Physics 121 Common Exam 1, Sample Exam 4 (Fall 2011)

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1 Physics 11 Common Exam 1, Sample Exam 4 (Fall 011) Name (Print): 4 Digit ID: Section: Honors Code Pledge: For ethical and fairness reasons we are all pledged to comply with the provisions of the NJIT Academic Honor Code. You must answer the exam questions entirely by yourself. Turn off all cell phones, pagers, or similar electronic devices. Use only your own calculator. Instructions: First, write your name and section number on both the Scantron card and this exam sheet. Use the formula sheet (last exam sheet) and no other materials. udget your time to about 4.4 minutes/question. (80 minutes/18 questions). The questions are all worth the same. You may skip a question that you find hard, since the professors will drop wrong answers: 16 correct answers will earn a score of 100%. The last two questions may be harder than the others. You must show how you got your answer on this set of exam sheets, including using the back if necessary. The professors may assume that answers with no work are guesses and indicate no understanding. Answer each question on the Scantron card using # pencil. Do not hesitate to ask for clarification of any exam question, if needed, from your proctor or Professor. 1. If vector A = 1i - 16j and = -4i + 10j, what is the magnitude of the vector C = A? A) 4 ) C) 90 D) 64 E) 13. The dot (scalar) product A. of two vectors A = 0i + 3j 4k and = 4i - 7j + 0k is: A) -16 ) 40 C) 3 D) 0 E) A vector is added to the vector C = 3i + 4j to yield a resultant vector that points in the + y direction and has a magnitude equal to that of C. What is the magnitude of? A) 6.3 ) 9.5 C) 18 D) 5 E) 3. Page 1 of 7

2 4. Which statement accurately describes the forces acting on the objects shown in the sketch? Object A has a negative charge of -Q coulombs. Object has a positive charge of Q coulombs. A - + A) The force on A is to the right and is the half the magnitude of the force acting on. ) The force on A is to the left and twice the magnitude of the force acting on. C) The force on A is to the right and twice the magnitude of the force acting on. D) The net force on A is to the left and has the same magnitude as the force acting on. E) The net force on A is to the right and has the same magnitude as the force acting on 5. In the sketch, Q = 30 µc, q = 5.0 µc, and d = 30 cm. What is the magnitude of the electrostatic force on q? d d A) 15 N ) 3N C) 38 N D) 7.5 N. E) 0 N, the charge is in equilibrium Q q Q 6. Two charges of +3µC and -3µC are located at a distance 1 cm from each other. The force between them is (approximately): A) 9 x 10 9 N, attraction ) 0.56 N, attraction C) 5.6 N, attraction D) N, repulsion E) 0.56 N, repulsion 7. The sketch at right represents field strength near a pair of charges. Which choice below lists the field magnitudes at points A,, C in ascending order? A) C,, A ), A, C C) C,, A D) A,, C E) Insufficient information Page of 7

3 8. A particle whose mass is 6.0 grams and whose charge is 5 nc is released from rest when it is 0 cm from a second particle of charge 8 µc. Find the magnitude of the particle s initial acceleration. A).0 m/s ) m/s C) 1.5 m/s D) 15 m/s E) 0.75 m/s 9. An electric dipole consists of two equal and opposite charges located on the horizontal axis, as shown in the figure. The direction of the electric field at point P on the equator of the dipole is: A) The electric field is zero at point P. a a ) C) D) E) -Q θ θ +Q P 10. For the electric dipole of the previous problem, assume that the angle θ = 45 degrees, a = 40 cm and the value of Q is 30µC. Calculate (approximately) the magnitude of the electric field at point P, A) The electric field is zero at point P. ) 6.0 x 10 5 N/C C) 8.5 x 10 5 N/C D) 1. x 10 6 N/C E) 1.7 x 10 6 N/C Page 3 of 7

4 11. Referring to the figure, what angle does the electric field vector at the center of the square make with the x -axis? The angle is measured counter-clockwise from the x-axis, as usual. Hint: the problem is simple to solve if you look for cancellations. A) 315 degrees ) 135 degrees C) 90 degrees D) 45 degrees E) Other +Q a a a -Q x -Q a +4Q 1. For the square configuration of the preceding problem, find the magnitude of the electric field in the center of the square. Use the following numbers for approximate calculations: a= cm., Q=1 µc, k e = 9 x 10 9 A) 7.0 x N/C ) 0 N/C C) 9.0 x 10 7 N/C D) 7.0 x 10 7 N/C E) 1.8 x 10 7 N/C 13. An electron moving along the x axis with a constant speed v = 5x10 6 m/s i enters a region of uniform electric field of E = -50 N/C j pointing in the negative y direction. When the electron leaves the field 40 nano-seconds later, what is its vertical displacement y relative to its value on the left in the sketch as it enters the field? The mass of an electron is (approximately) kg. Neglect the acceleration of gravity. A) 3. cm ) 1. cm C) 4. cm D) 5. cm E). m v y Page 4 of 7

5 Use the sketch at right for the following two problems. A spherical shell has a net positive charge of Q = 1.0 C. on it. Its radius R is 1.0 m. The charge is uniformly distributed over the surface of the sphere. 14. What is the magnitude of the electric field at a distance d = 1.0 m. from the surface of the sphere. A) 3.0 x 10 9 N/C ).5 x 10 9 N/C C) 1.0 x 10 1 N/C D) 1.0 x 10 9 N/C E) 0.0 N/C d r R 15. For the same sphere as above, find the magnitude of the electric field at a point that is a distance r = 0.5 m. from the center of the sphere. A) 9.0 x N/C ) 4.8 x N/C C) 3.6 x N/C D) 0.0 N/C E) 1.6 x N/C 16. A small conducting ball has a mass of 5.0 x 10 - kg and a negative charge of 50 µc. What electric field is needed to exactly balance the weight of the ball? (g = 9.8 m/s ) A) 9800 N/C down ) 10 3 N/C down C) 1,000 N/C up D) 9800 N/C up E) 9.8 X 10-3 N/C down Page 5 of 7

6 Extra Credit Problems: 17. Two tiny non-conducting balls have identical charges of q = 1 nc (10-9 C) and identical unknown masses m. They are hanging in equilibrium from non-conducting threads of length L = 1.0 m. Due to electrostatic repulsion the particles are separated from each other by a horizontal distance x = cm. Calculate m, the mass of each ball. A) 4.6 x 10-3 grams ) 0.3 grams C) 0.57 grams D) 1. x 10-3 kg E) 4.6 x 10-4 kg 18. Consider a thin rod of total length L with uniform charge per unit length λ on it as shown in the figure. Using the integral formula for the total electric field due to a continuous charge distribution (see formula sheet), calculate the total electric field a distance D from the right edge of the rod (as shown in the figure). The magnitude of the field is given by: 1 λ A) E = πε o D 1 λl ) E = 4 πε o D( D + L) 1 λ C) E = πε L o 1 D + L D) E = λ ln 4πε o D E) None of the above λ L D Page 6 of 7

7 Physics 11 Common Exam 1 Formulas Area of circle = πr Circumference of circle = πr 1 meter = 1000 mm = 100 cm 1 kg = 1000 g Surface area of sphere = 4πr Volume of a sphere = (4/3)πr 3, 1 µc = 10-6 C ε o = 8.85 x 10-1 C /N-m, e = x C, m e = 9.11 x kg r 1 q q r 1 1 Q Point charges: F = rˆ E = rˆ where rˆ is 4 πε 0 r 4 πε 0 r 1 nc = 1 nano-coulomb = 10-9 C a unit vector 1 k e = 9x10 4 πε Superposition: contributions to the field or force from point r n r r r r F net on 1 = F1,i = F1, + F1, 3 + F1, 4+ charges add as vectors at a point of interest i= Shell Theorem (spheres only): mimics point charge outside, inside E or F is zero E = force per unit test charge at a point F = qe F net = ma p 1 Dipole moment: p = qd τ dipole = pxe U dipole = -p.e Eon dipole axis + πε 3 r 0 z dq For continuous charge distributions: E = k e. rˆ (integrate over the distribution) dist r σ = surface charge density E conducting sheet = σ/ε o E non-conducting sheet = σ/ε o E line of charge = (1/πε o )λ/r λ = linear charge density Φ E = electric flux = q enc /ε o = E. da over a Gaussian surface Prefixes: n (nano) = 10-9, µ (micro) = 10-6, m (milli) = for large z v = v o + at x x o = v o t + ½at v = v o + a(x x o ) x x o = ½(v + v o )t Centripetal acceleration: a c = v /r F net = ma τ = I α τ = r X F Vectors: Components: a x = a cos(θ) a y = a sin(θ) a = a x i + a y j a = sqrt[ a x + a y ] θ = tan -1 (a y /a x ) Addition: a + b = c implies c x = a x + b x, c y = a y + b y Dot product: a b = a b cos(φ) = a x b x + a y b y + a z b z unit vectors: i i = j j = k k = 1; i j = i k = j k = 0 Cross product: a x b = a b sin(φ); c = a x b = (a y b z a z b y ) i + (a z b x a x b z ) j + (a x b y a y b x ) k a x b = b x a, a x a = 0 always; c = a x b is perpendicular to a-b plane; if a b then a x b = 0 i x i = j x j = k x k = 0, i x j = k j x k = i k x i = j Page 7 of 7

8 ANSWER KEY: Fall 011 Physics 11 Common Exam 1, Version 1. D 64. E E E The net force on A is to the right and has the same magnitude as the force acting on 5. D 7.5 N 6. C 5.6 N, attraction 7. A C,, A 8. C 1.5 m/s 9. E 10. D 1. x 10 6 N/C degrees 1. C 9.0 x 10 7 N/C 13. A 3. cm x 10 9 N/C 15. D 0.0 N/C 16. A 9800 N/C down grams 1 λl 18. E = 4 πε D( D + L) o Page 8 of 7