Electricity and Magnetism Guide

This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.9 g. Find the number of atoms in a nickel coin. Each mole (6.02 × 1023 atoms) has a mass of about 58.1 g. Correct answer: 5.07711 × 1022 atoms. Explanation: 003(part3of3)10.0points Find the magnitude of the charge of all these electrons. Correct answer: 2.27764 × 105 C. Explanation: Let : qe = −1.60218 × 10−19 C/electron . The total charge q for the ne electrons is q = n e qe = (1.42159 × 1024 electrons) Let : 23 Na = 6.02 × 10 atoms , M = 58.1 g , and m = 4.9 g . Mass is proportional to the number of atoms in a substance, so for m grams in N atoms in the nickel coin and M grams in Na atoms in one mole, we have m N = N a M m 4.9 g N = Na = (6.02 × 1023 atoms) M 58.1 g = 5.07711 × 1022 atoms . 002(part2of3)10.0points Find the number of electrons in the coin. Each nickel atom has 28 electrons/atom. Correct answer: 1.42159 × 1024 electrons. Explanation: Let : nNi = 28 electrons/atom . If nNi electrons are in each Nickel atom, then the total number of electrons ne in the coin is × −1.60218 = −2.27764 × 105×C10, −19  C/electron with a magnitude of 2.27764 × 105 C . 004 10.0points A sphere is charged with electrons to −9 × 10−6 C. How many electrons make up this charge? The elemental charge is 1.6 × 10−19 C. Correct answer: 5.625 × 1013. Explanation: Let : q = −9 × 10−6 C and qe = −1.6 × 10−19 C . The total charge is q = n qe q −9 × 10−6 C n= qe = −1.6 × 10−19 C = 5.625 × 1013 . ne = N nNi  × 1022 atoms = ×5.07711 (28 electrons/atom) 24 = 1.42159 × 10 electrons . 005 10.0points A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of −4.9 µC. How many electrons are transferred from the wool shirt to the plastic rod? The elemental charge is 1.6 × 10−19 C. 5) While Y and X remain in touch, the ball carring the positive charge is removed. X Correct answer: 3.0625 × 1013 . Y Explanation: 6) Then ball Y and X are separated. Let : q = −4.9 µC = −4.9 × 10−6 C e = −1.6 × 10−19 C . and X Y The total charge is q = ne e q −4.9 × 10−6 C ne = = e −1.6 × 10−19 C = 3.0625 × 1013 . 006 10.0points 1) Two uncharged metal balls, Y and X, each stand on a glass rod and are touching. X Y 2) A third ball carrying a positive charge, is brought near the first two. + X Y 3) While the positions of these balls are fixed, ball Y is connected to ground. + X Y 4) Then the ground wire is disconnected. + X Y After these procedures, what are the signs of the charge qY on Y and qX on X? 1. qY is positive and qX is neutral 2. qY is neutral and qX is negative 3. qY is positive and qX is positive 4. qY is neutral and qX is neutral 5. qY is negative and qX is negative correct 6. qY is positive and qX is negative 7. qY is negative and qX is positive 8. qY is negative and qX is neutral 9. qY is neutral and qX is positive Explanation: When the ball with positive charge is brought nearby, the free charges inside Y and X rearrange themselves. The negative charges are attracted and go to the left (i.e., move to X), leaving positive charges on the right hand side of the system YY, i.e.,in Y. When we ground Y, electrons flow from the ground to Y (making it neutral), whereas the negative charges in X are still held enthralled by the positive charge on the third ball. We break the ground. Now we remove the third ball with positive charge. The charge on X is redistributed in the system YY; i.e., they share the negative charge (equally if identical). Finally we separate Y and X. The signs of the charge on Y and that on X are both negative. 007 10.0points Two metal spheres that are initially uncharged are mounted on insulating stands, as shown. − − − − X SphereY 1.Negative Positive 008 10.0points Imagine a charge in the middle between two parallel plate conductors. There is no net charge on the plates, and the plates are not connected to ground. Y A negatively charged rubber rod is brought close to but does not make contact with sphere X. Sphere Y is then brought close to X on the side opposite to the rubber rod. Y is allowed to touch X and then is removed some distance away. The rubber rod is then moved far away from X and Y. What are the final charges on the spheres? SphereX electrons will be pushed to sphere Y. If X and Y are separated before the rod moves away, those charges will remain on X and Y, X is positively charged and Y is negatively charged. 2.Positive Negative correct 3.Positive Zero 4.Zero Positive 5.Zero Negative 6.Negative Negative 7.Zero Zero 8.Negative Zero 9.Positive Positive Explanation: When the negatively charged rod moves close to the sphere X, the negatively charged q What will happen if the charge is moved a little away from the middle? 1.The charge will remain stationary. 2.There is not enough information to tell. 3.The charge will return to the middle. 4.All of these can happen, depending on the size of the charge. 5.The charge will move away from the middle. correct Explanation: There will be an image-charge attracting it toward each metal surface. It will move toward the closest metallic surface because the closest image-charge attraction will be stronger. Any charge (free to move) will move toward the closest conductor it can find. 009 10.0points A particle of mass 62 g and charge 85 µC is released from rest when it is 52 cm from a second particle of charge −23 µC. Determine the magnitude of the initial acceleration of the 62 g particle. Correct answer: 1048.06 m/s2 . Explanation: Let : Coulomb’s law (in vector form) for the electric force exerted by a charge q1 on a second charge q3 , is m = 62 g , q = 85 µC = 8.5 × 10−5 C , d = 52 cm = 0.52 m , q1 q3 ~ rˆ13 , F13 = ke r2 Q = −23 µC = −2.3 × 10−5 C , ke = 8.9875 × 109 N · m2 /C2 . and where rˆ13 is a unit vector directed from q1 to q3 ; i.e., ~r13 = ~r3 − ~r1 . The force exerted on the particle is |q| |Q| F = ke d2 = m a |q| |Q| a = k e m d2 = (8.9875 × 109 N · m2 /C2 )    m2 ) −5  8.5 (0.062 −2.3 × 10 C  × 10−5 kg) C (0.52 × 2 = 1048.06 m/s . 010 10.0points A particle with charge 5 µC is located on the x-axis at the point 6 cm , and a second particle 5 µC −3 µC −6 µC with charge −6 µC is placed on the x-axis at −4 cm . −10 −8 −6 −4 −2 2 4 6 8 10 x → (cm) What is the magnitude of the total electrostatic force on a third particle with charge −3 µC placed on the x-axis at −2 cm ? The Coulomb constant is 8.9875 × 109 N · m2 /C2 . Explanation: q1 = 5 µC = 5 × 10−6 C , q2 = −6 µC = −6 × 10−6 C , q3 = −3 µC = −3 × 10−6 C , x1 = 6 cm = 0.06 m , x2 = −4 cm = −0.04 m , x3 = −2 cm = −0.02 m . ~ F = ~F13 + ~F23 q2  r 2 rˆ13 + r 2  23 q131 = ke rˆ232 2q3 9 = 8.9875 × 10 N · m /C  × (−ˆı) 2 (5(−0.08 × 10−6m)C) −6 (−6 × 10 C) +  (+ˆ ı ) (0.02 m)2×(−3 × 10−6 C) = 425.502 N , with a magnitude of 425.502 N . Correct answer: 425.502 N. Let : x13 = x3 − x1 = (−2 cm) − (6 cm) = −0.08 m x23 = x3 − x2 = (−2 cm) − (−4 cm) = 0.02 m x3 − x1 x̂13 = p = −ˆı (x − x )2 x3 3− x2 1 x̂23 = p = +ˆı ( x3 − x2 ) 2 Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles: and 011 10.0points A molecule of DNA (deoxyribonucleic acid) lies along a straight line. It is 2.606 µm long. The ends of the molecule become singly ionized; negative on one end, positive on the other. The helical molecule acts as a spring and compresses 0.8% upon becoming charged. Determine the effective spring constant of the molecule. Take into account the compressed length when calculating the distance between the ends of the molecule. The Coulomb constant is 8.99 × 109 N · m2 /C2 . Correct answer: 1.65632 × 10−9 N/m. Explanation: Let : ℓ = 2.606 µm = 2.606 × 10−6 m , ∆ℓ = 0.008 ℓ = 2.0848 × 10−8 m , kC = 8.99 × 109 N · m2 /C2 , and qe = 1.60218 × 10−19 C . What is the magnitude of the resulting force on the 3 nC charge at the origin? The Coulomb constant is 8.98755 × 109 N · C2 /m2 . Correct answer: 14.3622 nN. Explanation: Let : The compressed length is r = ℓ − ∆ℓ = 2.58515 × 10 −6 F = kC qb = −2 × 10−9 C , rb = 2 m , and m and the compression force is qe qe , r2 so the effective force constant is ke = 8.98755 × 109 N · C2 /m2 . Applying Coulomb’s Law for qo and qa , = −(8.98755 × 109 N · C2 /m2 ) (3 × 10−9 C) (9 × 10−9 C) × (7 m)2 = −4.95232 × 10−9 N directed along the x-axis. Applying Coulomb’s Law for qo and qb , Fob = −ke qo qa r 2ob = −(8.98755 × 109 N · C2 /m2 ) (3 × 10−9 C) (−2 × 10−9 C) × (2 m)2 012 10.0points Three charges are arranged in the (x, y) plane as shown. y (m) 10 9 8 7 6 5 4 3 2 1 0 qo qa r2oa Foa = −ke kC q 2 F = 2 e k= ∆ℓ r ∆ℓ 8.99 × 109 N · m2 /C2 = (2.58515 × 10−6 m)2 2 −19 1.60218 ×× 10 C −8 m 2.0848 10 × = 1.65632 × 10−9 N/m . qo = 3 × 10−9 C , qa = 9 × 10−9 C , ra = 7 m , = 1.34813 × 10−8 N directed along the y-axis. The magnitude of the resultant force is k~ q x + F 2y Fk = F2 h 2 −4.95232 × 10−9 N = −2 nC 3 nC + 9 nC x (m) 0 1 2 3 4 5 6 7 8 9 10 2 i1/2 1.34813 × 10−8 N 1 × 109 nN × 1 N = 14.3622 nN . 013 10.0points Three charges are arranged in the (x, y ) plane as shown. y (m) 10 9 8 7 6 5 4 3 2 1 0 × (9 × 10−9 C) (2 × 10−9 C) (4 m)2 = −1.0111 × 10−8 N . The angle measured counter-clockwise from the positive x axis is  Fy F  x  −1.0111 × 10−8 N = arctan ◦ −8 = 318.814 1.15554 . × 10 N  θ = arctan 2 nC x −7 nC 9 nC (m) 0 1 2 3 4 5 6 7 8 9 10 What angle does the resultant force on the 9 nC charge at the origin make with the positive x axis? Quadrant I lies between 0◦ and 90◦ , quadrant II between between 90◦ and 180◦ , etc. Correct answer: 318.814◦. Explanation: qo = 9 × 10−9 C , qa = −7 × 10−9 C , ra = 7 m , Let : qb = 2 × 10 rb = 4 m . −9 C, and Applying Coulomb’s Law to qo and qa , qo qa r 2oa Foax = −ke = −(8.98755 × 109 N · C2 /m2 ) × (9 × 10−9 C) (−7 × 10−9 C) (7 m)2 = 1.15554 × 10−8 N . 014 10.0points Four point charges, each of magnitude 2 µC, are placed at the corners of a square 22.9 cm on a side. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C2 . Correct answer: 1.31226 N. Explanation: Applying Coulomb’s Law to qo and qb , Foby = −ke qo qa r 2oa = −(8.98755 × 109 N · C2 /m2 ) Let : ke = 8.98755 × 109 N · m2 /C2 , d = 22.9 cm = 0.229 m , and Q = 2 µC = 2 × 10−6 C . 1 + Explanation: 2 + + 3 The forces are F34 F24 4 |q| = 7.3 C . and k = 8.9875 × 109 N · m2 /C2 . Each force has a magnitude Q2 d2 F34 = F24 = ke 8.98755 × 109 N · m2 /C2 2 2(0.229 × 10−6 m) C2 × = 0.685536 N and = F14 = F21 = F31 =  Fx = (−F31 cos θ − F21) ıˆ k q2 ◦  a22 cos 60 −  k q = − k q2 k q2 ıˆ a2  2 a2 −  1 = 3− ıˆ q2 a2 = − k 2 ıˆ 2 a The vector sum of F24 and F34 is in the same direction as F14 , so 2 + F24 2 F == F114 + F34   q 1 √ F34 + 2 F34 2 = 2 + 2 F34 2   1 √ = 2 + 2 (0.685536 N) = 1.31226 N . 015(part1of2)10.0points Three point charges are placed at the vertices of an equilateral triangle as shown. +7.3 C ̂ 4. 4 ıˆ m 60 ◦ +7.3 C +7.3 C What is the magnitude of the electric force on the +7.3 C charge at the bottom left-hand vertex of the triangle due to the other two charges? The value of the Coulomb constant is 8.9875 × 109 N · m2 /C2 . Correct answer: 4.28489 × 1010 N. k q2 . a2 The x-component of the net force is 2 ke Q 2 1 1 ke Q √ 2 = 2 d2 = 2 F34 . 2d q θ = 60 ◦ , a = 4.4 m , Let : F14 and the y-component is   k q2 2 sin a ◦ + F2 r q2 √ 3 k a2 ̂ , 2 ̂ = 60 Fy = so the magnitude of the net force is k~ q Fk = F2 x y =k 9 4 + √  q 2 =k 3 a = 8.9875 × 109 N · m2 /C2 = 4.28489 × 1010 N . 3 ! 2 qa2 4 √  3 7.3 C 4.4 m 2 016(part2of2)10.0points What angle does the resultant force on the +7.3 C charge at the bottom left-hand vertex of the triangle due to the other two charges? Quadrant I lies between 0◦ and 90◦ , quadrant II between between 90◦ and 180◦ , etc. Explanation: Correct answer: 210◦ . Explanation: The force between charges of the same sign is repulsive and between charges with opposite signs is attractive. +q ̂ Let : L = 12.4 cm = 0.124 m , m = 0.25 g = 0.00025 kg , θ = 4.6◦ . and a ıˆ L +q +q The resultant force is the sum of these two vectors. The angle θ defined counterclockwise from the positive x-axis is θ q a m m From the right triangle in the figure, q a L a = L sin θ , √ 2  − 3 k 2q  2a    3.0 k q2  Fy   = arctan  − 2 a2  θ = arctan Fx   = arctan  √  − 3 = 210◦ . 017 10.0points Two small metallic spheres, each of mass 0.25 g are suspended as pendulums by light strings from a common point as shown. The spheres are given the same electric charge, and it is found that the two come to equilibrium when each string is at an angle of 4.6 ◦ with the vertical. sin θ = and the separation of the spheres is r = 2 a. Consider the forces acting on one of the spheres: T Fe mg The sphere is in equilibrium horizontally: X 12. m 4 c 4.6◦ 0.25 g 0.25 g If each string is 12.4 cm long, find the magnitude of the charge on each sphere. The Coulomb constant is 8.98755 × 109 N · m2 /C2 and the acceleration of gravity is 9.81 m/s2 . Correct answer: 2.94706 nC. θ Fx = T sin θ T − Fsin e =θ 0= Fe and vertically: X Fy = T cos θ −Tmcos g =θ 0= m g , tan θ = T sin θ Fe = T cos θ mg Fe = m g tan θ. so From Coulomb’s law, the electric force between the charges has magnitude |Fe | = ke |q|2 , r2 where |q| is the magnitude of the charge on each sphere, so s |q| = r |Fe |ke r = (2 a) a 2 − 2 d2 = 0 s 2 /C4. 2 6◦ (0.00025 kg) (9.81 8.98755 × 109m/ Ns·2 )mtan × 9 10 nC × 1 C = 2.94706 nC . 018 10.0points Charge 2 × 10−18 C is on the y axis a distance 8 m from the origin and charge 3 × 10−18 C is on the x axis a distance d from the origin. What is the value of d for which the x component of the force on 3 × 10−18 C is the greatest? The Coulomb constant is 8.98755 × 109 N · m2 /C2 . Correct answer: 5.65685 m. Explanation: Q = 2 × 10−18 C , q = 3 × 10−18 C , a=8m . and The force on charge q on the x axis due to charge Q on the y axis is ~ F = For maximum x component of the force, ∂Fx = 0 is required, so ∂d ∂ Fx q Q a 2 − 2 d2 = = 0 ∂d 4 π ǫ0 (a2 + d2 )5/2 ke θ m g tan r ke θ m g tan = 2(2 (0.124 L sin θ) m) sin 4.6◦ Let : 1 q Q d 4 π ǫ0 a 2 + d 2 √ a 2 + d 2 qQ d 1 . = 4 π ǫ0 (a2 + d2 )3/2 = 1 q Q rˆ , 4πǫ0 r 2 p the force where r =ona2q+isd2 , so the x component of 1 q Q Fx = cos θ 4 π ǫ0 r 2 8 m a d = √ = √ = 5.65685 m . 2 2