Physics 107 TUTORIAL ASSIGNMENT #8

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1 Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type of poblem. Two small objects, A and, ae fixed in place and sepaated by 3.00 cm in a vacuum. Object A has a chage of +.00 µc, and object has a chage of.00 µc. How many electons must be emoved fom A and put onto to make the electostatic foce that acts on each object an attactive foce whose magnitude is 68.0 N? * An electically neutal model aiplane is flying in a hoizontal cicle on a 3.0-m guideline, which is nealy paallel to the gound. The line beaks when the kinetic enegy of the plane is 50.0 J. Reconside the same situation, except that now thee is a point chage of + on the plane and a point chage of at the othe end of the guideline. In this case, the line beaks when the kinetic enegy of the plane is 5.8 J. Find the magnitude of the chages. **3 ssm A small spheical insulato of mass 8.00 x 0 - kg and chage µc is hung by a thin wie of negligible mass. A chage of µc is held 0.50 m away fom the sphee and diectly to the ight of it, so the wie makes an angle θ with the vetical (see the dawing). Find (a) the angle θ and (b) the tension in the wie. *39 ssm A ectangle has a length of d and a height of d. Each of the following thee chages is located at a cone of the ectangle: + (uppe left cone), + (lowe ight cone), and (lowe left cone). The net electic field at the (empty) uppe ight cone is zeo. Find the magnitudes of and. Expess you an swes in tems of. 76 Concept Questions The dawing shows a positive point chage +, a second point chage that may be positive o negative, and a spot labeled P, all on the same staight line. The distance d between the two chages is the same as the distance between and the point P. With pesent, the magnitude of the net electic field at P is twice what it is when is pesent alone. (a) When the second chage is positive, is its magnitude smalle than, eual to, o geate than the magnitude of? Explain you easoning. (b) When the second chage is negative, is its magnitude smalle than, eual to, o geate than that in uestion (a)? Account fo you answe. Poblem Given that, detemine when it is (a) positive and (b) negative. Veify that you answes ae consistent with you answes to the Concept Questions.

2 Chapte 9 *9 ssm The potential at location A is 45 V. A positively chaged paticle is eleased thee fom est and aives at location with a speed v. The potential at location C is 79 V, and when eleased fom est fom this spot, the paticle aives at with twice the speed it peviously had, o v. Find the potential at. **0 A paticle is unchaged and is thown vetically upwad fom gound level with a speed of 5.0 m/s. As a esult, it attains a maximum height h. The paticle is then given a positive chage + and eaches the same maximum height h when thown vetically upwad with a speed of 30.0 m/s. The electic potential at the height h exceeds the electic potential at gound level. Finally, the paticle is given a negative chage. Ignoing ai esistance, detemine the speed with which the negatively chaged paticle must be thown vetically upwad, so that it attains exactly the maximum height h. In all thee situations, be sue to include the effect of gavity. **4 A positive chage + is located to the left of a negative chage. On a line passing though the two chages, thee ae two places whee the total potential is zeo. The fist place is between the chages and is 4.00 cm to the left of the negative chage. The second place is 7.00 cm to the ight of the negative chage. (a) What is the distance between the chages? (b) Find /, the atio of the magnitudes of the chages. **5 Chages and ae fixed in place, being located at a distance d to the ight of. A thid chage 3 is then fixed to the line joining and at a distance d to the ight of. The thid chage is chosen so the potential enegy of the goup is zeo; that is, the potential enegy has the same value as that of the thee chages when they ae widely sepaated. Detemine the value fo 3, assuming that (a) and (b) and. Expess you answes in tems of. **6 One paticle has a mass of 3.00 x 0-3 kg and a chage of µc. A second paticle has a mass of 6.00 x 0-3 kg and the same chage. The two paticles ae initially held in place and then eleased. The paticles fly apat, and when the sepaation between them is 0.00 m, the speed of the 3.00 x 0-3 kg paticle is 5 m/s. Find the initial sepaation between the paticles.

3 SOLUTIONS Chapte 8 5. REASONING The electons tansfeed incease the magnitudes of the positive and negative chages fom.00 µc to a geate value. We can calculate the numbe N of electons by dividing the change in the magnitude of the chages by the magnitude e of the chage on an electon. The geate chage that exists afte the tansfe can be obtained fom Coulomb s law and the value given fo the magnitude of the electostatic foce. SOLUTION The numbe N of electons tansfeed is N afte e befoe whee afte and befoe ae the magnitudes of the chages afte and befoe the tansfe of electons occus. To obtain electostatic foce: afte, we apply Coulomb s law with a value of 68.0 N fo the afte F k o afte Using this esult in the expession fo N, we find that F k N ( 68.0 N)( m) F C befoe 9 k N m / C e C. REASONING This is a poblem that deals with motion in a cicle of adius. As Chapte 5 discusses, a centipetal foce acts on the plane to keep it on its cicula path. The centipetal foce F c is the name given to the net foce that acts on the plane in the adial diection and points towad the cente of the cicle. When thee ae no electic chages pesent, only the tension in the guideline supplies this foce, and it has a value T max at the moment the line beaks. Howeve, when thee is a chage of + on the plane and a chage of on the guideline at the cente of the cicle, thee ae two contibutions to the centipetal foce. One is the electostatic foce of attaction between the chages and, since the chages have the same magnitude, its magnitude F is given by Coulomb s law (Euation 8.) as F k /. The othe is the tension T max, which is chaacteistic of the ope and has the same value as when no chages ae pesent. Whethe o not chages ae pesent, the centipetal foce is eual to the mass m times the centipetal acceleation, accoding to Newton s second law and stated in Euation 5.3, F c mv /. In this expession v is the

4 speed of the plane. Since we ae given infomation about the plane s kinetic enegy, we will use the definition of kinetic enegy, which is KE mv /, accoding to Euation 6.. SOLUTION Fom the definition of kinetic enegy, we see that mv (KE), so that Euation 5.3 fo the centipetal foce becomes F c mv KE Applying this esult to the situations with and without the chages, we get Centipetal foce max Centipetal foce k KE KE chaged unchaged Tmax + T Subtacting Euation () fom Euation () eliminates T max and gives Solving fo gives k KE KE chaged unchaged KE KE chaged unchaged 3.0 m 5.8 J 50.0 J C k N m / C 3. SSM REASONING The chaged insulato expeiences an electic foce due to the pesence of the chaged sphee shown in the dawing in the text. The foces acting on the insulato ae the downwad foce of gavity (i.e., its weight, W mg), the electostatic foce / F k (see Coulomb's law, Euation 8.) pulling to the ight, and the tension T in the wie pulling up and to the left at an angle θ with espect to the vetical as shown in the dawing in the poblem statement. We can analyze the foces to detemine the desied uantities θ and T. SOLUTION. a. We can see fom the diagam given with the poblem statement that and Tx Ty F which gives T sin θ k / W which gives T cosθ mg

5 Dividing the fist euation by the second yields Solving fo θ, we find that k θ tan mg T sinθ k tan / θ T cosθ mg ( N m /C )( C)( C) tan 5.4 ( kg)(9.80 m/s )(0.50 m) b. Since T cosθ mg, the tension can be obtained as follows: T mg ( kg) (9.80 m/s ) cosθ cos N 39. SSM WWW REASONING The dawing shows the aangement of the thee chages. Let E epesent the electic field at the empty cone due to the chage. Futhemoe, let E and E be the electic fields at the empty cone due to chages + and +, espectively. + E E E d 5 d θ + d Accoding to the Pythagoean theoem, the distance fom the chage to the empty cone along the diagonal is given by (d ) + d 5d d 5. The magnitude of each electic field is given by Euation 8.3, electic fields at the empty cone ae given as follows: E k /. Thus, the magnitudes of each of the E k k k 5d ( d 5 ) E k k k and E ( d ) 4d d

6 The angle θ that the diagonal makes with the hoizontal is θ tan ( d / d) Since the net electic field E net at the empty cone is zeo, the hoizontal component of the net field must be zeo, and we have k k cos 6.57 E E cos o 0 4d 5d Similaly, the vetical component of the net field must be zeo, and we have E k k sin 6.57 sin o 0 d 5d E These last two expessions can be solved fo the chage magnitudes and. SOLUTION Solving the last two expessions fo and, we find that 4 5 cos sin CONCEPT QUESTIONS a. The dawing at the ight shows the electic fields at point P due to the two + P E + chages in the case that the second d d E chage is positive. The pesence of the second chage causes the magnitude of the net field at P to be twice as geat as it is when only the fist chage is pesent. Since both fields have the same diection, the magnitude of E must, then, be the same as the magnitude of E. ut the second chage is futhe away fom point P than is the fist chage, and moe distant chages ceate weake fields. To offset the weakness that comes fom the geate distance, the second chage must have a geate magnitude than that of the fist chage. b. The dawing at the ight shows the + electic fields at point P due to the two E P E chages in the case that the second d d chage is negative. The pesence of the second chage causes the magnitude of the net field at P to be twice as geat as it is when only the fist chage is pesent. Since the fields now have opposite diections, the magnitude of E must be geate than the magnitude of E. This is necessay so that E can offset E and still lead to a net field with twice the magnitude as E. To ceate this geate field E, the second chage must now have a geate magnitude than it did in uestion (a).

7 SOLUTION a. The magnitudes of the field contibutions of each chage ae given accoding to k Euation 8.3 as E. With pesent, the magnitude of the net field at P is twice what it is when only is pesent. Using Euation 8.3, we can expess this fact as follows: k k k k k + o ( d ) ( d ) d d d Solving fo gives µ C.0 µ C Thus, the second chage is +.0 µc, which is consistent with ou answe to Concept Question (a). b. Now that the second chage is negative, we have k k k k k o 3 ( d ) ( d ) d d d Solving fo gives 0.50 µ C 6.0 µ C Thus, the second chage is 6.0 µc, which is consistent with ou answe to Concept Question (b). Chapte 9 9. SSM WWW REASONING The only foce acting on the moving chage is the consevative electic foce. Theefoe, the total enegy of the chage emains constant. Applying the pinciple of consevation of enegy between locations A and, we obtain mva + EPE A mv + EPE Since the chaged paticle stats fom est, v A 0. The diffeence in potential enegies is elated to the diffeence in potentials by Euation 9.4, EPE EPE A ( V VA ). Thus, we have A ( V V ) mv ()

8 Similaly, applying the consevation of enegy between locations C and gives Dividing Euation () by Euation () yields C ( V V ) m ( v ) () This expession can be solved fo V. VA V V V 4 C SOLUTION Solving fo V, we find that 4 VA VC 4(45 V) 79 V V V 0. REASONING The gavitational and electic foces ae consevative foces, so the total enegy of the paticle emains constant as it moves fom point A to point. Recall fom Euation 6.5 that the gavitational potential enegy (GPE) of a paticle of mass m is GPE mgh, whee h is the height of the paticle above the eath s suface. The consevation of enegy is witten as mv EPE EPE A mgha A mv mgh () We will use this euation seveal times to detemine the initial speed v A of the negatively chaged paticle. SOLUTION When the negatively chaged paticle is thown upwad, it attains a maximum height of h. Fo this paticle we have: v A? EPE A ( ) V A h A 0 (gound level) v 0 (at maximum height) EPE ( )V h h (the maximum height) Solving the consevation of enegy euation, Euation (), fo v A and substituting in the data above gives va mgh ( V VA ) m + () Euation () cannot be solved as it stands because the height h and the potential diffeence (V V A ) ae not known. We now make use of the fact that a positively chaged paticle,

9 when thown staight upwad with an initial speed of 30.0 m/s, also eaches the maximum height h. Fo this paticle we have: v A 30.0 m/s EPE A (+) V A h A 0 (gound level) v 0 (at maximum height) EPE (+)V h h Solving the consevation of enegy euation, Euation (), fo the potential diffeence (V V A ) and substituting in the data above gives V ( 30.0 m/s) V A m mgh + (3) Substituting Euation (3) into Euation () gives, afte some algebaic simplifications, va 4gh 30.0 m/s (4) Euation (4) cannot be solved because the height h is still unknown. We now make use of the fact that the unchaged paticle, when thown staight upwad with an initial speed of 5.0 m/s, also eaches the maximum height h. Fo this paticle we have: v A 5.0 m/s v 0 (at maximum height) EPE A V A 0 (since 0) EPE V 0 (since 0) h A 0 (gound level) h A h Solving Euation () with this data fo the maximum height h yields ( 5.0 m/s) ( 5.0 m/s) h g 9.80 m/s 3.9 m Substituting h 3.9 m into Euation (4) gives v A 8. 7 m / s. 4. REASONING AND SOLUTION a. Let d be the distance between the chages. The potential at the point x 4.00 cm to the left of the negative chage is k k V 0 d x x which gives d () x

10 Similaly, at the point x 7.00 cm to the ight of the negative chage we have which gives V k k 0 x + d x d + () x Euating Euations () and () and solving fo d gives d 0.87 m. b. Using the above value fo d in Euation () yields REASONING AND SOLUTION The electical potential enegy of the goup of chages is so a. If, then b. If and then EPE k /d + k 3 /(d) + k 3 /d 0 + (/) (/) o (/) o 3 6. REASONING The only foce acting on each paticle is the consevative electic foce. Theefoe, the total enegy (kinetic enegy plus electic potential enegy) is conseved as the paticles move apat. In addition, the net extenal foce acting on the system of two paticles is zeo (the electic foce that each paticle exets on the othe is an intenal foce). Thus, the total linea momentum of the system is also conseved. We will use the consevation of enegy and the consevation of linea momentum to find the initial sepaation of the paticles. SOLUTION Fo two points, A and, along the motion, the consevation of enegy is k mv, A + m v, A + m v, + m v, + A Initial kinetic enegy of the two paticles Initial electic potential enegy Final kinetic enegy of the two paticles k Final electic potential enegy Solving this euation fo / A and setting v,a v,a 0 since the paticles ae initially at est, we obtain

11 ( mv, mv, ) + + k () A This euation cannot be solved fo the initial sepaation A, because the final speed v, of the second paticle is not known. To find this speed, we will use the consevation of linea momentum: m v, A + m v, A m v, + m v, Initial linea momentum Setting v,a v,a 0 and solving fo v, gives v m 3, v, m 3 Final linea momentum kg ( 5 m/s ) 6.5 m/s kg Substituting this value fo v, into Euation () yields m A 9 6 ( N m / C )( C) 3 3 ( kg)( 5 m/s) ( kg)( 6.5 m/s) + A.4 0 m

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