Prep 1. Oregon State University PH 213 Spring Term Suggested finish date: Monday, April 9

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1 Oregon State University PH 213 Spring Term 2018 Prep 1 Suggeste finish ate: Monay, April 9 The formats (type, length, scope) of these Prep problems have been purposely create to closely parallel those of a typical exam (inee, these problems have been taken from past exams). To get an iea of how best to approach various problem types (there are three basic types), refer to these sample problems.

2 1. a. Evaluate each statement (T/F/N). As always, you must fully explain your reasoning an answers. (i) An electrically charge object can exert a net electrical force on an electrically neutral object. True. The charge object can polarize the neutral object an thereby exert a net attraction force on the neutral object, because it will be more strongly attracte to its nearer sie than it will be repelle by the farther sie. (ii) An electrically neutral object can exert an electrical force on another electrically neutral object. True. If the first object is polarize, it will o likewise to the secon object an the nearer ens will be more attracte than the farther ens will be repelle. (iii) An electrically neutral insulator can be polarize. True. If a charge object is brought near to a neutral insulator, the iniviual molecules of the insulator material will polarize in place become slight ipoles, that then orient themselves with the opposite charge sign facing the nearby charge object. The macroscopic effect of all these molecular ipoles is to create a cumulative ipole over the entire object. (iv) If you place a charge object near to (but not touching) an unknown object, an there is a repelling force exerte by each on the other, the unknown object coul be electrically neutral. Depens on your assumptions (so, N). If a charge object is hel near a neutral object, the neutral object becomes polarize, so that the nearer en is of the opposite charge which always attracts to the charge object (e.g. static cling ). So, if you mae the assumption that there are only two objects to consier, then repulsion coul not occur. However, if there were a thir object, also charge (same sign as the other) but more strongly, brought near the opposite en of the neutral object, that object s greater charge woul govern the polarization of the neutral object, so the neutral object woul have to present its like-charge polarize en towar the weaker charge object, thus causing repulsion between those two. (v) A proton exerts a positively irecte force on another proton. Not enough information. It all epens on the choice of coorinate axes. For example, if proton 1 is locate at the origin an proton 2 is locate along the positive x-, y- or z- axis, then F E.12 is inee in a positive axial irection, an the statement is true. But if proton 2 is locate along the negative x-, y- or z- axis (or at any point not on a coorinate axis), then F E.12 is not in a positive irection. The positive/negative esignations of the axial irections have nothing to o with the positive/negative esignations of charge. (vi) A proton exerts a positively irecte force on an electron. Not enough information. The reasoning is similar to item (v), above. b. X an Y are two uncharge metal spheres sitting at rest on ajacent insulating stans. The spheres are initially in contact with each other. A positively charge ro, R, is then brought close to X (without touching it). Sphere Y is then move away from X. Then R is move far away from both X an Y. What are the final net charges (positive, negative or neutral) of X an Y? As always, explain your answers use rawings as neee to help you reason an explain. The two spheres initially in contact are, electrically speaking, one object, XY. When R is brought close to the X en, this polarizes XY, so that the X en is negatively charge, while the Y en is positively charge. Then separating the two spheres strans those net charges, respectively, so that X now carries a net negative charge; Y carries a net positive charge. 2

3 2. a. Explain briefly but fully: How oes an object usually acquire a net positive charge? It loses electrons, which are either scrape off via friction (by a surface with a higher charge affinity) or chase off via polarization. b. If a metal object becomes positively charge, oes the mass of the object increase, ecrease, or stay the same? Explain briefly but fully. Its mass ecreases, because it has lost electrons which are loosely boun, especially in the outer valence orbitals. (The object has not gaine protons, which are tightly boun in nuclei). c. Explain briefly but fully: If a conuctive object carries a net charge, how is the excess charge istribute on the object an why is this so? Use one or more sketches if necessary. The excess charge resies on the outsie of the object, because these charges are free to move anywhere (this is a conuctor), an they repel one another. Thus, arraye along the outsie of the object is, by simple geometry, the farthest they can get from one another.. Explain briefly but fully: How can an electrically neutral object be attracte to a charge object? Use one or more sketches if necessary. The neutral object is polarize see item 1a(i), above. e. Explain briefly but fully: How can an initially neutral object acquire a positive charge without touching any charge object? Use one or more sketches if necessary. This woul be the case of sphere Y in item 1b, above. (Y touche only X, which was neutral.) f. How many electrons woul you nee in orer to have 1.50 nc of negative charge? As always, show all work, reasoning, an calculations. ( 1.50 x 10-9 C)/( 1.60 x C/electron) = 9.38 x 10 9 electrons 3

4 3. a. Evaluate the following statement (T/F/N), an as always, fully explain your reasoning: If you reuce by half the istance between two point charges, you ouble the magnitue of electrical force that each exerts on the other. b. Two point charges were previously 18.0 cm apart. Then they were move so that the force on each has become three times what it was before. How far apart are the charges now? c. A point charge (q 1 = +2e) is locate at (0, 0). Another point charge (q 2 = 3e) is locate at ( 2,0). A thir point charge (q 3 = 6e) is locate at (3,0). Fin the irection of the total electrical force exerte on q 1.. Let q > 0 an > 0. A particle with charge +q is stationary an locate at the point (0, ). A particle with charge q is stationary an locate at the point (0, ). In which irection woul an electron accelerate if it were locate at (L, 0), where L > 0? e. Evaluate (T/F/N) the following statement. Justify your answer fully with any vali mix of wors, rawings an calculations. Three protons are locate in the x-y plane: q A is at (0, 0); q B is at (1, 0); q c is at (0, ). If the net force on q A is in the 150 irection (angle measure conventionally), then f. A point charge, q 1, is locate at the origin, (0, 0). It exerts a force of known magnitue F 12 on an electron, q 2, locate at the point (4,0). Write an expression for the magnitue F 13 of the force that q 1 exerts on another charge, q 3 = +2e, locate at (0, 2). g. A point charge is locate at the origin. It exerts a force of known magnitue F 12 on an electron (q 2 ) locate at the point (4,0). Write an expression for the magnitue F 13 of the force it exerts on a charge q 3 = 3e, locate at (0,1). 4

5 4. a. Four ientical particles of charge +Q are equally space along a horizontal line. The istance between ajacent charges is. What is the magnitue of the electric force on the charge on the far left en? b. Three point charges, q 1, q 2 an q 3, are locate along the x-axis, with the same istance between q 1 an q 2 as between q 2 an q 3. q 2 = 3.40 nc. Fin the value of q 1 that will put q 3 into static equilibrium. c. Three charge particles are positione along a line. Particle B is positione a istance to the right of particle A. Particle C is positione an equal istance to the right of particle B. Particles A an C are electrons. Particle B is a proton. If the magnitue of the net electrical force on particle C is 5.00 x N, calculate.. Point charges are fixe at the corners of an equilateral triangle, as shown. q B is locate at x = 2.50 cm. The value of q C is known: q C = mc, an the net electrical force on q C is in the negative y-irection an has a magnitue of 450 N. Fin the magnitues an signs of q A an q B. q C q A q B 5

6 e. The current stanar moel of particle physics states that a proton is comprise of two up quarks, each of charge +(2/3)e, an one own quark, of charge (1/3)e. Assume that all three quarks are equiistant from each other at the istance of 1.50 x10-15 m. (i) What is the total force magnitue exerte on one of the up quarks by the other two quarks? (ii) What is the total force magnitue exerte on the own quark by the other two quarks? In the iagram, let q C be the own quark; q A an q B are the up quarks. (i) F E.Atotal = F E.BA + F E.CA = k q B q A / k q C q A / 2 60 = (4/9)ke 2 / 2 [cos180, sin180 ] + (2/9)ke 2 / 2 [cos60, sin60 ] = (4/9)ke 2 / 2 [ 1, 0] + (2/9)ke 2 / 2 [(1/2), ( 3/2)] = [(2/9)ke 2 /(2 2 ) (4/9)ke 2 / 2, 3(2/9)ke 2 /(2 2 )] = [(2/9)ke 2 /(2 2 ) (8/9)ke 2 /(2 2 ), 3(2/9)ke 2 /(2 2 )] = [ke 2 /(2 2 )][ (2/3), (2 3/9)] Therefore: F E.Atotal = [ke 2 /(2 2 )][ (2/3) 2 + (2 3/9) 2 ] 1/2 = [ke 2 /(2 2 )][4/9 + 12/81] 1/2 = [ke 2 /(2 2 )][36/ /81] 1/2 = [ke 2 /(2 2 )][48/81] 1/2 = 2 3ke 2 /(9 2 ) = 2 3(8.99 x 10 9 )(1.60 x ) 2 /[9(1.50 x ) 2 ] = 39.4 N q A q C q B (ii) F E.Ctotal = F E.AC + F E.BC = k q A q C / k q B q C / = (2/9)ke 2 / 2 [cos240, sin240 ] + (2/9)ke 2 / 2 [cos300, sin300 ] = (2/9)ke 2 / 2 [ (1/2), ( 3/2)] + (2/9)ke 2 / 2 [(1/2), ( 3/2)] = [0, 3(2/9)ke 2 / 2 ] Therefore: F E.Ctotal = 3(2/9)ke 2 / 2 That is: F E.Ctotal = F E.Atotal = 39.4 N 6

7 5. a. Evaluate each statement (T/F/N). As always, you must fully explain your reasoning an answers. (i) Assuming no other charges (or fiels) in the area, the electric fiel miway between a proton an electron is zero. False. See item 6b. (ii) The magnitue of the electric fiel cause by a point charge is the same at any point on a circle whose center is occupie by that charge. True. The E-fiel strength epens (spatially) only on the raial istance from the charge to the point in question: E = k q /r 2 (iii) The electric fiel is uniform in the region surrouning a single point charge. False. Uniform means ientical at every point. The E-fiel surrouning a point charge varies in both magnitue (closer points vs. farther) an irection (raially outwar can mean any q an f angles of spherical coorinates). Don t confuse uniform with symmetric. The E-fiel aroun a point charge is inee spherically symmetric (looks similar even after rotating), but the vector values at many points are efinitely ifferent. b. An electron is locate at the origin. Another electron is locate at (3,0). Fin the E-fiel irection at (1,0), using the positive x-axis as 0. c. An electron is locate at the origin. A proton is locate at (2,0). Fin the E-fiel irection at ( 1,0).. An electron is locate at the origin. A proton is locate at (2,0). Fin the electric fiel irection at (1,1). e. What s the irection of the electric fiel cause at the point ( 2,3) by an electron locate at the origin? 7

8 6. a. The net electric fiel prouce by two charges is shown to the right. Fully explain your reasoning an answer for each of the following: (i) What can you conclue about each of the two charges, P an Q? (P is the charge on the left; Q is on the right, an each charge s fiel lines emerge from it. Sorry for the poor resolution here.) Both P an Q are positive charges; their fiel lines raiate outwar. An Q > P ; Q creates more fiel lines at any given raial istance from it (this is use to represent a stronger fiel). (ii) Where in the iagram is the electric fiel magnitue strongest an where it is weakest? Again, a greater number of fiel lines shows a greater total fiel strength, so very near Q (on sies not next to P) woul be areas of strongest fiels; the region miway between P an Q woul be the location of the weakest fiel. (iii) How oes the force of P on Q compare to the force of Q on P? Newton s thir law applies to any kin of force. Thus: F E.PQ = F E.QP The force magnitues are equal; the force irections are opposite. b. In the rawing, q 1 is locate at point A, an q 2 is locate at point B. q 2 Let E 1B mean the electric fiel cause at point B by charge 1; an let E 2A means the electric fiel cause at point A by charge 2. A B Evaluate each statement (T/F/N). As always, you must fully explain your reasoning an answers. (i) If both q 1 an q 2 are negative, the irection of E 2A is the same as the irection of F 12. True. If both charges are negative, then: E 2A = (k q 2 / 2 AB ) 0 (The fiel vector at A points towar q 2 that s to the right.) F 12 = (k q 2 q 2 / 2 AB ) 0 (q 1 repels q 2, pushing it to the right) (ii) Regarless of their signs (±), if q 1 an q 2 are of equal magnitue, then E 2A = E 1B, always. False. Suppose that q 1 is positive an q 2 is negative. Then the two vectors are ientical (not opposites): E 2A = (k q 2 / 2 AB ) 0 (The fiel vector at A points towar q 2 that s to the right.) E 2A = (k q 1 / 2 AB ) 0 (The fiel vector at B points away from q 1 that s also to the right.) c. Write an expression for the E-fiel magnitue miway between an electron an proton that are a istance apart. q 1 8

9 6.. Two electrons are locate along the y-axis at points (0, s) an (0, s). You may consier the following as known values (so they may be inclue in your answers): k, e 0, e, s (i) Write an expression for the electric force magnitue exerte on one electron by the other. The basic Coulomb force equation: F E.12 = k q 1 q 2 /r 12 2 In this situation: An: q 1 = q 2 = e r 12 = 2s Thus: F E.12 = ke 2 /(2s) 2 = ke 2 /(4s 2 ) (ii) Write an expression for the net electric fiel (magnitue an irection) at the point (r, 0) on the positive x-axis. (You may consier r as a known value, too, so your answer may inclue r, k, e 0, e, s.) Be sure to use iagrams if they help! First, note from the iagram that only the x-components of each fiel will be significant; ue to the symmetry of the situation, their y-components will cancel each other. We nee only E x. So we nee only E 1.x an E 2.x. q 1 s s q 2 E total r 1 r q q r 2 E 2 E 1 The basic fiel equation: E 1 = k q 1 /r 1 2 Take the x-component only: E 1.x = (k q 1 /r 1 2 )(cosq) = (k q 1 /r 1 2 )(cosq) In this situation: r 1 = r 2 = (r 2 + s 2 ) An: cosq = r/r 1 An, again: q 1 = q 2 = e Therefore: E 1.x = (ker/r 3 1 ) = (ker)/(r 2 + s 2 ) 3/2 Double this to a E 2.x : E x = (2ker)/(r 2 + s 2 ) 3/2 (iii) Referring to the fiel expression from part b, how you woul calculate the value of r where (along the x-axis) that fiel strength is a maximum? You o not have to actually o the calculation; just fully explain your reasoning an the proceure. The expression from part b is a function, E x (r). To fin a local maximum, we woul therefore take the erivative of that function an set it equal to zero: E x /r = 0 In other wors, we solve ( (2ker)/(r 2 + s 2 ) 3/2 )/r = 0 for r. 9