CHAPTER 22. Answer to Checkpoint Questions

Transcription

1 60 CHAPTER ELECTRIC CHARGE CHAPTER Answer to Checkpoint Questions. C and D attract; B and D attract. (a) leftward; (b) leftward; (c) leftward 3. (a) a, c, b; (b) less than 4. 5e (net charge of 30e is equally shared) Answer to Questions. No, only for charged particles, charged particle-like objects. all tie 3. a and b 4. (a) between; (b) positively charged; (c) unstable 5. two points: one to the left of the particles and one between the protons 6. q = r, upward 7. 6q = d, leftward 8. a and d tie, then b and c tie 9. (a) same; (b) less than; (c) cancel; (d) add; (e) the adding components; (f) positive direction of y; (g) negative direction of y; (h) positive direction of x; (i) negative direction of x 0. (a) positive; (b) negative. (a) A, B, and D; (b) all four; (c) connect A and D, disconnect them, then connect one of them to B (there are two other solutions). (a) neutral; (b) negatively 3. (a) possibly; (b) denitely

2 CHAPTER ELECTRIC CHARGE 6 4. (a) Ground A with the wire, bring the positively charged rod near A but not touching it, remove the wire (A is now negatively charged), remove the rod, connect the wire between A and B (both are now negatively charged by the same amount), remove the wire. (b) Method : Connect the two spheres with the wire; bring the rod near sphere A (A now has excess negative charge and B has just as much excess positive charge); remove the wire; remove the rod. Method : Ground A with the wire, bring the positively charged rod near A but not touching it, remove the wire (A is now negatively charged), remove the rod, ground B with the wire, bring A near B (A is negatively charged and B is now positively charged by the same amount), remove the wire. 5. same 6. Once enough of the electrons have moved to the opposite end, any other conduction electron is repelled by the electrons at the opposite end as much as it is repelled by the negatively charged rod at the near end. 7. D 8. no (the person and the conductor share the charge) Solutions to Exercises & Problems E The charge transferred is q = it = (:5 0 4 A)(0 0 6 s) = 0:50 C : E (a) (b) F = q q r = (:00 C) (8: Nm =C ) (:00 m) = 8: N : F = q q r = (:00 C) (8: Nm =C ) (: m) = 8: N : 3E The magnitude of the force is given by F = q q r ;

3 6 CHAPTER ELECTRIC CHARGE where q and q are the magnitudes of the charges and r is the distance between them. Thus F = (8:99 09 Nm /C )(3: C)(: C) (:0 0 m) = :8 N : 4E The magnitude of the force that either charge exerts on the other is given by F = q q r ; where r is the distance between them. Thus r q q r = F s (8:99 0 = 9 Nm /C )(6:0 0 6 C)(47:0 0 6 C) = :38 m : 5:70 N 5E (a) From Newtons's third law m a = m a, so m = m a a = (6:3 0 7 kg)(7:0 m/s ) 9:0 m/s = 4:9 0 7 kg : (b) Since F = q = r = m a, q = r p m a = (3: 0 3 m) s (6:3 0 7 kg)(7:0 m/s ) 8: Nm =C = 7: 0 C : 6E Use jf AB j = jq A q B j=( d ), etc. (a) Since q A = Q and q C = +8Q, jf AC j = j( Q)(+8Q)j d = 4Q 0 d : After making contact with each other, both A and B now have a charge of [ Q + ( 4Q)]= = 3Q. When B is grounded its charge is zero. After making contact with

4 CHAPTER ELECTRIC CHARGE 63 C, which has a charge of +8Q, B has a charge of [0 + ( 8Q)]= = 4Q, so does charge C. Thus nally Q A = 3Q and Q B = Q C = 4Q. Therefore (b) (c) jf AC j = jf BC j = j( 3Q)( 4Q)j d j( 4Q)( 4Q)j d = 3Q 0 d ; = 4Q 0 d : 7E Let the initial charge on either spheres and be q. After being touched by sphere 3, sphere retains only half of q. After being touched by sphere 3, sphere has (q + q=)= = 3q=4 left. So F 0 = c(q=)(3q=4) = (3=8)cq = (3=8)F; where c is a constant. 8P Since q 3 is in equilibrium F 3 + F 3 = 0, i.e., which gives q = 4q : q q 3 (d) + q q 3 d = 0 ; 9E (a) The magnitude of the force is F = q q d = (8: Nm =C ) (0:0 0 6 C) (:50 m) = :60 N : (b) The magnitude of the force on q F = q F + F 3 + F F 3 cos = F p + cos now becomes = (:60 N) p + cos 60 = :77 N ; where we have used F = F 3 = kq q =d = kq q 3 =d : F F F 3 θ q 0P Put the origin of a coordinate system at the lower left corner of the square. Take the y

5 64 CHAPTER ELECTRIC CHARGE axis to be vertically upward and the x axis to be horizontal. The force exerted by the charge +q on the charge q is y F F = q(q) a ( j) ; -q F 3 x the force by the charge q to q is F F = (q)( q) ( p a) i + j p = q i p + a p j ; and the force by the charge +q to q is F 3 = (q)( q) a ( i) = 4q a i : Thus the horizontal (x) component of the resultant force on the charge F x = F x + F x + F 3x = q p a + 4 = (8: Nm =C ) (:0 0 7 C) (5:0 0 m) q is p + 4 = 0:7 N; while the vertical (y) component is F y = F y + F y + F 3y = q a + p = 0:046 N : P (a) Let the force F on Q be and solve for q : q = 9q. (b) Now which gives q = 5q : F = q Q ( a a=) + q Q = 0 (a a=) F = q Q ( a 3a=) + q Q (a 3a=) = 0 ;

6 CHAPTER ELECTRIC CHARGE 65 P Let the two charges be q and q. Then q + q = 5:0 0 5 C. Also F = q q = r, i.e., :0 N = q q (8: Nm =C ) (:0 m) : Solve for q and q : q, q = : 0 5 C and 3:8 0 5 C. 3P Assume the charge distributions are spherically symmetric so Coulomb's law can be used. Let q and q be the original charges and choose the coordinate system so the force on q is positive if it is repelled by q. Take the distance between the charges to be r. Then the force on q is F a = q q r : The negative sign indicates that the spheres attract each other. After the wire is connected the spheres, being identical, have 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 + q )=. The force is now one of repulsion and is given by F b = (q + q ) 4r : Solve the two force equations simultaneously for q and q. The rst gives and the second gives q q = r F a = (0:500 m) (0:08 N) 8: Nm /C = 3:00 0 C q + q = p s 4(0:0360 N) r 4( )F b = (0:500 m) 8: Nm /C = : C : Thus and Multiply by q q = (3:00 0 C ) q q 3:00 0 C to obtain the quadratic equation q = : C : q (: C)q 3:00 0 C = 0 :

7 66 CHAPTER ELECTRIC CHARGE The solutions are q = : C q (: C) + 4(3:00 0 C ) If the plus sign is used q = :000 6 C and if the minus sign is used q = 3:000 6 C. Use q = ( 3:000 )=q to calculate q. If q = :000 6 C then q = 3:000 6 C and if q = 3: C then q = : C. Since the spheres are identical the solutions are essentially the same: one sphere originally had charge : C and the other had charge 3: C. Another solution exists. If the signs of the charges are reversed the forces remain the same, so a charge of : C on one sphere and a charge of 3: C on the other also satises the conditions of the problem. : 4P Let the third charge q 3 be located on the line joining the two charges, a distance x from q = :0C. Then the net force exerted on q 3 is F 3 = q 3 q x + q = 0 : (r x) Substitute q = +:0C, q = 3:0C and r = 0 cm into the above equation and solve for x to obtain x = 4 cm. 5P (a) F = jq q j r = jq q j (x x ) + (y y ) = (8:99 09 Nm =C )j(3:0 0 6 C)( 4:0 0 6 C)j [( :0 3:5) + (:5 0:5) ](0 4 m ) = 36 N : The direction of F is such that it makes an angle with the positive x axis, where y y = tan :5 cm 0:5 cm = tan = 0 : x x :0 cm 3:5 cm (b) Let the third charge be located at (x 3 ; y 3 ), a distance r from q. Since q, q and q 3 must be on the same line, we have x 3 = x r cos and y 3 = y r sin. The force on q exerted by q 3 is F 3 = q q 3 = r, which must cancel with the force exerted by q on q : F 3 = jq q 3 j r = F ;

8 CHAPTER ELECTRIC CHARGE 67 which gives s r jq q 3 j (4:0 0 6 r = = C)(4:0 0 6 C)(8: Nm =C ) = 6:3 cm : F 36 N So x 3 = x r cos = :0 cm (6:3 cm) cos( 0 ) = 8:3 cm and y 3 = y r sin = :5 cm (6:3 cm) sin( 0 ) = :6 cm: 6P (a) If the system of three charges is to be in equilibrium the force on each charge must be zero. Let the third charge be q 0. It 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 0 could not be in equilibrium. Suppose q 0 is a distance x from q, as shown on the diagram to the right. The force acting on q 0 is then given by F 0 = qq0 4qq 0 = 0 ; x (L x) x L-x q q 0 4q where the positive direction was taken to be toward the right. Solve this equation for x. Canceling common factors yields =x = 4=(L x) and taking the square root yields =x = =(L x). The solution is x = L=3. The force on q is F q = qq0 x + 4q = 0 : L Solve for q 0 : q 0 = 4qx =L = (4=9)q, where x = L=3 was used. The force on 4q is F 4q = 4q L + 4qq 0 = 4q 4( 4=9)q + (L x) L (4=9)L = 4q 4q = 0 : L L With q 0 = (4=9)q and x = L=3 all three charges are in equilibrium. (b) If q 0 moves toward q the force of attraction exerted by q is greater in magnitude than the force of attraction exerted by 4q and q 0 continues to move toward q and away from its initial position. The equilibrium is unstable. 7P (a) The magnitudes of the gravitational and electrical forces must be the same: q d em = G M mm e d em ;

9 68 CHAPTER ELECTRIC CHARGE where q is the charge on either body, d em is the center-to-center separation of the Earth and Moon, G is the universal gravitational constant, M e is the mass of the Earth, and M m is the mass of the Moon. Solve for q: q = p GM m M e : According the text, M e = 5: kg, and M m = 7:36 0 kg, so q = s (6:67 0 Nm /kg )(7:36 0 kg)(5: kg) = 5:7 0 3 C : 8: Nm /C Notice that the distance r cancels because both the electric and gravitational forces are proportional to =r. (b) The charge on a hydrogen ion is e = : C so there must be q e = 5:7 C 03 :6 0 9 C = 3:6 ions : 03 Each ion has a mass of : kg so the total mass needed is (3:6 0 3 )(: kg) = 6:0 0 5 kg : 8P The magnitude of the force of either of the charges on the other is given by F = q(q q) r ; where r is the distance between the charges. You want the value of q that maximizes the function f(q) = q(q q). Set the derivative df=dq equal to zero. This yields q Q = 0, or q = Q=. 9P (a) Let the side length of the square be a. Then the magnitude of the force on each charge Q is F Q = p Qq + Q a ( p : a)

10 CHAPTER ELECTRIC CHARGE 69 Let F Q = 0, we get Q = p q. (b) For both F Q and F q to vanish we would require Q = p q and q = p Q. These two equations cannot be simultaneously valid unless Q = q = 0. So this is impossible. 0P (a) A force diagram for one of the balls is shown to the right. The force of gravity mg acts downward, the electrical force F e of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newtons second law yields T cos mg = 0 and the x component yields T sin F e = 0. Solve the rst equation for T (T = mg= cos ) and substitute the result into the second to obtain mg tan F e = 0. Now F e y θ mg T x tan = x= p L (x=) : If L is much larger than x we may neglect x= in the denominator and write tan x=l. This is equivalent to approximating tan by sin. The magnitude of the electrical force of one ball on the other is F e = q x : When these two substitutions are made in the equation mg tan = F e, it becomes so and mgx L = q x ; x 3 = q L x = 0 mg q L mg ; =3 (b) Solve x 3 = (= )(q L=mg) for q: r s 40 mgx q = 3 (0:00 kg)(9:8 m/s )(0:050 m) = 3 L (8: Nm /C )(:0 m) = :4 0 8 C : : P If one of them is discharged, there would no electrostatic repulsion between the two balls and they would both come to the position = 0, making contact with each other. A

11 630 CHAPTER ELECTRIC CHARGE redistribution of charges would then occur, with each of the balls getting q=. They would then again be separated due to electrostatic repulsion, which results in the new separation x 0 (q=) L = 0 mg =3 = =3 x = 4 =3 (5:0 cm) = 3: cm : 4 P (a) Since the rod is in equilibrium the net force acting on it is zero and the net torque about any point is also zero. Write an expression for the net torque about the bearing, equate it to zero, and solve for x. The charge Q on the left exerts an upward force of magnitude (= )(qq=h ), at a distance L= from the bearing. Take the torque to be positive. The attached weight exerts a downward force of magnitude W, at a distance L= x from the bearing. This torque is positive. The charge Q on the right exerts an upward force of magnitude (= )(qq=h ), at a distance L= from the bearing. This torque is negative. The equation for rotational equilibrium is The solution for x is qq h L + W x = L L x + qq h W qq h : L = 0 : (b) The net force on the rod vanishes. If N is the magnitude of the upward force exerted by the bearing then qq qq W N = 0 : h h Solve for h so that N = 0. The result is h = r 3qQ W : 3E The magnitude of the force is F = e r = (8:99 09 Nm =C ) (: C) (:8 0 0 m) = : N : 4E F = q r = ( e=3) r = (8:99 09 Nm =C )(: C) 9(:6 0 5 m) = 3:8 N :

12 CHAPTER ELECTRIC CHARGE 63 5E The mass of an electron is m e = 9: 0 3 kg so the number of electrons in a collection with total mass M = 75:0 kg is N = M m e = 75:0 kg 9: 0 3 kg = 8:3 03 electrons : The total charge of the collection is q = Ne = (8:3 0 3 )(: C) = :3 0 3 C : 6E Q = N A q = (6:0 0 3 )()(: C) = :9 0 5 C = 0:9 MC : 7E (a) The magnitude of the force between the ions is given by F = q r ; where q is the charge on either of them and r is the distance between them. Solve for the charge: q = p s 3:7 0 r F = (5:0 0 0 m) 9 N 8: Nm /C = 3: 0 9 C : (b) Let N be the number of electrons missing from each ion. Then Ne = q, or N = q e = 3: 0 9 C : C = : 8E (a) the number of electrons to be removed is (b) The fraction is frac = n = q e = :0 0 7 C : C = 6:3 0 : n NZ = 6:3 0 (:95 0 )(9) = 7:4 0 3 :

13 63 CHAPTER ELECTRIC CHARGE 9E (a) The magnitude of the force is given by F = q r ; where q is the magnitude of the charge on each drop and r is the center-to-center separation of the drops. Thus F = (8:99 09 Nm /C )(: C) (:00 0 m) = 8: N : (b) If N is the number of excess electrons (of charge e each) on each drop then N = q e = : C : C = 65 : 30E Let q = r = mg, we get r s r = q mg = (:60 0 8: C) 9 Nm =C (: kg)(9:8 m/s ) = 0:9 m : 3E The second electron must be placed a distance d under the rst one to produce an upward force that balances the weight of the rst one. So W = m e g = e d ; which gives r s d = e m e g = (:6 0 8: C) 9 Nm =C (9: 0 3 kg)(9:80 m/s ) = 5: m : 3P The current intercepted would be q i = (4R ) = 4(6: m) (500= s)(:6 0 9 C) = 0: A : t

14 CHAPTER ELECTRIC CHARGE P If charge of magnitude q passes through the lamp in time t the current is i = q=t. The charge is the total charge on one mole of electrons, or q = N A e, where N A is the Avogadro constant. Thus i = N A e=t, or t = N Ae i = (6:0 03 )(: C) 0:83 A = : 0 5 s : This is equivalent to :3 da. 34P The numer of moles of H O molecules in a glass of water (of volume V ) is n = V=M, where is the density of water and M is its molar mass. Each H O molecule has 0 protons. Thus Q = 0enN A = 0eV N A M = 0(:6 0 9 C)(: kg/m 3 )( m 3 )(6:0 0 3 = mol) :8 0 kg/mol = :3 0 7 C : 35P (a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair exerts forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. (b) Rather than remove a cesium ion, superpose charge e at the position of one cesium ion. This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion. The forces of the 8 cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge. The length of a body diagonal of a cube is p 3a, where a is the length of a cube edge. Thus the distance from the center of the cube to a corner is d = ( p 3=)a. The force has magnitude F = e d = e (3=4)a = (8:99 09 Nm /C )(: C) (3=4)(0: m) = :9 0 9 N :

15 634 CHAPTER ELECTRIC CHARGE Since both the added charge and the chlorine ion are negative the force is one of repulsion. The chlorine ion is pushed away from the site of the missing cesium ion. 36P In this case the net charge on each penny would be q net = 0:0000%. The force is then = q, where q = C and F = q net r = [(0:0000%)(37000 (8:99 C)] 0 9 Nm =C ) = :7 0 8 N : (:0 m) Therefore the magnitudes of the positive charge on the proton and negative charge on the electron cannot possibly dier by as much as 0:0000%. 37P The net charge carried by John whose mass is m is roughly q = (0:0%) mn AZ e M = (0:0%) (00 lb)(0:4536 kg/lb)(6:0 03 = mol)(8)(:6 0 9 C) 0:08 kg/mol = C ; and the net charge carried by Mary is half of that. So the electrostatic force between them is estimated to be F q(q=) d = (8: Nm =C ) (9 05 C) (30 m) = N : 38E Apply the principle of charge conservation. The results are: (a) an electron; (b) a proton. 39E None of the reactions given include a beta decay, so the number of protons, the number of neutrons, and the number of electrons are each conserved. Atomic numbers (numbers of protons and numbers of electrons) and molar masses (combined numbers of protons and neutrons) can be found in Appendix F of the text. (a) H has proton, electron, and 0 neutrons and 9 Be has 4 protons, 4 electrons, and 9 4 = 5 neutrons, so X has + 4 = 5 protons, + 4 = 5 electrons, and = 4 neutrons. One of the neutrons is freed in the reaction. X must be boron with a molar mass of = 9 g= mol: 9 B.

16 CHAPTER ELECTRIC CHARGE 635 (b) C has 6 protons, 6 electrons, and 6 = 6 neutrons and H has proton, electron, and 0 neutrons, so X has 6 + = 7 protons, 6 + = 7 electrons, and = 6 neutrons. It must be nitrogen with a molar mass of = 3 g=mol: 3 N. (c) 5 N has 7 protons, 7 electrons, and 5 7 = 8 neutrons; H has proton, electron, and 0 neutrons; and 4 He has protons, electrons, and 4 = neutrons; so X has 7 + = 6 protons, 6 electrons, and = 6 neutrons. It must be carbon with a molar mass of = : C. 40E (a) (b) F = q r = (8:99 09 Nm =C )[(: C)] (9:0 0 5 m) = N : a = F m = N 4(: kg) = :7 07 m/s : 4 (a) F = (Q = d )( (c) 0.5; (d) 0.5 and 0.85 ); 4 (a) Let J = qq= d. For < 0, F = J[= + =( + jj) ]; for 0 < <, F = J[= =( ) ]; for <, F = J[= + =( ) ]