Physics 222 Spring 2009 Exam 1 Version D (808898)

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1 Physics 222 Spring 2009 Exam 1 Version D (808898) Question Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly. 1. Question Details Electric Fields (5) [735027] A point charge of 10 µc is located at x = -3.0 cm, and a second point charge of -35 µc is located at x = +4.0 cm. Where should a third charge of +6.0 µc be placed so that the electric field at x = 0 is zero? x*03*0.55 cm w*05*1.82 cm z*03* cm y*05*0.70 cm v*10*1.35 cm The total electric field at x = 0 is given by Coulomb's law for each point charge added as vectors. (8.99e9 Nm 2 /C 2 ) (10e-6 C)/(9 cm 2 ) (i) + (8.99e9 Nm 2 /C 2 ) (-35e-6 C)/(16 cm 2 ) (-i) + (8.99e9 Nm 2 /C 2 ) (6e-6 C)/(x cm) 2 (-i). The direction of each of these vectors r is from the charge to the point at which the electric field is evaluated. The direction of the electric field from the +6.0 µc charge is unknown. However, it is easy to find. Since both of the other charges impart an electric field in the positive x direction, then the electric field here must be in the negative x direction to be able to get a sum of zero. The rest is simply algebra. Note that since r = -i for the +6.0 µc charge, then it must be to the right of zero. Thus, the value of x is positive. 2. Question Details Electric Fields (4) [735021] The diagram shows two point charges. The charge on 1 is 6.9 C and the charge on 2 is 7.6 C. What is the resultant electric force on a 3 nc charge placed at the point r = (5 m) i + (-4 m) j?

2 z*02* N y*03*( N) i + ( N) j x*04*( N) i + ( N) j v*10*( N) i + ( N) j w*05*( N) i + ( N) j The electric force from each charge has a magnitude given by the formula F = k q q 0 /r 2, where q is the charge causing the force and q 0 is the charge receiving the force. The components are then F x = F cos = F x / r and F y = F sin = F y / r. You then simply add the components of the electric force from each charge. Check the direction. The electric force is repulsive for like charges.

3 3. Question Details Electric Force (3) [586136] An electron just above the surface of the earth experiences an electric force of 4.47 it have? (Note: the mass of an electron is 9.11x10-31 kg.) N which points north. What acceleration will v*10*(9.81 m/s 2 ) down + (49.05 m/s 2 ) north w*07*(49.05 m/s 2 ) north y*02*58.86 m/s 2 45 o down and to the north x*03*(9.81 m/s 2 ) down z*01*58.86 m/s 2 You must include both gravitational and electrical force and add them as vectors. The gravitational force has a magnitude of mg and points down. 4. Question Details Electric Potential Difference (3) [742615] Three charges, +q, +Q, and -Q, are placed at the corners of an equilateral triangle as shown in Figure If the potential at the center of the triangle is 0 V, what is the value of +q? x*04*- 3/2Q v*10*0 y*02*+ 2Q w*04*+ 3/2Q z*02*- 2Q Figure The potential from the bottom two charges is V = k (+Q) / r + k (-Q) / r = 0 V. Any charge on the +q point would add to the potential. Therefore, it must be neutral. 5. Question Details Electric Potential Energy (4) [735034] Two spheres are mounted on identical horizontal springs and rest on a frictionless table, as in the drawing. When the spheres are uncharged, the spacing between them is m, and the springs are unstrained. When each sphere has a charge of µc, the spacing doubles. Assuming that the spheres have a negligible diameter, determine the spring constant of the springs. v*10* N/m x*06*3.60 N/m w*07*35.96 N/m

4 y*02*14.38 N/m z*01*16.00 N/m Conservation of energy works well for this solution. Initially, the energy is zero. Thus, the sum of the final energies must also be zero. There are only two types of energy present - elastic potential energy (U elastic ) and electric potential energy (U elastic ). As the balls move, the electric potential energy decreases; whereas, the elastic potential energy increases. The change is such that, k q 2 / r = 2(1/2 K x 2 ) where x is the displacement of each spring and r is the final distance between the balls. Remember that there are 2 springs. Notice that I have made the spring constant K to distinguish it from k = 8.99e9 Nm 2 /C 2. To find x and r, we must carefully consider what happened. Since the spacing of m doubles, then r = 2 ( m) = 0.1 m. The displacement of each spring on the other hand is given by the change in r or x = m. But wait! This displacement takes place on 2 springs each changing an equal amount. Thus each spring compresses x = ( m) / 2 = m. Thus, k q 2 / (0.1 m) = k (0.025 m) 2. We can now solve for k. 6. Question Details Electric Charge (2) [ ] Dr. Mike rubs his hair on a balloon, giving the balloon 2.10 charge on the balloon when he is done? excess electrons. If the balloon was initially neutral, what is the y*01*13.11 nc. w*06*33.64 nc. z*00*2.10 nc x*03* nc. v*10* nc. Take the number of excess electrons (N e = ) and multiply by the charge of each electron (-e = ) C to get a total of Q = nc = nc. Note the charge on an electron is always negative! 7. Question Details Electric Charge (1) [586135] There are 2.7x10 15 protons and 6.3x10 15 electrons in a very small metal ball. What is the total charge of the ball? v*10* C x*03* C z*01*it would depend on how many neutrons are present. y*02* C w*05* C The equation is Q = N p e + N e (-e).

5 8. Question Details Electric Force (2) [ ] Two charges q 1 = 83 nc and q 2 = 122 nc? separated by a distance of 15 mm. What is the electric force of one particle on the other? z*02*i cannot do this without knowing which charge you are finding the force on. w*06* N v*10* N y*05*4.50 x*05* N 10-3 N Use Coulomb's Law. F = k q 1 q 2 / r 2 F = (8.99x10 9 Nm 2 /C 2 ) (83 x 10-9 C) (122 x 10-9 C) / (15 x 10-3 m) 2 = N. 9. Question Details Electric Force (2) [735020] Two charged particles are separated by a distance r and the magnitude of the force between them is F. If I take the same two charged particles and move them apart until the distance between them is 2r, what is the magnitude of the force between them now? w*04*1/2 F v*10*1/4 F z*01*f x*04*4 F y*02*2 F The magnitude of the force between the charged particles is given by Coulomb's Force Law. F 12 = k q 1 q 2 / r This tells us that the force is inversely proportional to the distance between the particles squared. In other words, a larger distance means a smaller force and the force is not just twice as small but 2 2 times as small. Thus, it is 1/4 F. 10. Question Details Electric Charge (2) [ ] A metal conductor has C charge. How many excess protons does it have? v*10* protons. z*00*3.40 w*04*5.45 x*02*-2.12 y*01* protons protons protons protons.

6 Take the total change (Q = C) and divide it by the charge of each proton (e = ) C to get a total of N p = excess protons. Assignment Details Name (AID): Physics 222 Spring 2009 Exam 1 Version D (808898) Submissions Allowed: 100 Category: Exam Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsu.edu ) Last Saved: Feb 8, :54 PM MST Permission: Protected Randomization: Person Which graded: Last Feedback Settings Before due date Nothing After due date Nothing

Physics 212 Spring 2009 Exam 1 Version A (814723)

Physics 212 Spring 2009 Exam 1 Version A (814723) Question 1 2 3 4 5 6 7 8 9 10 Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form.