Physics 114 Exam 1 Fall 015 Name: For grading purposes (do not write here): Question 1. 1... 3. 3. Problem Answer each of the following questions and each of the problems. Points for each question and problem are indicated in red with the amount being spread equally among parts (a,b,c etc). Be sure to show all your work. Use the back of the pages if necessary.
Question 1. (10 points) (a) Which one of the diagrams below is not a possible electric field configuration for a region of space which does not contain any charges? Why? a. b. c. d. e. d. Electric Field lines cannot cross (b) Consider a constant electric field given by E=10i N/C. What is the electric field at the point x = 5m, y = 5 m? It is still E=10i N/C (c) Say you now put a 10 nc charge in the same field E=10i N/C at that point x = 5m, y = 5 m. What is the (vector) force on that charge? 9 7 F = qe = 100 10 i C = 10 i C (d) Say you now put a -10 nc charge in the same field E=10i N/C at that point x = 5m, y = 5 m instead of the positive charge. What is the (vector) force on that charge. 9 7 F = qe = -100 10 i C = -10 i C
Question. (10 points). (a) Three pith balls supported by insulating threads hang from a support. We know that ball X is positively charged. When ball X is brought near balls Y and Z without touching them, it attracts Y and repels Z. Since pith is an insulating material, what can we conclude (if anything) about the charges of Y and Z? A positive charge attracts a negative one and repels a positive one. A positively charged object can also attract a neutral object by polarizing the object. Here, molecules will be polarized with more negative charge towards the positive one. Thus, Z is positively charged, but Y can be negatively charged or neutral. (b) How do you charge something by induction using a positively charged rod and what charge does the object have after it is charged? Bring the rod close to the object and it will polarize. Ground the object now electrons flow into the object due to attraction to the positive of the object. Remove the ground Now the object has a net negative charge.
Question 3. (10 points). Consider the charges and surface shown below where Q1 and Q are within the closed surface and Q3 is outside. (a) Which charges contribute to the total flux through the surface? (b) Which contribute to the electric field? Q1 Q3 Q (a) Q1 and Q contribute to total flux through the surface. (b) All three charges contribute to the electric field. For flux Φ E = E da, but we can also think of it in terms of penetrating field lines. Here the flux from Q3 that enters the surface also exits. E =E1 +E +E3, and these are simply vector additions.
Problem 1. (15 points) Two point charges lie along x-axis. The charge q1 = C at x = 0.1 m and the charge q = - 4 C is at x = 0 m (that is, at the origin). (a) Find the resultant electric field at the point P which is at (0, 0.)m. (b) Calculate the force on a nc charge if it were placed at point P. (0,0.) y P q1 q (0,0) (0.1,0) x E q 4 0r. Use SOHCAHTOA. See 1 is 180 degress - arctan(/1) = 116.56 degrees. = 70 degrees r1 = 0.4 m and r = 0. m 9 (9 10 )() 11 E1 3.6 10 N/C 0.05 9 (9 10 )(4) 11 E 9 10 N/C 0.04 (a) E total = E 1 + E = (E) 1 cos(116.56) + E cos(70))(i ) + (E 1 sin(116.56) + E sin(70))j 11 11 1.61 10 i (3. 9) 10 j N/C 11 11 1.61 10 i 5.78 10 j (b) F = q test E = 10 9 C( 1.61 10 11 (i ) 5.78 10 11 (j ))N = [ 3. 10 (i ) 1.16 10 3 (j )]N
Problem. (15 points) (a)the electric field in the region of space shown is given by E = 3i N/C where y is in m. What is the magnitude of the electric flux through the top face of the cube shown? (b) Now say the field is replaced with a new one given by E = 3j N/C. Now what is the magnitude of the electric flux through the top face of the cube? (c) Now say the field is replaced with a new one given by. Now what is the magnitude of the electric flux through the top face of the cube? (a) E. da EAcos( ). The normal (and A vector) points in the positive y-direction. Thus, is 90 degrees so the flux is zero. (b) Now is 0 degrees so the flux is EA = (3)(9) = 7 Nm /C (c) Φ E =E A =(8i +j ) (9j )=18 Nm /C. You can also figure out what is (the angle of E with the normal). Tan( ) = 4, degrees. The flux is ( 64 4)(9)cos(76) 18 Nm /C
Problem 3. (15 points) A metal spherical shell has an inner diameter of 0.8 m (radius 0.4 m) and an outer diameter of 1 m. 10 nc of charge are added to the spherical shell. A 5 nc point charge is located at the center of the sphere. Use Gauss s law to determine the electric field at (a) 0. m, (b) 0.45 m, and (c) m from the center of the spherical shell. (d) How much charge is on the inner surface of the spherical shell? (e) How much charge is on the outer surface of the spherical shell. E. da qenc / 0. We choose spherical Gaussian surfaces for parts a-c. Due to the symmetry we see that the integral will always give EA = (E)(4 r ), where r is the radius of the Gaussian sphere which will be that where we want to find E. 9 5 10 1 (a) q enc= 5 10-9 C (the point charge) so E = 45 / 0.04 115 N/C 4 0 0. (b) E = 0 since this point is withn the metal (conducting) shell 9 15 10 1 (c) q enc= 15 10-9 C (the point charge and the sphere) so E = 135 / 4 33.75 N/C 4 0 (d) Sine E = 0 in the metal, if we draw a spherical Gaussian surface of radius (say) 0.45 m, we need to have q enclosed = 0. Thus, -5 nc has to be on the inner surface. (e) Since charge is conserved, and we started with 10 nc on the shell, there must be 15 nc on the outer surface.
Possibly Useful Information 1 q1 q F 885. X 10-1 ( C / N m ) 0 4 0 r e = 1.6 X 10-19 C E F q 0 q E 0 0 E. da q enc 4 0r 1 9 k e 9 10 N m / C 4 r = xi + yj + zk 0