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1 8.02 Spring 2013 Exam Two Solutions Problem 1 (25 points): answers without work shown will not be given any credit. A uniformly charged slab of charge per unit volume ! , extends from x = !2 m to x = +2 m (see upper figure). A uniformly charged sheet of charge per unit area ! is located at x = +2 m . The electric potential V (x) for this distribution of charge is plotted in the lower figure Analytically, the potential function is given by expression #(6 V ! m "1 )x + 12 V ; x < "2 m % V (x) = $"(1 V ! m "2 )x 2 + (2 V ! m "1 )x + 8 V ; " 2 m < x < +2 m . %"(6 V ! m "1 )x + 20 V ; x > 2 m & a) Determine a vector expression for the electric field in the regions: (i) x < !2 m , (ii) ! 2 m < x < +2 m , and (iii) x > 2 m . 1

2 ! "V x < !2 m E = ! i = !6 V # m !1i "x ! "V !2 m < x < +2 m E = ! i "x ( = 2 V # m !2 x ! 2 V # m !1 i ) ! "V x > +2 m E = ! i = 6V # m !1 i "x b) Determine ! , the charge per unit volume in the slab. Your answer can include ! 0 . [Hint: use Gausss Law]. We apply Gausss law to a cylinder of height x with its left end at the origin x = 0, with cross sectional area A (see top cylinder in the figure to the right). Since the charge density is constant, the charge within the cylinder is xA . Using the expression above for the electric field in the region 2 m < x < +2 m , we have ! E x=0 = !2 V " m !1i ! ( E x=x = 2 V ! m "2 x " 2 V ! m "1 i ) If we integrate over the surface of the cylinder, we get ! ! ! ! " E ! d A = A(E ! i # E x=0 ! i) = A((2 V ! m #2 x # 2 V ! m #1 ) + (2 V ! m #1 )) = (2 V ! m #2 x) A x=x So Gausss Law gives us (2 V ! m "2 x) A = xA# / $ o % # = (2 V ! m "2 )$ o c) Determine ! , the charge per unit area on the sheet. Your answer can include ! 0 . [Hint: use Gausss Law]. Now apply Gausss law to a cylinder going all the way through the slab, with cross sectional area A (see bottom cylinder in the figure above). The charge inside this volume is A + A d = A ( d + ). Using the expression above for the electric field in the region x < 2 m and x > 2 m , we have ! ! ! ! "# E ! d A = A(" E x

3 So Gausss Law gives us (12V ! m "1 ) A = A( # d + $ ) / % o & # d + $ = (12V ! m "1 )% o Therefore ! = (12V " m #1 )$ o # % d But from above we know that ! = (2 V " m #2 )$ o and d = 4 m , so we must have ! = (12V " m #1 )$ o # ((2 V " m #2 )$ o )(4 m) = (4V " m #1 )$ o . 3

4 Problem 2 (25 points): answers without work shown will not be given any credit. A negatively charged particle of mass m and charge !q is at the origin at t = 0 and ! moving downward with velocity v = !v j . Its subsequent trajectory is shown in the ! sketch. The magnitude of the velocity v = v is always the same, although the direction ! of v changes in time. For the region y < 0 , the magnetic field is uniform with magnitude B1 . For y > 0 , the magnetic field is also uniform but the magnitude is B2 . Ignore any gravitational field. In the region x > 11R , y > 0 , there is also an electric field in addition to the magnetic field. When x = 14R and y > 0 , the particle travels in a straight vertical direction ( + y -direction). (a) Is the direction of the magnetic field for the region y < 0 into or out of the page? Look !at the very beginning of the trajectory, at the origin. There the velocity is down and ! qv ! B1 must be to the right for the charge to move in the circle shown. Therefore, since ! ! ! this is a negative charge, v ! B1 must be to the left, and thus we must have B1 out of the page. (b) Derive an expression for the magnitude B1 of the magnetic field for the region y < 0 in terms of the given quantities, that is in term of q , m , R , and v . In general the relation between q , m , v and the radius a of the circular orbit is set by equating mv 2 / a to qvB , giving mv / a = qB or B = mv / aq In this particular case we have a = 3R so that B1 = mv / 3qR (c) Is the magnetic field in the region y > 0 into or out of the page? ! By the same arguments that we used in (a), we see that we must have B 2 into the page. 4

5 (d) What is the magnitude B2 of that magnetic field in that region in terms of in term of q , m , R , and v ? Here we have a = R so that B2 = mv / qR (e) How long does it take the charged particle to move from the origin to point P located at x = 8R (see figure above) along the x-axis? Give your answer in terms of the given quantities q , m , R , and v as needed. We can do this two ways. One way is to say that the speed of the particle remains constant at v and that the total distance traveled from the origin to P is 2 (3R) / 2 + 2 ( R) / 2 = 4 R . To travel this distance at speed v takes a time TP = 4! R / v . The other way to do this problem is to say that the period of one cyclotron orbit is general is given by T = 2! a / v but from above we have v = qaB / m , so the period is T = 2! m / qB , and thus the half period of the first orbit is T1 / 2 = ! m / qB1 and of the second orbit T2 / 2 = ! m / qB2 , so that the total time to get from the origin to P is T1 T2 m m TP = + = + . 2 2 qB1 qB2 Using our expressions for the magnetic fields from above, this is T1 T2 m m 4 R TP = + = + = , 2 2 q ( mv / 3qR ) q ( mv / qR ) v as before. (f) In the region x > 11R , y > 0 , there is also an electric field in addition to the magnetic field. Determine an expression for the magnitude and direction of the electric field in that region. Express your answer in terms of in terms of the given quantities, that is in term of q , m , R , and v . 5

6 We know that the magnetic field for y > 0 is given by (see (c) and (d) above) ! B 2 = !(mv / qR)k . Because the particle is moving in a straight line, the force must be zero, so we must have ! ! ! ! ! ! mv 2 ( ) q E + v ! B 2 = 0 " E = # v ! B 2 = #$ j ! (#mv / qR)k = qR i so that ! mv 2 E= i . qR 6

7 Problem 3 (25 points): answers without work shown will not be given any credit. A spherical capacitor consists of two thin conducting spherical shells and a spherical shell of dielectric material partially filling the space between the conducting shells. The inner conducting shell has radius a and the outer conducting shell has radius c . The space between is partially filled (from a out to b ) with a material of dielectric constant ! . The space between b and c is empty (vacuum). a) Determine the capacitance. We put a charge +Q on the inner conducting shell and Q on the outer conducting shell. To find the electric field in the region a < r < b we take a Gaussian sphere of this radius and apply Gausss Law in the presence of a dielectric ! ! Q "# ! E " d A = $ o , yielding Q ! Q 4! r 2" Er = $ E a

8 b) When the conducting surfaces of the capacitor are connected to a battery with a potential difference V , determine an expression for the stored energy U . 1 2 oV 2 U = CV = 2 . 2 1 1 1 1 1 b c + a b c) While the battery is still connected, the entire region between the spherical shells is filled with the material of dielectric constant ! . By what amount has the magnitude of the free charge changed on either conducting shell !Q " Q f # Qi ? The original charge with battery connected is 4 oV Qi = CV = . 1 1 1 1 1 + b c a b The final charge with battery connected is 4 oV Q f = CV = . 1 1 a c The difference in the charge 4 oV 4 oV Q Q f Qi = . 1 1 1 1 1 1 1 a c b c + a b 8

9 Problem 4 of 4: (25 points) Concept Questions (Parts A through E) Part A (5 points). A parallel plate capacitor, initially with vacuum between the plates, is connected to a battery and charged to +Q on the upper plate and !Q on the lower plate respectively. The battery is then disconnected and afterwards the area of each plate is increased by the same amount. Which of the following statements is true? a) The electric field in the capacitor increases and the energy stored increases. b) The electric field in the capacitor remains the same and the energy stored increases. c) The electric field in the capacitor decreases and the energy stored increases. d) The electric field in the capacitor increases and the energy stored remains the same. e) Electric field in the capacitor remains the same & the energy stored remains the same. f) The electric field in the capacitor decreases and the energy stored remains the same. g) The electric field in the capacitor increases and the energy stored decreases. h) The electric field in the capacitor remains the same and the energy stored decreases. i) The electric field in the capacitor decreases and the energy stored decreases. Q Solution: The electric field is given by E = = . Since the battery is o o A disconnected, the charge Q remains constant, and with an increase in the area A the Q2 electric field thus goes down. The total energy in the capacitor is U = , with 2C C = o A / d , so since the area goes up the capacitance goes up and at constant Q that means the energy goes down. The correct answer is ___________i_____________. 9

10 Part B (5 points) Two hollow conducting shells are concentric. The outer shell 2 is negatively charged with charge !2Q . The inner shell is neutral. We then place a negatively charged point- like object with charge !Q at the center of the shells. Let VO1 be the potential at O1 and VO2 be the potential at O2. After equilibrium is reached, a) the surface charge at O2 is positive and VO1 > VO2 . b) the surface charge at O2 is positive and VO1 < VO2 . c) the surface charge at O2 is positive and VO1 = VO2 . d) the surface charge at O2 is negative and VO1 > VO2 . e) the surface charge at O2 is negative and VO1 < VO2 . f) the surface charge at O2 is negative and VO1 = VO2 . g) the surface charge at O2 is zero and VO1 > VO2 . h) the surface charge at O2 is zero and VO1 < VO2 . i) the surface charge at O2 is zero and VO1 = VO2 . The correct answer is __________e______________. 10

11 Solution: When we place a charge !Q at the center of the shells, then +Q appears on I1, !Q appears on O1, +Q appears on I2. Because the outer shell 2 is negatively charged with charge !2Q , !3Q appears on O2. The electric field points inward between the shells and so VO1 < VO2 . Because the electric field points towards lower potential. 11

12 Part C (5 points) Consider a hollow cylinder of length L and inner radius a and outer radius b , as shown in the figure above. The material has resistivity !r . Suppose that b = 2a and the length L is much greater than the radius a , ( L >> a ). Note that ln(b / a) = ln 2 = 0.69 . Consider the following two cases. (I) Suppose a potential difference is applied between the ends of the cylinder and produces a current flowing parallel to the axis. (II) Suppose the same potential difference is applied between the inner and outer surfaces so that current flows radially. a) The resistance is larger in case (I). b) The resistance is larger in case (II). c) The resistance is the same in both cases. d) It is impossible to determine without more information. r l r L b r dr b In case I, RI = = . In case II, RII = = r ln , so A (b a 2 2 ) a 2 rL 2 L a r b ln RII 2 L a 1 b 3a 2 = = 2 ( b 2 a 2 ) ln = 2 ln 2 < 1 since a

13 Part D (5 points) Two charged particles of identical charge move in circular orbits in the same uniform magnetic field B = Bo z . Particle 1 has twice the speed of particle 2, v1 = 2v2 . Particle 2 has four times the mass of the particle 1, m2 = 4m1 . a) The radius of the orbit of particle 2 is four times the radius of the orbit of particle 1. b) The radius of the orbit of particle 2 is twice the radius of the orbit of particle 1. c) The radius of the orbit of particle 2 is equal to the radius of the orbit of particle 1. d) The radius of the orbit of particle 2 is one half the radius of the orbit of particle 1. e) The radius of the orbit of particle 2 is one quarter the radius of the orbit of particle 1. Solution: Newtons Second Law for each charged object is m1v12 mv qv1 B = ! r1 = 1 1 r1 qB m2 v22 mv qv2 B = ! r2 = 2 2 r1 qB Therefore r2 m2 v2 = r1 m1v1 Because v1 = 2v2 , and m2 = 4m1 , the ratio of the radii of the orbits is r2 4m1v2 = =2 r1 m1 2v2 The correct answer is _________b______________. 13

14 Part E (5 points) At the instant shown in the figure to the right, an electron at the position shown with charge q = !e is moving with ! velocity v = !vj . A very long wire carries a current in the direction shown in the figure. At the instant depicted, the force on the electron is 1. Zero. 2. in the + i -direction. 3. in the ! i -direction. 4. in the + j -direction. 5. in the ! j -direction. 6. in the +k -direction. 7. in the !k -direction. Solution: The direction of the force on the electron is ! ! dir(Fe ) = dir(!ev(! j) " B wire ) . Only the horizontal portion of the wire contribute to the magnetic field at the position of the electron. The direction of the magnetic field of the wire at the position of the electron is given by ! ! dir(dB wire ) = dir(Id s ! r = dir(Ids(" i) ! ((cos# (" i) + sin # ( j)) = "k . Therefore ! dir(Fe ) = dir(!ev(! j) " (!k)) = ! i . The correct answer is ___________3_____________. 14

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