Solutions to practice problems for PHYS117B.02 Exam I

Transcription

1 Solutions to practice problems for PHYS117B.02 Exam I 1. Shown in the figure below are two point charges located along the x axis. Determine all the following: y +4 µ C 2 µ C x 6m 1m (a) Find the magnitude and direction of the electric field at the origin. This problem reduces to one dimension since the charges themselves and the point at which we make our calculation are all along a single line. Electric field points away from positive charges and toward negative charges. So both of our charges produce electric field at the origin pointing to the right, or in the +î direction. ( E = k Q 1 + k Q ) 2 î (1) E = r 2 1 ( [ Q1 k [ 4 10 E = ( r 2 1 r Q 2 r ]) î (2) ]) î (3) E = ( [ ]) î (4) E = ( [ ]) î (5) E = ( ) V mî (6) Please note that since we took care of the directions of the electric field by hand, that we used the absolute value of the charges so that we would not accidentally account for the signs of the charges twice. (b) Find the magnitude of the potential at the origin. 1

2 This is easy since potential is a scalar (no vectors!) V = k Q 1 + k Q 2 (7) r 1 r [ 2 ] 4 10 V = (8) 6 1 V = [ ] (9) V = 11970V (10) (c) Find the place on the x axis at which the field is zero. OK, this is a tricky one. First we look between the two charges. Well, in between they both point their field to the right. This means that these two fields will never cancel. So we know that the zero point is to the left of the left charge or to the right of the right charge since only at these places are the directions of the fields due to the two charges opposite. Now, we know that to be equal in strength, we need to be close to the weak charge and far from the strong charge. This only happens when we are to the right of the right charge. OK, in this region, we can calculate the results: E = 0 = k (x + 6) 2 k (x 1) 2 (11) [ ] 0 = k (x + 6) 2 2 (x 1) 2 (12) [ 0 = 4 (x + 6) 2 2 ] (13) (x 1) 2 4 (x + 6) 2 = 2 (x 1) 2 (14) 4 (x 1) 2 = 2 (x + 6) 2 (15) 4 (x 2 2x + 1) = 2 (x x + 36) (16) 4x 2 8x + 4 = 2x x + 72 (17) 2x 2 32x 68 = 0 (18) x 2 16x 34 = 0 (19) x = 17.9m (20) (d) Find the places on the x axis at which the potential is zero. Now, with potential, the quantity is a scalar. This means that the positive charge always contributes a positive amount, and the negative charge always contributes a negative amount. All you need to satisfy is that you get the same size contribution from each and the potential will be zero. This occurs when you are closer to the smaller charge and happens both in between the charges and to the right of the small charge. 2

3 In Between: Q 1 0 = k (x + 6) + k Q 2 (1 x) (21) 4 0 = k (x + 6) + k 2 (1 x) (22) 4 (x + 6) = 2 (1 x) (23) 4 (1 x) = 2 (x + 6) (24) 4 4x = 2x + 12 (25) 0 = 6x + 8 (26) x = 1.333m (27) To the Right: Q 1 0 = k (x + 6) + k Q 2 (x 1) (28) 4 0 = k (x + 6) + k 2 (x 1) (29) 4 (x + 6) = 2 (x 1) (30) 4 (x 1) = 2 (x + 6) (31) 4x 4 = 2x + 12 (32) 2x 16 = 0 (33) x = 8.0m (34) 2. Shown in the figure below are two point charges located along the x axis. Determine all the following at the point x = 0 and y = 8m: y +4 µ C +4 µ C x 6m 6m 3

4 (a) The electric field. Again, this is the hard(er) one since the electric field is a vector. However, there is some simplification to the problem since we understand that the answer will lie along the y axis by symmetry. So, all we have to calculate is the y component of each field and then sum them up. First, the magnitude of the field due to one cherge: E = k Q r 2 (35) r = = 10 (36) E = (37) E = 360V/m (38) To get the y component, we need to multiply this by the cos θ. Well, since we have all sides of the triangle we can get this easily: cos θ = adj hyp = 8 = 0.8 (39) 10 E y (onecharge) = E cos θ (40) E y (onecharge) = = 288V/m (41) E y (twocharges) = 2 E y (onecharge) = 576V/m (42) (b) The electric potential. As always, no vectors, no problem. We just add up the scalar potentials: V = k Q r + k Q r V = 2k Q r (43) (44) V = (45) V = 7200V (46) 4

5 3. Point Charges (a) Find the magnitude and direction of the electric field, E at the point P. To solve this problem we will need to add the electric fields from each of the two point charges, E 1 and E 2. We will need to write these in component format so that they can be added to produce the final result. First, however, let us find the magnitude of each field: E 1 = k Q 1 r 2 1P (47) E 1 = µC (48) E 1 = E 1 = 878 V m E 2 = k Q 2 r 2 2P (49) (50) (51) E 2 = µC (52) E 2 = E 2 = 621 V m (53) (54) OK, now we have the magnitude of each of these two fields, but these are vectors and not scalars, so we need to add them as vectors. Using the angles labeled in the figures we can find the following: E 1 = E 1 cos θ 1 î + E 1 sin θ 1 ĵ (55) 5 cos θ 1 = = (56) 4 sin θ 1 = = (57) E 1 = î ĵ (58) E 1 = 686 V mî V mĵ (59) E 2 = E 2 cos θ 2 î E 2 sin θ 2 ĵ (60) 5

6 sqrt(5*5 + 4*4) E 1 Θ 1 (x=4, y=4) 4 Q = +4 µ C (x= 1, y=0) Θ 1 5 Q = 2 µ C (x=2, y= 1) θ 2 Point P (x=4, y=4) Q = +4 µ C (x= 1, y=0) 5 θ 2 2 sqrt(2*2 + 5*5) Q = 2 µ C (x=2, y= 1) 6

7 2 cos θ 2 = = (61) 5 sin θ 2 = = (62) E 2 = î ĵ (63) E 2 = 231 V mî 576 V mĵ (64) Now that each of these is in component format, they are very easy to add: E T OT = E 1 + E 2 (65) E T OT = ( )î + ( )ĵ (66) E T OT = 455 V mî 28 V mĵ (67) Then, we simply get the magnitude and direction by: E T OT = (68) E T OT = 456 V m (69) tan θ T OT = (70) θ T OT = 3.5 o (71) (b) What would be the magnitude of the force on a particle of charge q = 4µC placed at point P. This one is REALLY easy since F = q E. So, using the total field from the previous step, we can easily get the total force: F = (72) F = N (73) (c) What is the potential, V, at the point P? This one is much easier to calculate than the field simply because the potential is a scalar and the field is a vector. We calculate it in two parts: V 1 = k Q 1 r 1P (74) V 1 = (75) V 1 = 5622V olts (76) 7

8 V 2 = k Q 2 r 2P (77) V 2 = 9 10 (78) V 2 = 3343V olts (79) V T OT = V 1 + V 2 = 2279V (80) (d) How much work is required to bring the q = 4µC charge from infinity to the point P. Ah ha! Again an easy one. The particle has a total energy of U = qv (P ) when placed at the point P. We solved for the potential at point P (V (P ) = 2279V olts) in the previous step. So, all we do to find the potential energy at point P is multiply: U = qv (81) U = = 9.12mJ (82) Then, the potential energy is 0 at infinity and for the work done by the field we get: W = U U P = 9.12mJ (83) The minus sign tells us that WE need to do work against the field, if we want to bring the charge from infinity to point P. (e) Assume that the q = 4µC charge has a mass of m = 12µg and is released from rest starting at the point P. What will be the velocity of this mass when it reaches infinity? Again, very easy. All of the 9.12 milli-joules will turn into kinetic energy when the particle is released and moves to infinity. Note that the other charges will not move, because they are fixed in the positions given in the problem. If they were released as well, then the potential energy of the system will be converted to kinetic energy of all charges. But here we have only released the charge from point P, so using conservation of energy we find: 1 2 mv2 = U (84) 2U v = (85) m v = (86) 9 v = m s (87) 8

9 4. Shown in the figure below is a sphere of charge +Q and radius a. Use I a II +Q Gauss Law to find the electric field in each of these regions: The charge distribution has a spherical symmetry. This means that if we rotate it about any axis going through the center of the sphere, the charge distribution will coincide with itself. The electric field MUST have the same symmetry as the charge distribution that created it. So it has to point along the radius of the sphere and have the same magnitude at points that are equal distance away from the center of the sphere. So OK, we ll choose the Gauss surface for a spherical charge to be a sphere concentric with the charge distribution!!! Then evaluating the flux through the surface is easy. The electric field is constant along the surface and it is always parallel to the surface vector (no angles to worry about in the flux calculation). (a) Region I (r > a). EA = Q in (88) A = 4πr 2 (89) Q in = Q (90) E4πr 2 = Q (91) E = 1 4π Q r 2 (92) (b) Region II (r < a). This one is a little harder since we only contain a fraction of the total charge inside our sphere: EA = Q in (93) A = 4πr 2 (94) Q in = Q V taken V total (95) 4 3 Q in = Q 4 3 πa3 (96) Q in = Q r3 a 3 (97) 9

10 E4πr 2 = Q r 3 a 3 (98) E = 1 4π Q a 3 r (99) 5. Shown in the figure below is an infinitely long cylinder of radius a filled with a uniform change density +ρ Coulombs m. Use Gauss Law to find the 3 I a II +ρ electric field in each of these regions: In this problem we have an infinite object. To have a non-zero field, an infinite object must have an infinite total charge. This is why we DO NOT deal with the total charge of the object, we only deal with charge density!! Also, since we will have to make a closed Gaussian surface, we will need to make this finite. Doing so will introcude a non-physical parameter into the calculation. This non-physical parameter will cancel in the answer (unless we make a mistake). The charge distribution has cylindrical symmetry. This means that we can rotate it along the axis going through the center, translate it along the axis, reflect it with respect to any plane going through the axis or perpendicular to the axis and nothing will change. the only way the electric field will not change with these transformations is if the field is oriented along the radius of the cylinder and has equal magnitude at equal distance from the axis of the cylinder. We choose as out Gaussian surface, a cylinder with radius r and length L. Note that only the barrel of the cylinder counts for Gauss Law since the electric field is along the surface at the end caps. (a) Region I (r > a). EA = Q in (100) 10

11 A = 2πrL (101) Q in = ρ V taken = ρ πa 2 L (102) E2πrL = ρ πa2 L (103) E2r = ρ a2 (104) E = ρ a2 1 2 r (105) (b) Region II (r < a). This time the derivation is very similar. The only difference is the volume taken: EA = Q in (106) A = 2πrL (107) Q in = ρ V taken = ρ πr 2 L (108) E2πrL = ρ πr2 L (109) E2r = ρ r2 (110) E = ρ r 2 (111) 11

12 6. Shown in the figure below are two concentric spheres of charge. A charge +Q is uniformly distributed over the outer sphere between r=b and r=c. A charge -Q is uniformly distributed over the inner sphere r < a. The space between the spheres, a < r < b, is empty. Find the electric field in I II III IV Q a b c +Q each of these regions: The problem involves an object with spherical symmetry. We ll use Gauss s law and make all our Gauss surfaces spheres concentric with the charged spheres. By symmetry we know the direction of the field ( along the radius). We just need to find the magnitude in each region. (a) Region I (r > c). SO EASY! Q in = 0, thus E=0. (b) Region II (b < r < c). OK, now we have a challenge (the biggest one of this problem). In this case, the Q in will include all of the inner charge and only a fraction of the outer charge. OUCH!!! EA = Q in (112) A = 4πr 2 (113) Q in = Q V taken V total Q (114) V total = 4 3 πc3 4 3 πb3 (115) V taken = 4 3 πr3 4 3 πb3 (116) V taken V total = 4 3 πr3 4 3 πb3 4 3 πc3 4 3 πb3 (117) 12

13 V taken V total = r3 b 3 c 3 b 3 (118) Q in = Q r3 b 3 c 3 b 3 Q = Qr3 c 3 c 3 b 3 (119) E4πr 2 = Q r 3 c 3 c 3 b 3 (120) E = Q r 3 c 3 1 4π c 3 b 3 r 2 (121) (c) Region III (a < r < b). Oh, sweet simplicity!!! We take all the inner charge and none of the outer charge. So, Q in = Q. EA = Q in (122) A = 4πr 2 (123) Q in = Q (124) E4πr 2 = Q (125) E = 1 4π Q r 2 (126) Hmmm...seem familiar...yes, this is the same as two problems ago. (d) Region IV (r < a). Deja vu...this is the same as two problems ago... EA = Q in (127) A = 4πr 2 (128) Q in = Q V taken V total (129) 4 3 Q in = Q 4 3 πa3 (130) Q in = Q r3 a 3 (131) E4πr 2 = Q r 3 a 3 (132) E = 1 4π Q a 3 r (133) 13

14 7. Shown in the figure below are two concentric cylinders of charge. The cylinders are uniformly filled with charge densities ±ρ Coulombs m as indicated in the figure. The space between the cylinders, a < r < b, is empty. 3 Find the electric field in each of these regions: I II III IV ρ a b c +ρ If we assume that the cylinders are infinitely long or at least very long, then we can use Gauss law to find the field.sorry that I forgot to say that the cylinders ARE very long when I posted the problem. In the general case, this problem is quite difficult to solve because it will require integration over the finite charge distribution using Coulomb s law. But I didn t mean to be mean... I just wanted you to use Gauss s law. So we proceed as before. First we determine from symmetry that the field can only have components along the radius. The we find the magnitude in each region using Gauss s law. (a) Region I (r > c). So, you know the drill...cylindrical charge cylindrical surface. The pain here is the volume of the outer cylinder. EA = Q in (134) A = 2πrL (135) Q in = Q outer + Q inner (136) Q inner = ρ V inner (137) Q inner = ρ πa 2 L (138) Q outer = ρ V outer (139) 14

15 Q outer = ρ [ πc 2 L πb 2 L ] (140) Q outer = ρπl [ c 2 b 2] (141) Q in = ρπl [ c 2 b 2] ρ πa 2 L (142) Q in = ρπl [ c 2 b 2 a 2] (143) E2πrL = ρπl [ c 2 b 2 a 2] (144) E2r = ρ [ c 2 b 2 a 2] (145) E2r = ρ [ c 2 b 2 a 2] 1 (146) 2 r (b) Region II (b < r < c). In this part of the problem, we take all of the inner charge, but only a fraction of the outer charge. EA = Q in (147) A = 2πrL (148) Q in = Q outer + Q inner (149) Q inner = ρ V inner (150) Q inner = ρ πa 2 L (151) Q outer = ρ V taken (152) Q outer = ρ [ πr 2 L πb 2 L ] (153) Q outer = ρπl [ r 2 b 2] (154) Q in = ρπl [ r 2 b 2] ρ πa 2 L (155) Q in = ρπl [ r 2 b 2 a 2] (156) E2πrL = ρπl [ r 2 b 2 a 2] (157) E2r = ρ [ r 2 b 2 a 2] (158) E2r = ρ [ r 2 b 2 a 2] 1 (159) 2 r (c) Region III (a < r < b). Hey, this is just like part a of a previous problem. EA = Q in (160) A = 2πrL (161) Q in = ρ V taken = ρ πa 2 L (162) 15

16 E2πrL = ρ πa2 L (163) E2r = ρ a2 (164) E = ρ a2 1 2 r (d) Region IV (r < a). This is just like part b of a previous problem. (165) EA = Q in (166) A = 2πrL (167) Q in = ρ V taken = ρ πr 2 L (168) E2πrL = ρ πr2 L (169) ρ r2 E2r = (170) E = ρ r 2 (171) 8. Shown in the figure below are two concentric CONDUCTING spheres of charge. Because the spheres are conducting, the charge will not be uniformly distributed over their volumes, but will instead accumulate on the surfaces. The inner sphere has a total charge -Q and the outer sphere has a total charge +2Q. Use Gauss Law to determine all the following: I II III IV Q a b c +2Q Boy, I love conducting spheres. The charges are all on the surfaces. So your Gauss surface always either contains all of the charge or contains none (no bogus fractions). Also, E is zero inside conductors, sweet. So we determine the charges first, then the fields. All your answers are point charge answers with the right charge in the center of the sphere. 16

17 (a) The electric field in Region I (r > c). Total charge = +1Q, (b) The electric field in Region II (b < r < c). IN THE CONDUCTOR, E=0. (c) The electric field in Region III (a < r < b). Total charge = -1Q, (d) The electric field in Region IV (r < a). IN THE CONDUCTOR, E=0. E = 1 4π Q r 2 (172) E = 1 4π Q r 2 (173) (e) The charge on the surface r=c. +Q (since you needed to send +1Q for shielding, that leaves +1Q here). (f) The charge on the surface r=b. +Q (since you like gotta shield out the inner guys charge to have zero field in the conductor). (g) The charge on the surface r=a. -Q (all the inner charge). 17

18 9. The figure below shows a set of equipotential lines that were created by some set of charges. The lines represent potentials ranging from -300 V to V in steps of 100 V. Also labelled in the plot are the points a = (0,0), b = (1.5 m, 3 m), and c = (3 m, 2.5 m). b c a 300 V 0V +300 V (a) Which point (a, b, or c) has the highest E? Point a. The density of equipotential lines is highest around that point. (b) Which point (a, b, or c) has the lowest E? Point c.the density of equipotential lines is lowest around that point. (c) Estimate the strength of the electric field at a point 1/2 way between b and c. These points are separated by 100 V. They are also separated by some distance. Field is volts per meter...and you can get the magnitude of the field from the derivative of the potential, which is to say that we have to see by how much the potential changes in the vicinity of the point where we want to calculate the field. V = 100 (174) d = ( x) 2 + ( y) 2 (175) d = (1.5) 2 + (0.5) 2 (176) d = (1.5) 2 + (0.5) 2 (177) d = 1.58 (178) E = V d = 63.24V/m (179) (d) Draw several (roughly 6) electric field lines on the plot. 18

19 b c a 300 V 0V +300 V (e) If these lines were generates by two point charges (one to the left and one to the right), determine which side of the graph contains which charge (i.e. either +left/-right or -left/+right). -left/+right The answer follows from the fact that the field always points in the direction of decreasing potential and is always away from a positive charge and toward a negative charge. 19

20 10. Shown in the figure below is a parallel plate capacitor. The area of each plate is 1 m 2 and the separation between the plates is 1 mm (0.001 m). The region between the plates is filled with a dielectric material of dielectric constant K=2. +Q Q d (a) Determine the capacitance of this device. OH NO!!! A dielectric. Don t panic. All we do is use ɛ = K in every place we would have used. C = ɛ A d (180) 1 C = (181) C = (182) C = F (183) (b) If the capacitor is charged to Q = 6µC, what is the voltage difference between the plates? V = Q C (184) (c) What is the electric field between the plates? V = (185) V = 339V olts (186) E = V x = (187) E = 33900V/m (188) 20

21 (d) What is the stored energy in the system? (e) In what form is the energy stored? Electric Field U = 1 QV (189) 2 U = 1 2 ( )339 (190) U = 0.001J (191) 21

22 11. Shown below is a circuit containing a capacitor network. Let the values of the labeled components be as follows: C 1 = 2µF, C 2 = 3µF, C 3 = 4µF, C 4 = 6µF, V = 10V (a) Find the equivalent capacitance of the network. To solve this problem we must reduce the capacitor network to an equivalent capacitor step by step. All we need to know is how to add capacitors in series and in parallel. C1 CA V C4 The first figure shows how we make the equivalent capacitor C A from C 2 and C 3 as follows: C A = C 2 + C 3 (192) C A = 3µF + 4µF = 7µF (193) Now we must combine C A and C 1 in order to form C B : 1 = C B C 1 C A (194) 1 = 1 C B 2µF + 1 7µF (195) 1 = C B µf (196) C B = 1.555µF (197) Now finally, we combine C B with C 4 in order to make C C, the equivalent capacitor to the whole network: 22

23 CB V C 4 C V 23

24 C C = C 4 + C B (198) C A = 6µF µF = 7.555µF (199) (b) Find the charge on each capacitor. We use the definition of capacitance to find V = Q C. Later, we would need to remember that the capacitance is a measure of how much energy you can store in the electric field of the capacitor. We use the formula U = 1 2QV, or any other equivalent formula that we ll get using the definition of capacitance. Here is what we know initially: C (µf ) V (Volts) Q (µc) U (Joules) C 1 2 C 2 3 C 3 4 C 4 6 C A 7 C B C C OK, 2 out of 3 on the bottom line and we we fill in: C (µf ) V (Volts) Q (µc) U (µjoules) C 1 2 C 2 3 C 3 4 C 4 6 C A 7 C B C C OK, now we know that since C C is made from the parallel combination of C 4 and C B that all three of these share the same voltage: C (µf ) V (Volts) Q (µc) U (µjoules) C 1 2 C 2 3 C 3 4 C C A 7 C B C C OK, 2 out of 3 for C 4 and C B, so we finish out the lines as: 24

25 C (µf ) V (Volts) Q (µc) U (µjoules) C 1 2 C 2 3 C 3 4 C C A 7 C B C C Now we notice that C B is made from C A and C 1 in series. Thus, all three of these have the same charge: C (µf ) V (Volts) Q (µc) U (µjoules) C C 2 3 C 3 4 C C A C B C C Yup, 2 out of 3 for C 1 and C A, so: C (µf ) V (Volts) Q (µc) U (µjoules) C C 2 3 C 3 4 C C A C B C C Finally, C 2, C 3, and C A all have the same voltage. Thus, we find: C (µf ) V (Volts) Q (µc) U (µjoules) C C C C C A C B C C (c) Find the voltage across each capacitor. in table (d) Find the energy stored in each capacitor. in table 25

26 12. Shown in the figure below is an electrical circuit. The values of the components are as follows: R 1 = 1kΩ, R 2 = 2kΩ, R 3 = 3kΩ, V 1 = 10V, V 2 = 20V. (a) Write the Kirchoff voltage law for the loop containing V 1, R 1, andr 2. I will assume the following three currents: i. I 1 flows to the right through R 1. ii. I 2 flows down through R 2. iii. I 3 flows up through R 3. In this case, we can solve for the voltage equation: 0 = V 1 R 1 I 1 R 2 I 2 (200) 0 = I I 2 (201) 10 = 1000I I 2 (202) (b) Write the Kirchoff voltage law for the loop containing V 2, R 3, andr 2. 0 = V 2 R 3 I 3 R 2 I 2 (203) 0 = I I 2 (204) 20 = 3000I I 2 (205) (c) Write the Kirchoff current law. Looking at the top junction, we find that the currents I 1 and I 3 flow in while I 2 flows out. In this case, I 1 + I 3 = I 2 (206) 26

27 (d) Find the current through R 1. The best first step is always to put the current equation into the voltage equation. In this case: 10 = 1000I I 2 (207) 10 = 1000I (I 1 + I 3 ) (208) 10 = 3000I I 3 (209) 20 = 3000I I 2 (210) 20 = 3000I (I 1 + I 3 ) (211) 20 = 5000I I 1 (212) OK, to eliminate I 3, let s multiply the first equation by 5 and the second equation by -2. We can then add them: 50 = 15000I I 3 (213) 40 = 10000I I 1 (214) 10 = 11000I 1 (215) I 1 = = 0.909mA (216) (e) Find the current through R 2. Now that we know I 1, we can plug this into a previous equation containing I 1 and I 2 : 10 = 1000I I 2 (217) 10 = I 2 (218) = 2000I 2 (219) I 2 = = mA (220) (f) Find the current through R 3. OK, Let s put the I 1 and I 2 into the current equation: I 1 + I 3 = I 2 (221) 0.909mA + I 3 = mA (222) I 3 = mA (223) 27

28 (g) Find the power dissipated by each resistor. We can calculate power any of the following ways: P = V I = I 2 R = V 2 R. Presently we know R and I for every resistor. So, this makes for the simplest calculation: R (Ω) I (A) P (Watts) R R R

29 13. Shown below is a circuit containing a resistor network. Let the values of the labeled components be as follows: R 1 = 2kΩ, R 2 = 3kΩ, R 3 = 4kΩ, R 4 = 6kΩ, V = 10V (a) Find the equivalent resistance of the network. To solve this problem we must reduce the resistor network step by step. Three figures were generated to show this reduction. In the first figure, we combine resistors R 2 and R 3 to form an equivalent resistor R A. R 1 RA R 4 V The calculation of R A is as follows: 1 = R A R 2 R 3 (224) 1 = 1 R A 3kΩ + 1 4kΩ (225) 1 = R A kω (226) R A = 1.714kΩ (227) Next we reduce the circuit further by combining resistor R 1 with resistor R A and thereby form a resistor that we will call R B This one is a simpler calculation: R B = R 1 + R A (228) R B = 2kΩ kΩ (229) R b = 3.714kΩ (230) 29

30 R B R 4 V Finally, we need to combine resistors R B and R 4 so as to make an equivalent resistor R C. This will end the reduction and answer this part of the problem: R C V 1 = R C R B R 4 (231) 1 1 = R C 3.714kΩ + 1 6kΩ (232) 1 = R C kω (233) 30

31 R C = 2.294kΩ (234) (b) Find the current through each resistor. OK, to solve all of the remaining steps of this problem, we will use a technique that is the equivalent of filling a table. The table will have 7 rows labeled as (R 1, R 2, R 3, R 4, R A, R B, R C ). There will be four columns labeled as (R, V, I, P). At this point, the table can be filled in as follows: R (Ω) V (Volts) I (A) P (Watts) R R R R R A 1714 R B 3714 R C OK, now here is the secret to solving the table. Since V = IR, whenever, we get any two of the three values (V, I, R) we can get the third. In the bottom row, we can find I of resistor C since we have R and V. After that P = V I,and we have finished filling in the bottom row: R (Ω) V (Volts) I (A) P (Watts) R R R R R A 1714 R B 3714 R C OK, now for the cool part. The resistor R C is made from the parallel combination of R 4 and R B. Since this is a parallel combination, the voltages in R C, R 4, and R B, are ALL THE SAME! So we simply copy the voltage on R C to R 4 and R B giving us: R (Ω) V (Volts) I (A) P (Watts) R R R R R A 1714 R B R C OK, now we see that we have 2 out of 3 for R 4, and R B, and so we can fill in the full lines giving us: 31

32 R (Ω) V (Volts) I (A) P (Watts) R R R R R A 1714 R B R C Fine, now we know that R B was made from the series combination of R 1 and R A. In that case, R 1, R A, and R B all have identically the same current, so we copy that value to multiple places: R (Ω) V (Volts) I (A) P (Watts) R R R R R A R B R C Cool, we have 2 out of 3 in the rows for resistors R 1 and R A and so we fill out these lines to have: R (Ω) V (Volts) I (A) P (Watts) R R R R R A R B R C OK here comes the last step! R A is made from the parallel combination of R 2 and R 3. So, these all have the same voltage that we copy to other parts of the table: R (Ω) V (Volts) I (A) P (Watts) R R R R R A R B R C Finally 2 out of 3 in the last two rows and we simply finish out the table: 32

33 R (Ω) V (Volts) I (A) P (Watts) R R R R R A R B R C (c) Find the voltage across each resistor. in table (d) Find the power dissipated by each resistor. in table Of course, you only need to write the table once. 14. The circuit below begins with the switch in the open position and the capacitor carrying zero charge. The battery has V B = 10V, the resistor has R = 5kΩ, and the capacitor has C = 12µF. At time=0, the switch in the circuit is closed. (a) Determine the time constant of this circuit. τ = RC = 5kΩ12µF = 60msec (b) Draw sketches of the time dependence of each of the following: i. The voltage on the capacitor as a function of time, V C (t). 10 V V C ii. The charge on the capacitor as a function of time, Q C (t). iii. The voltage on the resistor as a function of time, V R (t). t 33

34 Q C 120 µ C t V R 10 V t I R 2 ma t 34

35 iv. The current through the resistor as a function of time. To be counted for full credit the vertical axis of each sketch should be labeled with a numerical value indicating either the initial or asymptotic value of the quantity plotted. (c) Write an equation for each of the following: i. The voltage on the capacitor as a function of time, V C (t). The capacitor is charging. The voltage on the capacitor is a saturating exponential whose eventual ( voltage) is the full 10 V of the battery. Thus, V C (t) = 10V 1 e t 60msec ii. The charge on the capacitor as a function of time, Q C (t). Since the asymptotic voltage on the capacitor is 10 V, and V = Q C or Q = CV, the asymptotic ( charge on ) the capacitor is Q final = 120µC. Q C (t) = 120µC 1 e t 60msec iii. The voltage on the resistor as a function of time, V R (t). From Kirchoff s loop rule we can see that the resistor takes up the balance of the battery voltage that is not across the capacitor. This is a decaying exponential: V R (t) = 10V e t 60msec iv. The current through the resistor as a function of time. Well, V = IR implies that I = V R. Since that starting voltage on the resistor is 10 V, the starting I = 10V 5kΩ = 2mA. I R(t) = 2mAe t 60msec (d) At what time does the voltage on the capacitor reach 3 V? Using the capacitor voltage formula, we plug in 3 V for the voltage in the V C (t) formula. Then we have: ( 3V = 10V 0.3 = 1 e t 60msec ( 1 e t 60msec ) ) (235) (236) 0.7 = e t 60msec (237) t ln 0.7 = 60msec (238) t = 60msec ln 0.7 = 21.4msec (239) 35

36 15. Shown in the figure below is a mass analyzer system. This system operates by accelerating a charged particle across the gap between two capacitor plates, giving it energy. After it has been accelerated, it enters a magnetic field and is bent in a circular trajectory until it reaches the exit E = 25 MeV R=1.2 m B =? You are to design the mass analyzer system so that 12 C ions with charge, q = , and mass m = kg = kg are accelerated to an energy of 25 MeV. Answer all the following: (a) What should be the voltage difference across the capacitor? This one is very easy if you are good in the use of ev and MeV units. The charge of the carbon ion in this problem is 1 e (or C). First, let s solve this problem using e s as the charges. We know that the kinetic energy of the ion will come from converting the potential energy that s available in the system. U = qv (240) 25M ev = (1e)V (241) 25MV = V (242) V = 25, 000, 000V olts. (243) OK, if you really like Joules, we can do it that way also: U = 25MeV C = J (244) J = qv (245) J = CV (246) V = 25, 000, 000V olts (247) 36

37 (b) If the separation between the plates is 1 meter, what would be the electric field between the plates? Here the simplest way to get the solution is to realize that the electric field has units of V m. The full formula uses a derivative of the potential to make each of the vector components of E, ( E x = dv dx î). However, for a constant electric field (like in the capacitor plates, the formula is simpler ( E x = V x î). Thus, E x = V x î (248) E x = V x = V 1meter (249) E x = 25, 000, 000 V m (250) E y = E z = 0 (251) (c) If the area of each plate is 50m 2, what is the capacitance and charge of this device? For a simple parallel plate capcitor, the capacitance is given by: A C = d = ɛ 50m 2 0 1meter (252) C = F 50m 2 m 1meter (253) C = F = 443pF (254) OK, for any capacitor, the voltage across the capacitor is proportional to the charge it holds as given by: V = Q C (255) Q 25, 000, 000 = (256) Q = Coulombs = 11.07mC (257) (d) What will be the velocity of the ions after they are accelerated by the capacitor? This one is really easy. We know the energy of the ions when they leave the capacitor (25 MeV). All this energy is in the form of Kinetic Energy. 1 2 mv2 = 25MeV = ev C (258) 37

38 1 2 mv2 = J (259) J v = (260) m v = 12 J (261) kg v = m s (262) (e) What magnetic field should be used so that the ions travel in a circular path with radius, R = 1.2meters. To make the ions travel on a circular path, we need to apply a force that is perpendicular to their velocity. We can achieve that by introducing magnetic field that is perpendicular to the velocity. Then the magnitude of the force is: F = qvb sin 90 O (263) F = qvb = ma (264) a = v2 r (265) qvb = m v2 r (266) B = mv qr (267) B = kg (268) B = 2.08T esla (269) (f) How would this differ if the ions were instead 14 6 C ions? Answer in a few sentences, not a calculation. If the ions were 14 6 C, the energy would not change (since mass did not enter those calculations). However, the 14 6 C ions are heavier and it thus would be harder to bend their trajectory. They will move with a larger radius in the magnetic field. (g) What would be the usefulness of this device? You could use it to measure both the 14 6 C and 12 6 C content of some sample of once-living material and thereby date the material. (h) What electric field would you put inside the magnet so as to make the ions travel in a straight line instead of a curved path? In this case, we want an electric field strong enough that its force cancels out the magnetic force. Of course, for the cancellation to happen the two forces need to have opposite directions. 38

39 qe = qvb (270) E = vb (271) E = = V m (272) 39

40 16. A wire with radius a carries a current I. The current is uniformly distributed throughout the wire. I i ii a (a) Use Ampere s Law to calculate the magnetic field in region i. ANSWER: The wire has cylindrical symmetry with all the charges moving parallel to the axis of the wire. The magnetic field has to be tangent to circles that are concentric with the wire. We don t know what the strength of the field is as a function of the distance to the center of the wire. That s what we are going to find out using Ampere s law. Our Ampere s path will be a circular loop whose length is 2πr. This path contains all of the current. B dl = µ 0 I in (273) I in = I (274) BL = µ 0 I (275) B = µ 0I 2πr (276) (b) Use Ampere s Law to calculate the magnetic field in region ii. ANSWER: Our Ampere s path will be a circular loop whose length is 2πr. This path encloses only a fraction of the current. B dl = µ 0 I in (277) I in = I πr2 πa 2 = r2 a 2 (278) (279) BL = µ 0 I r2 a 2 (280) B = µ 0Ir 2 2πra 2 = µ 0I 2πa 2 r (281) 40

41 17. You have a solenoid, a battery and ammeter and a Hall probe. The Hall probe is a device which allows you to measure magnetic field. You decide to test Ampere s law. (a) Assuming the solenoid is very long, make a prediction of the strength of the magnetic field inside the solenoid. ANSWER: For a very long solenoid one can prove that the magnetic field outside the solenoid is 0 and inside, it is pointing along the axis. We have done this in class and I will not repeat it here, since the problem does not ask to prove that. Our Ampere s path will be a rectangular loop of length L that is partially inside and partially outside the solenoid. This line integral of B has no contribution outside the coil, since there B = 0 and no contribution on the sides (field perpendicular to path). So, the only contribution is BL from the side inside the coil. Also, if the solenoid carries a current I, the loop catches MORE than I since it contains more than one loop of the coil. We ll assume that the Ampere s path has N loops that cut through it. B dl = µ 0 I in (282) I in = NI (283) BL = µ 0 NI (284) N B = µ 0 L I (285) (b) Next you want to test your prediction. You can measure the magnetic field using the Hall probe, but to see if your calculation from (a) holds you need to measure the current that creates the magnetic field. So you build a circuit using the battery, the ammeter and the solenoid. Draw the circuit making sure that the ammeter is connected properly. We need to connect the ammeter in series with the coil. I tried to make a hand drawing for you... but it didn t come out nice, so I m not including it. This part was not hard to figure out anyway and has no calculation in it. (c) The ammeter measured current I = 1 Ampere, and you counted that the solenoid has 500 coils and is 0.2 m long. Calculate the magnetic field. Now you are ready to compare to what you measured with the Hall probe. ANSWER: N B = µ 0 L I (286) B = 4π (287) B = T esla (288) 41