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Solutions and Tests for Exploring Creation With Physics SOLUTIONS TO THE PRACTICE PROBLEMS T'ORMODULE #13 1. This is a simple application of Equation (13.1). We have the two charges and the distance between them, so we just plug those values into the equation to calculate the force: r-:, kq,q, 1 t" (9.0 x 10e Newton .m2 e' ) . (6.7 e) . (3.. 1 €) - 1.3 x 101r Newtons (r.2 n)'z Since one charge is negative and the other positive, this force is attractive. Thus, the charges exert 1.3 x 101l-Newton force towards each other. a and the force and we need to determine the distance between the charges. Equation (13.1) relates these variables. The only thing we have to watch is units. The charges are in ffiC, but "k" uses C. Thus, we must convert mC to C. 2.Inthis problem, we know the charges kq,q, ) rNewton .m2 (9.0 x 10e 2.5 x105 Newtons (s.0, ro, C2 - . ) (3 .2x 10-a C) . (5.5 x 10-' C) (r)' ryd) . (:.2, ro-oe) .(s.s * ro-o e) = 0.080 m 2.5 x 105 l$erntons The distance between the objects, then, is 0.080 m. are given everything but the charge of the two objects. You might thi* at fus that there's no way to solve the problem, because there are two things you don't know: both charges. However, the problem tells us that they are the same. So we can call them each "q." That will give ur 3. In this problem, we only one variable: kqrq, r2 (9.0 x 10e 2.2x105 Newtons = Newton .m2 C2 (0.45 m)'z )'q'q Ptactice Problem Solutions q- 2.2 x1 6s N"*ens . (0.45 m)2 - 2.2x 10-3 C 9.0 x 10e charge has a magnitude of 2.2 mC. taneous electrostatic force on the -3 .4-mC e is zero because the forces exerted of the other ch cel out. The other charges are equal in magnitude and the same distance from the charge interest. of They each exert an attractive force, pulling the charge of interest in 1ite directions. Thus, equal and opposite forces work on the -3.4-mC charge, making the net force At the same time, however, this is not a static system. Even though the net force on the -3.4-mC is zero, the net forces acting on the other two are not. The -3 .4-mC charge attracts each &e other two charges. The other two charges repel each other, but they are farther away from each than the -3 .4-mC charge is from them. As a result, the net force acting on each of the positive ges is NOT zero. Thus, the other two charges will move towards the -3 .4-mC charge. e* In this problem, we are only worried about the +6.4-mC charge. As a result, we only consider the which act on thatparticular charge. The +1 .2-mC charge exerts a repulsive force whose itude is: o kq,q, - r' 1 (9.0 x l0e F- Newtons .m2 ).(6.4x e2 10-3 e).(t .2x10-'e) (1.5 m)2 : 3. 1 x 104 Newtons The other force acting on the +6.4-mC charge is the repulsive force exertedby the o - kq,q, ) r- (9.0 x 10e F_ +9.l-mC charge. Newtons .m' ).(6.4x e2 10-3 q . (9.1x 10-3 e) - (2.0 m)2 1.3 x 105 Newtons The fact that both forces are repulsive will give us the directions ofthe force vectors, making our force diagram look like this: 1.3 x 105 N 3.1 x 104 N Since the force vectors both point in the same dimension, we car treat them as one-dimensional rectors. This means we can take care of direction with positives and negatives and then simply add the 91 92 Solutions and Tests for Explodng Creation $7ith Physics magnitudes together. Using the convention that vectors pointing to the left are negative, the total force is: Ftotr -1.3x 10sNewtons + 3.1x 1O4Newtons = -1.0x 105Newtons : A negative force means that tle final vector points to the left. Thus, the final instantaneous electrostatic force is 1.0 x 10) Newtons to the left. go through all of the trouble of adding the electric fields from each of the remaining charges together, or you could re alize that since you already calculated the force experienced by a charge placed at that point, you can use that information to calculate the electric field using Equation b. You could (13.2): E- F -1.0x10s Newtons eo 6.4x10-3 C - -1.6x112 Newtons C Notice what I did here. Equation (13.2) tells you that the electric field is the force experienced by a charged particle divided by that particle's charge. Thus, I took the force experienced by the 6-4-mC charge as calculated previously, and I divided by 6.4 mC (converted to C)' The result, then, is the electric field at that point. Since both "E" and "F" are vectors in the equation, I kep the negative sien for "F," since it denotes that the force is pointed to the left. Thus, the electric field is l'6 x 10' Newtons/Coulomb to the left. Since we are only interested in the - 1 .2-mC charge, we only need concem ourselves with forces which act on that charge. The +1.2-mC charge exerts an attractive force on it: 6. rr kq,q, l-) r' (9.0 x 10e Newtons .m2 e2 F_ ).(t.2x 10-3 e).(1 .2x10-',€) - 8. 1x104 Newtons (0.40 m)2 The +2.4-mC charge also exerts an attractive force on it: rr r= kq,q, , r- (9.0 x 10e F_ Newtons .m2 .(1.2x 10-3 q-Q.4x 10-3 e) ) e2 (0.20 m)'z Our force diagram, then, looks like this: 6.5 x 10s N 8.1 x 104 N - 6.5x10s Newtons Practice Problem lotal fo Solutions rttre vectors do not point in the same dimension, we will have to add them with trigonometry. utl*'ever, let's define the angles properly. Vector angles are always defined counterclockwise &e pnsitive x-axis. This means that the first angle is 341o, and the angle for the vector on the left f' \ou' we can add these vectors: lmg )ya uation A*: (8.1x 10a Newtons).cos(341o) : : (8.1 x lOa Newtons) . sin(341o) : - 2.6x 104 Newtons B* = (6.5 x 10s Newtons).cos(255") - - 1.7 x10s Newtons A, B, bya -mC : (6.5 x 105 Newtons) .sin(255o ) = 7.7 x10a Newtons 6.3x105 Newtons - C*: A^* Br: 7.7 Cy: A, + By: -2.6 x lOaNewtons + -6.3 x 105 Newtons x 10a Newtons + -I.7 x 105 Newtons: -9 x 1e 'e sign sf Newtons -6.6 x 105 Newtons ::.at's left to do now is convert these x- and y-components into vector magnitude and direction: il ES : 104 naagnitude 0= - C*'+C u' = Jf-g x 104 Newtons)2 +(-6.6x 105 Newtons)' - 6.7 x 105 Newtons ,/C.),/-e.e"totxe.r*e*s),-^ tan-'l -l l= tan 'l --------------l= 80" (-9xl0.Nev*ons/ l.C, l - S::ce both the x and y components ofthe vector are negative, we know that the vector is in quadrant This means that we need to add 180o to the angle above to properly defrne the vector angle. The electrostatic force on the -1.2 mC charge, then, is 6.7 x 10s Newtons at an anele of 260". I. =-<lantaneous -. \\hen drawing electric field lines, the lines go out ofpositive charges and into negative charges. ijs case, the negative charge is twice as large as the positive charge, so it has twice as many lines _:!rmg into it as the positive charge has going out of it. I will draw 6 lines going out of the positive :harge and into the negative charge. However, if 6 lines go out of the positive charge, go into the negative charge, because it has twice the charge. =ust xt additional In 6 93 94 Solutions and Tests for Exploring creation with physics . That,s where the field lines are the most dense. 8' To calculate the electric field, I will have to sum up the fields_created by each stationary charge. are asked to calculate the field at the midpoint between the charges, which is 1.25 m from each charge' Let',s start with the positive charge. ,,81.,, we will we call its electric field E Lr -kQ --;-= (9.0 x 10e Newton .m2 C2 f' ) .(1.0 e) - 5.g x l oe (1.25 m)2 Newtons C Now rememb er, electric field,lines alwayspoint directly awayfrom positive charges. Thus, this field is pointed directly to your tigt t. Now let's calculate the electric fielJ from the negative :lt# (9.0 x 10'g E2 Igyton'm2 )'(2.0 e) Ct F =*= t- - l.2x 1010 Newtons Since electric field lines always point towards negative charges, this electric field points directly to your right as well. Thus, the two electric fields look like this: 5.9 x lOe N/C o-+ 1.2 > x loto N/C since the two vectors arcpatallel, we don't need to use trigonometry to add them. we can just add them directly: E,r,r, Newtons =E, *Ez =5.8 x 1g' IYtons + l.2x 10,, Newtons 1.8 x 10'o = C: The electric field, then, is . Note that this answer with the qualitative diagram of the electriCn.t,o trrut *. d..*ln thf previous problem. The diagram has the electric field i, b.t*.en the two charges pointed to the right, as does this result. agrees To calculate the force experienced by aparticle, we could use Equation (13.1), but that won : be necessary' we already know the electric-field at thepoint directly inietween the two charges. Thus, we can quickly calculate the force using Equation (r3.2): E-I 9o F = eoE - (-6.6x 10-a e) . I .g x l0ro Newtons - -l .2x10' Newtons Practice Problem Solutions negative sign means that the force is pointed opposite the electric field. Thus, since the electric i is pointed to the right, the charge experiences a 1.2 x l07-Newton force that is pointed to the left. q Srnce all of the stationary charges have equal magnitudes, they all have the same number of lines umcln,e out of them. These lines bend away from each other because they cannot cross. ,\I the very center ofthe triangle, the magnitude ofthe electric field is zero, because the three particle's ields cancel out at that point. The ion in this case has two protons, so its nucleus has a positive charg e of 3.2 x l0-te C ltwice the :harge of a single proton). Since there is only one electron in the ion, however, we can solve this loblem by setting the centripetal force equal to the electrostatic force. - r-t. mv2 _kqrq2 r12 kq,qrr kq,q, -.2 \/v-1 r-m v = 3.lxl06 (9.0 x 10e rtn m sec This electron has a speed of 3.1 x 106 m/sec. Newton.m2 .(3 .2x10-" e) . (1.6 x l0-le e) ) e2 (5 .29 x 10-1'*) . (9. I x 10-3' kg; 95