As customary, choice (a) is the correct answer in all the following problems.

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1 PHY2049 Summer 2012 Instructor: Francisco Rojas Exam 1 As customary, choice (a) is the correct answer in all the following problems. Problem 1 A uniformly charge (thin) non-conucting ro is locate on the central axis a istance b from the center of an uniformly charge non-conucting isk. The length of the ro is L an has a linear charge ensity λ. The isk has raius a an a surface charge ensity σ. The total force among these two objects is (1) F = λσ (L+ a 2 +b 2 (b+l) 2 +a 2 )ˆk (2) F = λσa2 L ˆk 4ɛ 0 b 2 (3) F = λσa2 L ˆk 8ɛ 0 b 2 (4) F ( = σ 1 (5) F = λσl2 L 2 +a 2ˆk L L 2 +a 2 )ˆk We saw in class that the electric fiel create at any point along the central axis is given by E(z) = σ ( z 1 )ˆk a 2 +z 2 Breaking up the ro into an infinite number of infinitesimally small point charges q, we have that the net force on each tiny charge is F = q E(z). Summing up all these 1

2 contributions, an using the fact that q = λz gives F = q σ ( 1 b+l ( z 1 Problem 2 = λσ ˆk = λσ ˆk b z a 2 +z 2 )ˆk (1) (z a 2 +z 2 ) b+l ) z a 2 +z 2 b = λσ ( L a 2 +(b+l) 2 + a 2 +b 2 )ˆk (4) A uniformly charge (thin) non-conucting shell (hollow sphere) of raius R with the total positive charge Q is place at a istance away from an infinite non-conucting sheet carrying a uniformly istribute positive charge with a ensity σ. The istance is measure from shell s center (point O). What is the magnitue of the total electric fiel at the center of the shell? (2) (3) (1) σ (2) Q 4πɛ 0 R 2 + σ (3) Q 4πɛ 0 R 2 + σ ɛ 0 (4) Q 4πɛ 0 R 2 (5) Q 4πɛ 0 R + σ ɛ 0 This is problem is very easy to solve if one recalls the superposition principle. The total electric fiel at any point in space is equal to the sum of the iniviual contributions from each source. The electric fiel prouce by the sphere in its interior is always zero 1. The electric fiel prouce by a non-conucting infinitely long sheet is σ everywhere in space. Therefore, the sum of these two contributions at the center of the sphere is simply σ. 1 This can be seen as a consequence of Gauss Law for this spherically symmetric situation 2

3 Problem 3 A roun wastepaper basket with a 0.15 m raius opening is in a uniform electric fiel of 300 N/C, perpenicular to the opening. The total flux through the sies an bottom, in N m 2 /C, is: (1) - 21 (2) 4.2 (3) 0 (4) 280 (5) can t tell without knowing the areas of the sies an bottom Because the electric fiel is a constant everywhere, the electric flux through any close surface is zero. Thus, we can write Φ = E A = 0 Decomposing the entire surface of the basket into the sies, bottom, an top, yiels Φ = E A = E A+ E A+ E A = 0 sies bottom top thus Therefore sies E A+ sies E A = bottom E A+ bottom top E A (5) = EA top (6) = Eπr 2 (7) = = 300π (8) = 21 (9) E A = 21N/Cm 2 3

4 Problem 4 An electric fiel given by E = 10î 5(y 2 + 5)ĵ pierces the Gaussian cube of the figure, where the cube is 2 m on a sie. (E is in newtons per coulomb an y is in meters.) What is the net electric flux through the entire cube? (1) -80 N/C m 2 (2) 80 N/C m 2 (3) 0 (4) 20 N/C m 2 (5) -20 N/c m 2 We can break up the entire close surface into the six sies of the cube. The front an back sies o not contribute since E lies on the x,y plane, thus it is parallel to these sies. Also, the x component E x = 10 of the electric fiel is constant. This implies that the contributions from this component woul cancel in each other out among all sies. The same will happen with the constant part in E y = 5y Therefore, we en up with Problem 5 Φ = 5 y 2 xz +5 y 2 xz = 5 (2) 2 A = 5 (2) = 80 y=2 y=0 A graph of the x component of the electric fiel as a function of x in a region of space is shown in the figure. The scale of the vertical axis is set by E xs = 16.0 N/C. The y an z components of the electric fiel are zero in this region. If the electric potential at the origin is 10 V, what is the electric potential (in V) at x = 4.0 m? (1) 26 (2) -6 (3) 36 (4) 0 (5) 42 4

5 By efinition we have b V b V a = E s a Where a an b are just two points in which we measure the electric potential V. Since we are given that V(x = 0) = 10 V, we might as well use that as our point a, an x = 4 as b to fin V b. Since the y an z components of the electric fiel are zero everywhere, we have E s = E x x, an E x is given by the plot as a function of x. Therefore, we can write V b = V a b a E x x (10) But the integral above is simply the area uner the curve E x vs. x. We just nee to be careful with the sign of that area since it is negative in the range 0 x 3, an positive in the range x > 4. Thus, V(x = 4) = V(x = 0) 4 0 E x x (11) = 10 ( 24+8) = 26 (12) Problem 6 In the figure, a charge particle (either an electron or a proton; you nee to fin out which it is) is moving rightwar between two parallel charge plates. The plate potentials are V 1 = 25 V an V 2 = 35 V. The particle is slowing own from an initial spee of m/s at the left plate. What is its spee, in m/s, just as it reaches plate 2? (1) (2) (3) not possible to know without knowing the plates separation (4) (5) If we use conservation of energy (potential plus kinetic), this problem is really straighforwar. The potential energy of the particle when it starts from plate 1 is U 1 = qv 1 an 5

6 when it arrives at plate 2 is U 2 = qv 2. (Recall that the electric potential is a continuous function in space, therefore a particle very close to the plates will be at an electric potential equal to the one on the plate 2 ). Therefore, conservation of energy reas U 1 +K 1 = U 2 +K 2 (13) qv mv2 1 = qv mv2 2 (14) Solving for v 2 gives v 2 = v q m (V 1 V 2 ) (15) But now it comes a crucial point. What m o we use? The mass of a proton or that of an electron? This is very important since their masses iffer by a factor of almost 2000! To clarify this, we recall that E = V. This implies that the electric fiel, at a certain point in space, points in the opposite irection of V at that location. Imagine we raw an x axis going from plate 1 to plate 2. Since the electric fiel is constant everywhere between the plates, i.e. E = (Ex,0,0) where E x = constant, the potential V(x) is a monotonic function of x in going from one plate to the other. Thus, since the potential at plate 1 is V 1 = 25 V an ecreases to V 2 = 35 V at plate 2, V is negative along the entire range of x. In other wors, V points to negative irection of x (left). Therefore, from E = V we arrive at the conclusion that E points in the positive irection on x (right). Now, since the particle is slowing own, the total force on it must be in the opposite irection of motion. Since the particle is traveling to the right, the net force ought to point to the left. From F = q E, we see that this is only possible if q is negative since E points to the right. Therefore, the particle is an electron. Using then the electron s mass m e = kgs, an charge q = C, we have v 2 = v q m (V 1 V 2 ) = m/s (16) 2 Notice that since E = V, the electric potential V must be a continuous function. Otherwise it woul imply an infinite E. 6

7 Problem 7 The figure shows a parallel-plate capacitor of plate area A an plate separation 2. The left half of the gap is fille with material of ielectric constant κ 1 = 12; the top of the right half is fille with material of ielectric constant κ 2 = 20; the bottom of the right half is fille with material of ielectric constant κ 3 = 30. What is the capacitance in terms of ɛ 0, A, an? (1) 9 ɛ 0A (2) 62 ɛ 0A (3) 18 ɛ 0A (4) 31 ɛ 0A (5) none of these Assuming that the separation between the plates is much smaller than their extension, we can ignore fringe effects at the eges of the plates an at the junction of the two materials. Since the plates of a capacitor are mae of conucting material, the plates are equipotential surfaces. This an the first statement allows us to consier this system as mae of capacitor three capacitors C 1,C 2 an C 3 where C 2 an C 3 are in series, an this combination is in parallel with C 1. Thus With C eq = C 1 + C 2C 3 C 2 +C 3 (17) C 1 = κ 1 ɛ 0 A/2 2 C 2 = κ 2 ɛ 0 A/2 an C 3 = κ 3 ɛ 0 A/2 (18) we have C eq = ɛ 0A 2 = 9 ɛ 0A ( κ1 2 + κ ) 2κ 3 κ 2 +κ 3 (19) (20) 7

8 Problem 8 In the figure shown, a potential iference of V = 10 V is applie across the arrangement of capacitors with capacitances of C 1 = C 2 = 4µF, an C 3 = 6µF. What is the charge q 2 on capacitor C 2? (1) 20µC (2) 40µC (3) 60µC (4) 80µC (5) 10µC Capacitors C 1 an C 2 are in series, thus, share the same charge. If V 1 an V 2 are the voltages across each of them, we have V = V 1 +V 2 (21) 10 = Q/C 1 +Q/C 2 (22) Since C 1 = C 2 C = 4µF, we get Q = CV 2 = 20µC (23) Problem 9 What is the minimum mechanical work that has to be one on the charge q = 1µC in orer to bring it from point a to point b? In figure, the soli sphere of charge Q = 2µC with a raius R = 2m is hel fixe in space. Point a is locate at 12m from the center of the sphere an point b at 10m as shown. (1) J (2) J (3) J (4) J (5) J 8

9 Since both charges have the same sign, an we are bringing the charge q closer to Q, we immeiately know that we have to make a positive work ue to the electric repulsion. Inee, the work is W = q(v b V a ) = q ( kq kq ) r b r a = ( ) (24) (25) = J (26) Problem 10 The figure shows a non-conucting (thin) isk with a hole. The raius of the isk is b an the raius of the hole is a. A total charge Q is uniformly istribute on its surface. Assuming that the electric potential at infinity is zero, what is the electric potential at the center of the isk? (1) 2kQ b+a 2kQ (2) b a (3) 2kQ (4) 0 (5) kq b 2 a 2 b 2 We saw in class that the potential prouce by a charge isk of raius R, at a istance z from it, along its central axis was σ ( ) z 2ɛ 2 +R 2 z 0 By supersposition, we can think of the potential create by isk with the hole as the sum of two isks, with the same but opposite surface ensities: V(z) = σ ( ) z 2ɛ 2 +b 2 z 0 = σ ( z 2 +b 2 z 2 +a 2 ) Since we are only intereste at the center, we have σ ( z 2 +a 2 z ) (27) (28) (29) V(0) = σ (b a) (30) The total area of the isk with the hole is A = π(b 2 a 2 ), thus σ = Q π(b 2 a 2 ) 9

10 Finally V(0) = 1 (b a) 2kQ(b a) = (b a)(b+a) = 2kQ b+a Q π(b 2 a 2 ) (31) (32) (33) Problem 11 A wire segment of length L has constant linear charge ensity λ > 0. Which of the following expressions gives the magnitue of the electric fiel a istance D from the center of the wire (see figure)? L/2 (1) kλd L/2 (4) 0 (5) kλd x (D 2 +x 2 ) 3/2 (2) kλd L/2 L/2 x D +x L 0 x D 2 +x 2 L x (3) kλd D 2 +x 2 0 Putting the ro along the x axis with its center at the origin, the problem boils own to compute E at the point (x,y) = (0,D) The contribution from the segment x is E q = k D 2 +x 2 = kλ x D 2 +x 2 (34) 10

11 Also, we see immeiately that the x-component of the total electric fiel will be zero ue to mutual cancellations among mirror-symmetric segments of the ro. Thus, we only nee the y-component, therefore E y = E D D 2 +x = kλd x 2 (D 2 +x 2 ) 3/2 (35) To get the total electric fiel, we simply a up all of these contributions, hence Problem 12 E y = L/2 x E y = kλd L/2 (D 2 +x 2 ) 3/2 (36) A charge Q is place in the center of a shell of raius R. The flux of electric fiel through the shell surface is Φ 0. What is the new flux through the shell surface, if its raius is ouble? (1) Φ 0 (2) 2Φ 0 (3) 4Φ 0 (4) Φ 0 /2 (5) Φ 0 /4 Gauss law states that the electric flux through a close surface is Φ E s = Q enc (37) ɛ 0 from where we see that the flux only cares about the total charge enclose by the surface. By increasing the raius of the sphere we are merely increasing the size of the surface, but the enclose charge remains the same. Therefore the new flux is just the ol Φ 0. 11

12 Problem 13 Two very small spheres have equal masses m, carry charges of the same sign an value q, an hang on strings of length L as shown in figure. Due to the repulsive force, the spheres are separate by some istance. Fin this istance. Assume that L so that you can use the approximation tanα sinα α (1) 3 2L q2 k mg (2) 3 L q2 k mg (3) 2L q2 k mg (4) L q2 k Lq mg (5) 2 k 2mg To achieve equilibrium, we see from the figure that we nee Now, combining these two equations as T cosα = mg an T sinα = F e (38) tanα = F e mg (39) 12

13 an using the Coulomb s force among the two small spheres F e = kq 2 / 2, gives tanα = kq2 2 mg (40) Finally, for small values for α, tanα sinα = /2 L, we have from where we obtain Problem 14 /2 L kq2 2 mg 3 2L kq2 mg (41) (42) In figure, how much charge is store on the parallel-plate capacitors by the 10 V battery? One is fille with air, an the other is fille with a ielectric for which κ = 2.0; both capacitors have a plate area of m 2 an a plate separation of 1.00 mm. (1) 0.53 nc (2) 0.35 nc (3) 1.06 nc (4) 0.53 µc (5) 0.35 µc The charge store on capacitor C 1 is q 1 = C 1 V = κ ɛ 0A V = = an on capacitor C 2 is q 2 = C 2 V = ɛ 0A V = = Thus, the total is q tot = = 0.53nC 13