( )( 10!12 ( 0.01) 2 2 = 624 ( ) Exam 1 Solutions. Phy 2049 Fall 2011


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1 Phy 49 Fall 11 Solutions 1. Three charges form an equilateral triangle of side length d = 1 cm. The top charge is q =  4 μc, while the bottom two are q1 = q = +1 μc. What is the magnitude of the net force acting on q? (1) 64 N () 1 N () 7 N (4) 6 N (5) N By symmetry the net force will be aligned vertically downward (attractive force). So we need to project the force from each of the two lower forces onto the vertical axis: F net = k q 1q d cos! + k q q d cos! ( ) ( 4 )(!1"6 1!1 "6 ) = 9!1 9 (.1) = 64. Four charges are evenly spaced along the x axis with a separation distance d = cm. The values of the charges are: q1 = + μc, q =  1 μc, q = +1 μc, and q4 = + μc. What is the net electrostatic force along the x  axis acting on charge q1 due to the other charges? (1) 8. i N ()  6 i N () i N (4) i N (5) i N Carefully work out the magnitude and direction of the force between charge q1 and the others: F 1 = F 1 + F 1 + F 14 = +k q q 1 î! k q q 1 d d ( )( 1!1 ) = 9 "19. ( ) î! k q q 1 4 ( d) î # %! ( ) 4! 6 $ 9 & ' ( î = 8. î N
2 Phy 49 Fall 11. Two charged particles are fixed to the x  axis: particle 1 of charge q1 = μc at x =. m, and particle of charge q =  8 μc at x =.8 m. At what coordinate along the x axis is the net electric field produced by the particles equal to zero? (1) .4 m ().4 m () 1.4 m (4) m (5) .8 m..8 x The above diagram shows the direction and approximate magnitude of the electric field contribution from each charge. Given the relative strengths, the only place where the field can go to zero is to the left of x=.. E x =!k " q q 1 = q 1.! x " ± =.8! x.! x " x =!.4 ( ) + k q (.8! x) = (.8! x).! x ( ) 4. Two charges are separated by a distance d = 1 m along the y  axis, bisected by the x  axis. The upper charge has q1 = + x 19 C and the lower charge has q =  x 19 C. At position P along the x  axis, which is equidistant from each charge by distance d, what is the y component of the net electric field? (1) .5 N/C ()  45 N/C () N/C (4) 4.5 N/C (5) 7 N/C
3 Phy 49 Fall 11 The diagram shows the direction of the electric field contribution from each charge. The y component of the field from each charge is: E 1y =!k q 1 sin! d E y =!k q sin! d E y = E 1y + E y =! k ( q d 1 + q ) =!.5 5. A glass rod forms a semi circle of radius r = 5 cm with a charge of +q distributed uniformly along the upper quadrant and  q distributed along the lower quadrant, where q = 5 pc. What is the magnitude and the direction (as the polar angle relative to the direction of the x  axis) of the electric field at the center P of the semi circle. (1) N/C, θ = 7 () 6 N/C, θ = 18 () N/C (4) N/C, θ = (5) 6 N/C, θ = 9 The diagram shows the direction of the electric field contribution from each quadrant. By symmetry, the net field points down in the! ĵ direction, which is! = 7! The magnitude (not really needed to choose the correct answer!) is given by:
4 Phy 49 Fall 11! = q " R dq =!ds =!Rd# E +q " y = $ k dq R sin# = " % $ k!rd# " / % sin# " / R = +k! R cos# " " /= $k! R & E y tot = $k! R = $ 4k " q R = $ 6. An electron is released from rest from the surface of a large non conducting plate. The surface charge density of the plate is σ=  4 μc/m. Once released, what is the velocity of the electron when it reaches a distance. cm away from the plate? You may ignore gravity. The magnitude of the electric field from the surface of a large non conducting sheet is E =! " Since the sheet is negatively charged, the electron will be accelerated away from it: ma = F = ee a = ee m e From kinematics, we can find: v f! v i = ad v f = eed m e = e" d # m e v f = 1.54 $1 7 m/s 7. A copper wire with a resistivity of! = 1.7 "1 #8 $m has a circular cross section of radius r =.5 cm and a length L = 1 m initially. It is stretched to a length of m while maintaining a constant volume. What is the new total resistance of the stretched wire? Hint: the radius changes. (1).Ω (). Ω ().65 Ω (4).4 Ω (5) 9 Ω
5 Phy 49 Fall 11 V 1 =!r 1 L 1 = V =!r L =!r L 1 " r 1 = r " r = 1 r 1 " R = # L A = # L 1!r = # L 1!r 1 / = 9# L 1!r 1 " R =.$ 8. A light bulb with a tungsten wire filament of resistivity! = 5.5 "1 #8 $m is connected to an EMF source that maintains a potential difference of 1 V. If the light bulb is designed to dissipate 6 W in light and heat, what is the necessary resistance of the filament? (1) 4 Ω () 1 Ω () 6 Ω (4) Ω (5) 5.5x18 Ω We actually do not need to know the resistivity: P = 6 W =! " R =! P R = 1 #1 6 = 4$ 9. In the circuit shown, ε1 = ε = 1 V and R1 = R = R = 1 Ω. What is the magnitude of the current flowing through the middle resistor labeled R1? (1).67 A (). A ().5 A (4) A (5) i i i 1 i 1
6 Phy 49 Fall 11 By symmetry, we have from the junction rule: i = i 1 then by the loop rule (left or right loop):! " ir " i 1 R 1 =! " i 1 R " i 1 R 1 = # i 1 =! ( R 1 + R / ) = 1 15 = A 1. The diagram shows a network of resistors connected to an EMF source ε1. If all resistors have the same resistance R, what is the equivalent resistance Req of the circuit connected to ε1? (1) 1R=1 () 1R=7 () R=9 (4) R (5) 8R The resistors at the bottom left (7, 8) are in parallel and have resistance R/. That is in series with more (1, ) for resistance 5R/. That in turn is in parallel with one (6) for resistance 5R/7. That is in series with one resistor () for 1R/7. That covers the left side of the EMF. All that is in parallel with resistors in series (4, 5) with resistance R, giving an overall equivalent resistance of : 1 = 7 R eq 1R + 1 R = 1 1R! R eq = 1R 1
7 Phy 49 Fall A 1 μf capacitor with an initial potential difference of 1 V is discharged through a resistor when a switch between is closed at t =. At t = s the potential difference across the capacitor is 4 V. What is the resistance of the resistor? (1) 18 kω () 145 kω () 49 kω (4) 67 kω (5) 9 kω For a capacitor that is discharging in an RC circuit, we have: ( ) = q e!t/rc ( ) = V e!t/rc since q = CV ( ) ( ) = 1 4 = e = e/rc!/rc e q t " V t V V t " ln( ) = R = RC C ln( ) = 18k# 1. The flux of the electric field (4 N/C) iˆ+ ( N/C) ˆj+ (16 N/C) kˆ through a. m portion of the yz plane is: (1) 48 N m /C () 4 Nm /C () 4 Nm /C (4) Nm /C (5) 6 Nm /C S: Area A = (m )ˆ i. Electric flux Φ = E A = 4 = 48N m / C. 1. Charge Q is distributed uniformly throughout an insulating sphere of radius R. The magnitude of the electric field at a point R/4 from the center is: S: Make a gauss surface of radius r=r/4 and use gauss law E da = 4πr E 4π 4π r qenc = r ρ; Q == R ρ qenc = Q( ) R 1 Q r Q E = q enc = = 4πε r 4πε R R 16πε R
8 Phy 49 Fall A round wastepaper basket with a.15m radius opening is in a uniform electric field of N/C, perpendicular to the opening. The total flux through the sides and bottom, in N m /C, is: (1) 1 () 4. () (4) 8 (5) can't tell without knowing the areas of the sides and bottom S: There are no charges in the basket because of uniform electric field. Gauss law gives E da + E da = s+ b Φ = s+ b o E da = o E da = E πr = π (.15) = 1N m / C 15. Eight identical spherical raindrops each have a potential V on their surface, relative to the potential far away. They coalesce to make one spherical charged raindrop whose surface potential is: (1) 4V () V= () V (4) V=8 (5) 8V S: Initial spherical raindrop has radius R, charge q and surface potential V. Final large raindrop has radius R, charge 8q and surface potential V f. For spherical symmetry charge distribution, the surface potential is q V =. 4πε R Therefore V (8q) = = 4πε (R) f 4 V 16. Two particle with charges Q and Q are fixed at the vertices of an equilateral triangle with sides of length a. If k = 1/4πε, the work required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges is: (1) kqq/a () kqq/a () kqq/a (4) (5) kqq=a
9 Phy 49 Fall 11 S: Work required is equal to the change of potential energy kq kq kqq W = ΔU = q( V f Vi ) = q[( )] = ( a / ) a a 17. An electron goes from one equipotential surface to another along one of the four paths shown below. Rank the paths according to the work done by the electric field, from least to greatest. (1) 4 and tie, then, then 1 () 4,,, 1 () 1,, 4 and tie (4) 1,,, 4 (5) 4,, 1, S: Work done by the electric field is equal to minus the change of potential energy: W =!q"v = +e"v Electron has negative charge (e). Therefore the works done are +e, +1e, +e +1e for paths 1,,,4 respectively. 18. The plate areas and plate separations of five parallel plate capacitors are capacitor 1: area A, separation d capacitor : area A, separation d capacitor : area A, separation d / capacitor 4: area A /, separation d capacitor 5: area A, separation d / Rank these according to their capacitances, least to greatest. (1) 4, 1 and tie, then 5, () 5, 4,,, 1 () 5, and 4 tie, then 1, (4) 1,,, 4, 5 (5), 5, 1 and tie, then 1, 4
10 Phy 49 Fall 11 S: The capacitances for parallel plate capacitors is ε A/d. Hence the capacitances are ε A/d, ε A/d, 4ε A/d, ε A/4d, ε A/d for capacitors 1,,, 4, 5 respectively. 19. A V battery is connected across capacitors of capacitances C 1 = 4 µf, C =1 µf and C =6 µf as in the figure. Find the charge on capacitor C in µc. (1) 4 () 6 () 8 (4) 6 (5) 48 S: C and C are connected in parallel, C = C + C =16µF. C and C 1 are connected in series, C eq =(16/5) µf. Q eq =C eq V=64µC. Q Q 1 eq 64 V1 = = = = 16V C C1 4 V Q = V 1 = C V = V V 1 = 16 = 4V = 6 4 = 4µ C. An airfilled parallelplate capacitor has a capacitance of pf. The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 4 pf. The dielectric constant of the wax is: (1) 4. ().5 (). (4).5 (5) 8. S: Airfilled parallelplate capacitor has area A and plate separation d. ε A = pf d ε A pf κ = 4 pf κ = 4 pf κ = 4 d
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