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hernandez (uh286) – oldmidterm 01 – Turner – (59070) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. at (43 cm, 0.0). The Coulomb constant is 9 × 109 N m2 /C2 . Calculate the magnitude of the force on the charge at the origin. Correct answer: 7.4482 N. 001 10.0 points Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere “1” is charged with 3 nC, and sphere “2” is charged with 15 nC. Both objects have the same mass. 1 nC is equal to 1 × 10−9 C. As they repel, Explanation: 1. they have the same magnitude of acceleration. correct 2. sphere “2” accelerates 25 times as fast as sphere “1”. 3. sphere “1” accelerates 5 times as fast as sphere “2”. 4. they do not accelerate at all, but rather separate at constant velocity. 5. sphere “2” accelerates 5 times as fast as sphere “1”. 6. sphere “1” accelerates 25 times as fast as sphere “2”. Let : q = 7 µC = 7 × 10−6 C , (x1 , y1 ) = (0, 25 cm) = (0, 0.25 m) , (x2 , y2 ) = (43 cm, 0) = (0.43 m, 0) and (x0 , y0 ) = (0, 0) . The electric fields are ke q + y12 9 × 109 N m2 /C2 7 × 10−6 C = 0 + (0.25 m)2 = 1.008 × 106 N/C , and ke q Ex = 2 x2 + y22 9 × 109 N m2 /C2 7 × 10−6 C = (0.43 m)2 + 0 = 3.40725 × 105 N/C , Thus q Ey = E= x21 Ex2 + Ey2 Explanation: and The force of repulsion exerted on each mass q is determined by F = q E = q Ex2 + Ey2 1 Q1 Q2 = 7 × 10−6 C F = = ma q 4 π ǫ0 r 2 · (1.008 × 106 N/C)2 + (3.40725 × 105 N/C)2 where r is the distance between the centers of = 7.4482 N . the two spheres. ~ 12 k = kF ~ 21 k kF Since both spheres have the same mass and are subject to the same force, they have the same acceleration. 002 10.0 points Three equal charges of 7 µC are in the x-y plane. One is placed at the origin, another is placed at (0.0, 25 cm), and the last is placed 003 10.0 points Two identical small charged spheres hang in equilibrium with equal masses as shown in the figure. The length of the strings are equal and the angle (shown in the figure) with the vertical is identical. The acceleration of gravity is 9.8 m/s2 and the value of Coulomb’s constant is 8.98755 × 109 N m2 /C2 . hernandez (uh286) – oldmidterm 01 – Turner – (59070) vertical directions must separately add up to zero: X Fx = T sin θ − Fe = 0 X Fy = T cos θ − m g = 0 . 0. 2 5m 7◦ 0.04 kg 0.04 kg Find the magnitude of the charge on each sphere. Correct answer: 1.41013 × 10−7 C. Explanation: Let : L = 0.25 m , m = 0.04 kg , θ = 7◦ . and L q m θ a m q From the right triangle in the figure above, we see that a sin θ = . L Therefore a = L sin θ = (0.25 m) sin(7◦ ) = 0.0304673 m . The separation of the spheres is r = 2 a = 0.0609347 m . The forces acting on one of the spheres are shown in the figure below. T T cos θ Fe 2 From the second equation in the system mg above, we see that T = , so T can be cos θ eliminated from the first equation if we make this substitution. This gives a value Fe = m g tan θ = (0.04 kg) 9.8 m/s2 tan(7◦ ) = 0.0481315 N , for the electric force. From Coulomb’s law, the electric force between the charges has magnitude |q|2 |Fe | = ke 2 , r where |q| is the magnitude of the charge on each sphere. Note: The term |q|2 arises here because the charge is the same on both spheres. This equation can be solved for |q| to give s |Fe | r 2 |q| = ke s (0.0481315 N) (0.0609347 m)2 = (8.98755 × 109 N m2 /C2 ) = 1.41013 × 10−7 C . 004 10.0 points A circular arc has a uniform linear charge density of 6 nC/m. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C2 . y 14 ◦ 6 θ 2. 8 θ T sin θ mg Because the sphere is in equilibrium, the resultant of the forces in the horizontal and m x hernandez (uh286) – oldmidterm 01 – Turner – (59070) What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 36.835 N/C. Explanation: Let : λ = 6 nC/m = 6 × 10−9 C/m , ∆θ = 146◦ , and r = 2.8 m . θ is defined as the angle in the counterclockwise direction from the positive x axis as shown in the figure below. 73 ◦ ◦ r 73 θ ~ E First, position the arc symmetrically around the y axis, centered at the origin. By symmetry (in this rotated configuration) the field in the x direction cancels due to charge from opposites sides of the y-axis, so Ex = 0 . For a continuous linear charge distribution, Z ~ = ke dq r̂ E r2 In polar coordinates dq = λ (r dθ) , where λ is the linear charge density. The positive y axis is θ = 90◦ , so the y component of the electric field is given by dEy = dE sin θ . Note: By symmetry, each half of the arc about the y axis contributes equally to the electric field at the origin. Hence, we may just 3 consider the right-half of the arc (beginning on the positive y axis and extending towards the positive x axis) and multiply the answer by 2. Note: The upper angular limit θ = 90◦ . The lower angular limit θ = 90◦ − 73◦ = 17◦ , is the angle from the positive x axis to the right-hand end of the arc. ! Z ◦ λ 90 E = −2 ke sin θ dθ ̂ r 17◦ = −2 ke λ [cos (17◦ ) − cos (90◦ )] ̂ . r Since ke λ = (8.98755 × 109 N · m2 /C2 ) r (6 × 10−9 C/m) × (2.8 m) = 19.259 N/C , E = −2 (19.259 N/C) × [(0.956305) − (0)] ̂ = −36.835 N/C ̂ ~ = 36.835 N/C . kEk ~ in Alternate Solution: Just solve for kEk a straight forward manner, positioning the beginning of the arc on the positive x axis (as shown in the original figure in the question). θ is still defined as the angle in the counterclockwise direction from the positive x axis. ! ◦ Z ke λ 146 Ex = − cos θ dθ ı̂ r 0◦ ke λ [sin (146◦ ) − sin (0◦ )] ı̂ r = −(19.259 N/C) × [(0.559192) − 0.0] ı̂ = [−10.7695 N/C] ı̂ , ! ◦ Z ke λ 146 Ey = − sin θ dθ ̂ r 0◦ =− ke λ [cos (0◦ ) − cos (146◦ )] ̂ r = −(19.259 N/C) × [1.0 − (−0.829038)] ̂ = [−35.2255 N/C] ̂ , =− hernandez (uh286) – oldmidterm 01 – Turner – (59070) q ~ = E2 + E2 kEk y x h = (−10.7695 N/C)2 + (−35.2255 N/C)2 = 36.835 N/C . i1/2 005 10.0 points From the electric field vector at a point, one can determine which of the following? I) the direction of the electrostatic force on a test charge of known sign at that point; II) the magnitude of the electrostatic force exerted per unit charge on a test charge at that point; III) the electrostatic charge at that point. 1. I and II only correct 2. II and III only 3. III only 4. I, II and III Let : m = 9.6 × 10−5 g = 9.6 × 10−8 kg , Ex = 4.6 N/C , Ey = Ez = 0 , vy = 2.3 × 106 m/s , vx = vz = 0 , and t = 0.7 s . According to Newton’s second law and the definition of an electric field, ~ = m~a = q E ~. F Since the electric field has only an x component, the particle accelerates only in the x direction q Ex ax = . m To determine the x component of the final velocity, vxf , use the kinematic relation vxf = vxi + a (tf − ti ) = a tf . Since ti = 0 and vxi = 0, 006 10.0 points A particle of mass 9.6 × 10−5 g and charge 86 mC moves in a region of space where the electric field is uniform and is 4.6 N/C in the x direction and zero in the y and z direction. If the initial velocity of the particle is given by vy = 2.3 × 106 m/s, vx = vz = 0, what is the speed of the particle at 0.7 s? Correct answer: 3.68928 × 106 m/s. Explanation: q E x tf m (0.086 C) (4.6 N/C)(0.7 s) = (9.6 × 10−8 kg) = 2.88458 × 106 m/s . vxf = 5. I only Explanation: The definition of the electrostatic force is ~ ~ = F . In another way, F ~ = q E. ~ This means E q ~ is in the same direction of, or opposite diF ~ depending on the sign of the rection to E, charge. And if we only consider the magnitude F = qE for a unit charge, the force on it F is just = E. This is statement II. q 4 vxf No external force acts on the particle in the y direction so vyi = vyf = 2.3 × 106 m/s. Hence the final speed is given by q 2 + v2 vf = vyf xf 2 = 2.3 × 106 m/s 2 + 2.88458 × 10 m/s 6 = 3.68928 × 106 m/s . 1/2 Note: This is analogous to a particle in a gravitational field with the coordinates roπ tated clockwise by (90◦ ). 2 007 (part 1 of 2) 10.0 points hernandez (uh286) – oldmidterm 01 – Turner – (59070) A dipole (electrically neutral) is placed in an external field. (a) − + − + (b) + − − (c) + 5 Gauss’ law states I ~ · dA ~ = Q. ΦS = E ǫ0 Solutions: The electric dipole consists of two equal and opposite charges separated by a distance. In either situation (c) or (d), the electric field is uniform everywhere between the parallel infinite plates. Thus, the electric force on one charge is equal but opposite to that on another so that the net force on the whole dipole is zero. By contrast, electric fields are nonuniform for situations both (a) and (b). (d) For which situation(s) shown above is the net force on the dipole equal to zero? 1. Another combination 2. (c) and (d) correct 3. (a) and (c) 008 (part 2 of 2) 10.0 points For which situation(s) shown above is the net torque on the dipole equal to zero? 1. (a) only 2. (c) and (d) 3. other combination 4. (b) and (d) 4. (c) only 5. (c) only 5. (b) and (d) 6. (a) only 6. None of these 7. None of these Explanation: Basic Concepts: Field patterns of point charge and parallel plates of infinite extent. The force on a charge in the electric field is given by ~ = qE ~ F and the torque is defined as ~ = ~r × F ~ T ~ = k ∆q r̂ ∆E r2 X ~ = ~i. E ∆E Symmetry of the configuration will cause some component of the electric field to be zero. 7. (a) and (c) correct 8. (a) and (b) Explanation: A electric dipole can be regarded as a pair of charges of opposite sign. Only in figures (a) and (c), the electric fields are along the direction of ~r, where ~r is the vector between ~ is the pair of charges. Therefore the force F also along ~r . This will lead to zero torque, since ~ = ~r × F ~ ∝ ~r × ~r = 0 . T For figures (b) and (d), the torque on both charges are nonzero and the resultant torques are also nonzero. 009 10.0 points hernandez (uh286) – oldmidterm 01 – Turner – (59070) Five charges are placed in a closed box. Each charge (except the first) has a magnitude which is twice that of the previous one placed in the box. All charges have the same sign and (after all the charges have been placed in the box) the net electric flux through the box is 9.5 × 107 N · m2 /C. What is the magnitude of the smallest charge in the box? Correct answer: 27.1338 µC. 6 r4 r3 r2 r1 q3 q2 q1 R1 R2 R3 R4 R5 Explanation: Let : Φ = 9.5 × 107 N · m2 /C Let the first charge in the box be q. Then the other four charges have magnitude 2 q, 4 q, 8 q, and 16 q. We can figure out a value for q by considering Gauss’ law, Φ= qtotal . ǫ0 Our closed box is a Gaussian surface, and we know the total flux through this closed surface. Hence, Φ= (1 + 2 + 4 + 8 + 16) q 31 q = ǫ0 ǫ0 and solving for q, Hint: Under static conditions, the charge on a conductor resides on the surface of the conductor. What is the charge Qr3 on the inner surface of the larger spherical conducting shell? 1. Qr3 = +q1 − q2 2. Qr3 = 0 3. Qr3 = −q1 − q2 − q3 4. Qr3 = +q1 + q2 + q3 5. Qr3 = −q1 − q2 + q3 6. Qr3 = +q1 + q2 q= 1 (9.5 × 107 N · m2 /C) 31 × (8.85419 × 10−12 C2 /N · m2 ) 1 × 106 µC × 1C = 27.1338 µC . 010 10.0 points A point charge q1 is concentric with two spherical conducting thick shells, as shown in the figure below. The smaller spherical conducting shell has a net charge of q2 and the larger spherical conducting shell has a net charge of q3 . 7. Qr3 = +q1 8. Qr3 = −q1 − q2 correct 9. Qr3 = −q1 + q2 10. Qr3 = −q1 Explanation: The net charge inside a Gaussian surface located at r = R4 must be zero, since the field in the conductor must be zero. Therefore, the charge on the inner surface of the large spherical conducting shell must be −(q1 + q2 ). 011 10.0 points hernandez (uh286) – oldmidterm 01 – Turner – (59070) 1 . 4 π ǫ0 A uniformly charged sphere (an insulating sphere with radius R) is shown in the figure below. Given: k = Q is the total charge inside the sphere. R p R 2 If Q = 5.3 × 10−6 C, the magnitude of the ~ electric field at r = R is given by kEk 2 ~ = kQ . 1. kEk 16 R2 ~ = k Q . correct 2. kEk 2 R2 ~ = 2kQR. 3. kEk ~ = kQ . 4. kEk 8 R2 ~ = kQR. 5. kEk ~ = 4kQR. 6. kEk ~ = kQ . 7. kEk 4 R2 ~ = kQ. 8. kEk R2 ~ = kQ . 9. kEk 32 R2 ~ = 8kQR. 10. kEk Explanation: Pick aspherical Gaussian surface with ra R dius r, where r = concentric with the 2 sphere of charge. From Gauss’ law, we find Φ = E · 4 π r2 . ρ= 4 3 Q , so the net charge inside the Gausπ R3 7 R sian surface of radius is 2 4 3 πr qen = ρ 3 Q 4 3 = πr 4 3 π R3 3 r3 =Q 3. R Therefore the electric field is Φc E= 4 π r2 Q r3 1 = ǫ0 R3 4 π r 2 Qr = , 4 π ǫ0 R3 at r. At r = R we have 2 Q 1 E= 4 π ǫ0 2 R2 kQ . = 2 R2 012 (part 1 of 2) 10.0 points Consider the figure +Q A −Q + − y + − + − + − + − x + − + C D − + − + − + − + − #1 #2 B Of the following elements, identify all that correspond to an equipotential line or surface. 1. line AB only correct 2. neither AB nor CD 3. line CD only 4. both AB and CD Explanation: hernandez (uh286) – oldmidterm 01 – Turner – (59070) Consider the electric field +Q A −Q + − y + − + − + − + − x + C D − + − + − + − + − + − #1 #2 B An equipotential line or surface (AB) is normal to the electric field lines. 013 (part 2 of 2) 10.0 points Consider the figure C B + +q A − −q 8 Through what potential difference would an electron need to be accelerated for it to achieve a speed of 6 % of the speed of light (2.99792 × 108 m/s), starting from rest? Correct answer: 919.798 V. Explanation: Let : s = 6 % = 0.06 , c = 2.99792 × 108 m/s , me = 9.10939 × 10−31 kg , qe = 1.60218 × 10−19 C . and The speed of the electron is v = 0.06 c = 0.06 2.99792 × 108 m/s = 1.79875 × 107 m/s , By conservation of energy 1 me v 2 = −(−qe ) ∆V 2 v2 ∆V = me 2 qe = 9.10939 × 10−31 kg D Of the following elements, identify all that correspond to an equipotential line or surface. 1. both AB and CD 2 1.79875 × 107 m/s × 2 (1.60218 × 10−19 C) 2. line CD only correct 3. line AB only = 919.798 V . 4. neither AB nor CD Explanation: Consider the electric field: C A B − + D An equipotential line or surface (CD) is normal to the electric field lines. 015 (part 1 of 2) 10.0 points The figure below shows two particles, each with a charge of +Q, that are located at the opposite corners of a square of side d. Q + d + Q P What is the direction of the net electric field at point P ? 1. 014 10.0 points hernandez (uh286) – oldmidterm 01 – Turner – (59070) 2. 3. correct 017 10.0 points The two charges Q are fixed at the vertices of an equilateral triangle with sides of length a as shown. q a 4. 5. Q Explanation: At point P , the electric fields due to the two positive charged particles have the same magnitude. One points downward, and the other horizontally to the left. Thus the direction of the net electric field points left and downward, forming a 45◦ angle with the horizontal. 016 (part 2 of 2) 10.0 points What is the potential energy of a particle of charge q that is held at point P ? √ 2 qQ 1. U = 2 π ǫ0 d 2. U = 0 a a Q The work required to move a charge q from the other vertex to the center of the line joining the fixed charges is √ 2kQq 1. W = a kQq 2. W = a 4kQq 3. W = a 4. W = 0 6kQq a 2kQq correct 6. W = a Explanation: 5. W = √ 2 qQ 4 π ǫ0 d 1 qQ 4. U = correct 2 π ǫ0 d 1 qQ 5. U = 4 π ǫ0 d Explanation: The potential energy of the particle charged +q is the sum of the potential energies due to the two +Q charged particles. 3. U = 9 U = U1 + U2 1 qQ 1 qQ = + 4 π ǫ0 d 4 π ǫ0 d 1 qQ . = 2 π ǫ0 d Qq Qq Qq +k = +2 k a a a Qq Qq Qq U2 = +k a + k a = +4 k a 2 2 Qq W = U2 − U1 = 2 k . a U1 = +k 018 10.0 points If the potential in a region is given by the function V = 2 x − y 2 − cos(z) , what is the y-component of the electric field at the point P = (x′ , y ′ , z ′ )? hernandez (uh286) – oldmidterm 01 – Turner – (59070) 1. Ey = −2 y ′ Let : λ = 2.4 µC/m = 2.4 × 10−6 C/m , R = 2.4 m , 2 R = 4.8 m , and ke = 8.98755 × 109 N · m2 /C2 . 2. Ey = − sin(z ′ ) 3. Ey = 10 y ′3 3 4. Ey = 2 R 5. Ey = y ′2 p 2R 6. Ey = 2 x′ 7. Ey = cos(z ′ ) 8. Ey = 2 y ′ correct 9. Ey = sin(z ′ ) x′2 4 Explanation: The electric field is the gradient of the potential, so the y-component of the E-field evaluated at P is 10. Ey = ∂V Ey = − ∂y ∂ 2 x − y 2 − cos(x) =− ∂y = − [−2 y] = 2y. 2R Let p be the origin. Consider the potential due to the line of charge to the right of p. Z Vright = dV Z dq = ke r Z 3R λdx = ke x R 3R = ke λ ln x = ke λ ln 3 . R By symmetry, the contribution from the line of charge to the left of p is the same. The contribution from the semicircle is Z π λRdθ Vsemi = ke R 0 Z π = ke λ dθ 0 π 019 10.0 points A wire that has a uniform linear charge density of 2.4 µC/m is bent into the shape as shown below, with radius 2.4 m. 2.4 m 4.8 m p 4.8 m The Coulomb constant is 8.98755 × 109 N · m2 /C2 . Find the electrical potential at point p. Correct answer: 1.15159 × 105 V. Explanation: = ke λ θ 0 = ke λ π . Hence the electric potential at p is Vp = Vright + Vlef t + Vsemi = 2 ke λ ln 3 + ke λ π = ke λ (2 ln 3 + π) = (8.98755 × 109 N · m2 /C2 ) × (2.4 × 10−6 C/m) × (2 ln 3 + π) = 1.15159 × 105 V .