dipole force make the net dipole-dipole force between macroscopic objects too small to be detected. -54 ...
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