Conceptual Exercises 11. In order of increasing magnitude of the net force (from smallest net force to biggest net force), it s C, then A, then B.

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1 Physics 140 Conceptual Questions 3. Rubbing the comb through your hair strips trons off of your hair and deposits them on the comb so that the comb acuires a net negative charge. When this negatively charged comb is brought near the bits of (trically neutral) paper, trons in the atoms making up the paper are repelled by the comb. These trons are thus a little bit farther away from the comb, on average, than the positively charged nuclei of the atoms in the paper. Therefore, the trical force of attraction between the comb and the positively charged nuclei is a little bit stronger than the trical force of repulsion between the comb and the trons surrounding the nuclei in the comb. This leads to a net force of attraction between the bits of paper and the comb. It s important to point out here that neither the trons in the paper nor the positively charged nuclei in the paper actually move appreciably throughout the paper. In a solid, the nuclei are held in fixed positions by very strong interatomic and intermolecular forces. The trons surrounding a particular nucleus are still constrained to the volume around that nucleus. The force of repulsion that the comb exerts on the trons is not enough to allow them to break free from this nucleus and go wandering off throughout the paper. Also, it s important to observe here that nothing in this process changes the net charge of the bits of paper. They remain trically neutral. Conceptual Exercises 11. In order of increasing magnitude of the net force (from smallest net force to biggest net force), it s C, then A, then B. Charge A experiences a force of attraction from charge B and from C. The magnitude of the net force on Charge A is therefore 5 ( ) = + = A d ( d ) 4d Charge B experiences a force of attraction from Charge A and a force of repulsion from Charge C. The magnitude of the net force on B is therefore ( ) = + = B d d d Charge C experiences a force of attraction from Charge A and a force of repulsion from Charge B. The magnitude of the net force on C is therefore 3 ( ) = = C d ( d ) 4d Comparing these three results, we see that the net force on C is the smallest. The net force on A is larger. The net force on B is largest. 4. Gauss s law relates the net tric flux through any surface to the net charge enclosed by the surface: Qencl Φ = ε0 So we just need to look at the surfaces A, B, C, and D and figure out the net charge enclosed by each. Looking at the surfaces, I find that the net charges enclosed are as follows: Surface A: Qencl =+ + =+ Surface B: Qencl =+ =+ Surface C: Qencl =+ = 0 Surface D: Qencl = So in order of increasing net tric flux, it s D, then C, then B, then A. 1

2 Physics 140 Problems 7. (a.) Let N e be the number of trons and e N p be the number of protons. Then it must be true that N + N = 155 (1) p Furthermore, the net charge of the whole collection of particles would be Q = N + N, () net e e p p in which e is the charge of each tron and and so () becomes e p p is the charge of each proton. But 19 = e= C 19 =+ e= C, Q = e N N, (3) net p e 17 and we know that this must eual C. Euations (1) and (3) are a pair of simultaneous euations in the two unknowns N p and N e. By the usual procedure for solving a pair of simultaneous euations, I find: Q ( 155) C ( 155)( C net + e + ) N p = = e C and N = 59 p N = 933 e (b.) The mass of the system would be in which m e is the mass of an tron and m sys msys = Neme + Npmp, m p is the mass of a proton. So: ( 933)( kg) ( 59)( kg) = + m sys 5 = kg 8. r = F r = = k F 1 r 1 9 N m C C C 1.77 N r = 1.9 m

3 Physics The charge 1 experiences a force of attraction from and a force of repulsion from 3. The magnitude of the force that exerts on 1 is 1 F1 = k = r1 d The magnitude of the force that 3 exerts on 1 is F31 = k = = r13 ( d ) 4d So the net force on 1 is 3 5 = F1 F31 = = d 4d 4d 9 N m ( 1 10 C) C F net = = 3 N 40.1 m 1. Think of the line the charges are on as being the x axis. Then think of three regions: Region I, to left of 1 ; Region II, between 1 and 3 ; Region III, to the right of 3. If were place in Region I, the forces on due to 1 and 3 would both point to the left. Therefore, there is no way these forces could cancel to make the net force on zero. Similarly, if were placed in Region III, the forces that 1 and 3 exert on it would both point to the right. Therefore, there is no way the net force could be zero. The only region in which the net force can be zero is therefore in Region II. Now imagine that is placed in Region II so that the net force on it is zero. Let the distance that is from 1 be called x. Then the magnitude of the force that 1 exerts on would be F1 = x and the magnitude of the force that 3 exerts on would be If the net force on is to be zero then F = 3 ( 0.3 m x) F1 = F3 = x 0.3 m x 1 = 3 x ( 0.3 m x) and now I just have to do a little algebra to solve this for x. In general, this is a uadratic. So you could cross-multiply, gather all the terms on one side, then use the uadratic formula (keeping only the positive solution for x.) One trick to make the algebra a little easier, though, is the following. Rewrite (1) as 0.3 x = 3 x Then take the suare root of both sides. In general, there d be two solutions 0.3 x = ± 3 x (1) 3

4 Physics 140 But I know I want only the positive solution. (If x were negative, this would mean that is actually in Region I, and I know that the net force can t be zero there.) So 0.3 x = 3 x 0.3 x = 3x x ( 1+ 3) = x = = 0.1 m Let s define a coordinate system with the origin at the location of, the positive x axis to the right and the positive y axis upward. Then the force that 1 exerts on would be, in component form F1 = yˆ d The force that 3 exerts on would be F3 = xˆ d The force that 4 exerts on would be F4 = ( F4 cos 45 ) xˆ+ ( F4 sin 45 ) yˆ Because 1 cos 45 = sin 45 =, this becomes F4 F4 = ( xˆ+ yˆ) But the magnitude of F 4 is 8 F4 =, r4 and from the figure, So r 4 = d 8 4 F4 = = ( d ) d and 4 d 4 ( ) F4 = ( xˆ+ yˆ) = ( xˆ+ yˆ) = xˆ+ yˆ d d The net force exerted on 1 is then = F1 + F3 + F4 ( ) = yˆ xˆ ( xˆ yˆ ) + + d d d 4

5 Physics 140 ( ) ( ) = + xˆ+ + yˆ d d d d ( + ) ( ) = xˆ+ yˆ d d = { ( + ) xˆ+ ( ) yˆ} d Plugging in numbers, I get 9 ( )(.4 10 ) = ( + ) xˆ+ ( ) yˆ ( 0.33) = ( 4.0 N) xˆ+ ( N) yˆ The magnitude of the net force is therefore F = F, + F, = ( 4.0 N) + ( N) = 4. N net net x net y { } To specify the direction, let s call the angle that F net makes with the x axis φ. Then φ = tan = If we call θ the angle that F net makes with the +x axis, then θ = 180 φ = (a.) The tric fields due to 1 and are eual in magnitude and opposite in direction. Therefore, they cancel. The net tric field is then just the tric field due to 3. The magnitude of this field is Enet = E3 =, r3 in which r 3 is the distance from 3 to the midpoint between 1 and. From the figure, this distance is Therefore Plugging in numbers, I get d 3d r3 = d = 4 E net 4 = = 3d 3d 4 E net = N/C (b.) greater. At the point described, the tric fields due to and 3 do not cancel, as the fields due to 1 and did in Part (a.). (c.) Let s define a coordinate system with the +x axis pointing along the line from to 3 and the +y direction pointing along the line from 1 to the midpoint between and 3. Then the fields due to and 3 will be: 7 5

6 Physics 140 The field due to 1 will be 4 E = E3 = xˆ = xˆ ( d /) d E1 = yˆ, r 1 in which r 1 is the distance from 1 to the midpoint between and 3. From the figure, 3d r1 = d ( d /) = 4 So 4 E1 = yˆ = yˆ 3d 3d 4 And the net tric field is then Enet = E1+ E + E3 = xˆ yˆ xˆ yˆ + + = + d d 3d d 3d Plugging in numbers E ( N/C) ˆ ( net = x+ N/C) yˆ The magnitude of this is E = N/C N/C = N/C net 41. (a.) 1 and 3 are positive, because the field lines go away from them. (b.) There are 8 lines leaving 1 and 1 lines terminating on. Therefore = 1 = 1 1 = = 5.00 µ C 1 = µ C (c.) By exactly the same arguments I just made in Part (b.), 3 = 1 = µ C Gauss s law says Q encl Φ = = ε ( ) ( C C) + ( C) C Nm 5 Φ =.8 10 Wb 0 1 Q Φ = ε encl 0

7 Physics 140 In Gauss s law, Φ is the net tric flux through the entire closed surface. If the surface has six sides, then Φ is the sum of the fluxes through the individual sides: Φ =Φ 1+Φ +Φ 3 +Φ 4 +Φ 4 +Φ Plugging in the numbers they gave me, I get Φ = Wb So 1 C 9 Qencl = ε0φ = ( Wb) = C = 5.09 nc N m 55. Gauss s law says Q Φ = ε Φ means the net flux through the entire closed Gaussian cylinder shown. This is, strictly speaking, the flux through the lateral surface plus the flux through the encaps: Φ =Φ +Φ +Φ encl left lateral right endcap surface endcap But everywhere on the left and right endcaps, E has no component perpendicular to the area, so the flux is zero. This leaves Φ =Φ = EA = E π rl, lateral lateral surface surface 0 in which I ve used the fact that the area of the lateral surface of a right circular cylinder of radius r and height L is π rl. So Gauss s law becomes ( π rl) E = ε0 But what s Q encl? Well, I figure that it s the charge per unit length on the wire times the length of the wire enclosed: charge Qencl = ( length enclosed) = λl unit length Putting this into Gauss s law gives λl E( π rl) = ε0 So λ E =, πε 0r as we wanted to show. Q encl 7

8 Physics (a.) Consider the free-body diagram (FBD) for the negative charge: y T 0 70 F x mg From this picture, it follows that F = Tcos 70 But what s T? Well, thinking about the forces in the y direction, I get Tsin 70 = mg The tension is therefore mg ( kg)( 9.81 m/s ) 3 T = = = N sin 70 sin 70 (1) Plugging this back into (1), I get F 3 4 = N cos 70 = N (b.) (c.) I just answered this in Part (a.) F F = ( m) ( m) ( N)( m) = = = C k 9 N m C (a.) Imagine a coordinate system with the origin at the initial location of the tron, the +x axis to the right, and the +y axis downward. Once the tron enters the region between the plates of the capacitor, it experiences a downward force of magnitude F = ee, in which e is the magnitude of the charge of the tron. The tric field is constant; therefore the force is constant. Furthermore, this force has no x component! Therefore, there is no acceleration in the x direction and the x component of the tron s velocity is therefore constant. Let t be the time when the particle emerges from the other side of the capacitor. Call the tron s x component of velocity v x. Then 0.05 m v x =, t 8

9 Physics 140 and vx = v0x = m/s. So the time when the tron emerges from the other side is 0.05 m 9 t = = s m/s At this time, y = m. Now, recalling the euations of kinematics for motion with constant acceleration (from Physics 1401): 1 y = y0 + v0yt+ ayt Because of the way I chose the origin, y 0 = 0. And because particle was shot in horizontally, v 0 y = 0. So I have 1 y = a t From Newton s second law so I have Solving this for E, I get Plugging in y = m at t a y y = F ee m = m, 1 ee y = t m my E = et 9 = s, I get 31 ( ) kg m 3 E = = N/C ( C)( s) (b.) The speed of the tron when it leaves the cap is the magnitude of its velocity vector at that time x y v = v + v But we already know one of the two components of the velocity. Specifically, we know that vx = v0 x = m/s We just need to find v y. Well, recalling the euations of kinematics for constant acceleration from Physics 1401 again, ( C)( N/C) ee ( 9 vy = v0 y + at y = 0+ at y = t = s -31 m kg ) v y = m/s Plugging this into the expression for v, I get v =. 10 m/s 9