February 5, 2014 Last Name: First Name: Student number: PHYS 1420 6.0 – Physics with Applications to Life Sciences Test 3AB Answer all questions in the space provided on this test paper. Use pen to write. Calculators are allowed as aid. Appendix: Permittivity   8.85 1012 C 2 / N  m2 Coulomb constant ke  8.99 109 N  m2 / C 2 q1 q2 PE  WAB  qEx xB  xA  r2 Q A A A q PE V  ke C    0 V  VB  VA  r q V Ed   0 d d Capacitors in parallel combination: Ceq  C1  C2  C3  ... E  ke q r2 F  ke Capacitors in series combination: 1 1 1 1     ... Ceq C1 C2 C3 1 1 Q2 2 Energy stored in a capacitor: Energy  QV  CV  2 2 2C v y  v0 y  a y t 1 y  v0 y t  a y t 2 2 2 2 v y  v0 y  2a y y P  P0  gh , Atmospheric pressure: P0 = 1.013x105 Pa. 1 Question 1 (8 marks). Three point charges are placed on the y-axis: a charge q at y = a, a charge -2q at the origin, and a charge q at y = -a. Find the magnitude and a direction of the resulting electric field at the arbitrary point P located on the positive x-axis at the distance x from the origin. Draw the diagram and show all components of the field. Apply the principle of superposition and add the fields as vectors to find the net field. The fields due to each charge are shown below. cos  x x  a2 2 The components of the field:  q  E1  E2  ke  2 2  a x   2q  E3  ke  2  x  E1y  E1 sin ,E2 y   E2 sin so E y  E1y  E2 y  0 x  q  Ex1  E2 x   E1 cos   ke  2 2  2  a  x  a  x 2  , E3 x   E3   x  q  Ex  E1x  E2 x  E3 x  2ke  2 2  2  a  x  a  x 2  2q   ke  2   x   or  1 x E x   k e 2q 2  x a2  x2      k 2 q 1 e    1 32  2  x  1 a2  x2      (in the –x 32     direction). Question 2 (5 marks). How much work would it take to push two protons very slowly from a separation of 2.00x10 -10 m (a typical atomic distance) to 3.00x10-15 m (a typical nuclear distance)? If the protons are both released from rest, at the distance 3.00x10-15 m, how fast are they moving when they reach their original separation? 2 The work required is the change in electrical potential energy. The protons gain speed after being released because their potential energy is converted into kinetic energy. Using the potential energy of a pair of point charges relative to infinity, we have  e2 e2  W  PE  PE2  PE1  ke     r2 r1  Factoring out the e 2 and substituting numbers gives   1 1 14 W  (900  109 N  m2/C2 )(160 1019 C) 2   J   768 10 15 10 3  00  10 m 2  00  10 m   The protons have equal masses; they will have equal speeds and hence equal kinetic energy. 1  PE  KE1  KE2  2 KE  2 mv2   mv2 2   Solving for v gives v PE 7.68 1014 J   6.78 106 m / s 27 m 1.67 10 kg Question 3 (6 marks). A proton is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v0 and is directed at an angle α below the horizontal. (a) Find the maximum distance that the proton descends vertically below its initial elevation. (b) After what horizontal distance does the proton return to its original elevation? (c) Sketch the trajectory of the proton. The direction of the initial velocity is shown below (a is the acceleration due to the electric field). The force exerted by the electric field is constant since the field is uniform and gives the proton a constant acceleration. Apply the constant acceleration equations for the x- and y-components of the motion, just as for projectile motion. The electric field is upward so the electric force on the positively charged proton is upward and has magnitude F  eE. Use coordinates where positive y is downward. Then applying  F  ma to the proton gives that ax  0 and a y  eE/m. In these coordinates the initial velocity has components vx  v0 cos and v y  v0 sin  , as shown. (a)Finding hmax : At y  hmax the y-component of the velocity is zero. v y  0,v0 y  v0 sin  ,a y  eE/m, y  y0  hmax  ? v2y  v02y  2a y ( y  y0 ) y  y0  hmax  v 2y  v02y 2a y v02 sin 2  2(eE/m)  mv02 sin 2  2eE (b) Use the vertical motion to find the time t: y  y0  0,v0 y  v0 sin  ,a y  eE/m,t  ? y  y0  v0 yt  12 a yt 2 3 With y  y0  0 this gives t   2v0 y ay  2(v0 sin  ) 2mv0 sin   eE/m eE Then use the x-component motion to find d: ax  0,v0 x  v0 cos ,t  2mv0 sin  /eE,x  x0  d  ? 2 2  2mv0 sin   mv0 2sin  cos mv0 sin 2  x  x0  v0 xt  12 axt 2 gives d  v0 cos   eE eE eE   (c) The trajectory of the proton: NOTES: In part (a), a y  eE/m  48 1010 m/s2  This is much larger in magnitude than g, the acceleration due to gravity, so it is reasonable to ignore gravity. The motion is just like projectile motion, except that the acceleration is upward rather than downward and has a much different magnitude. hmax and d increase when  or v0 increase and decrease when E increases Question 4 (9 marks) (combined questions A and B versions). 1. A proton moves 10 cm along the direction of an electric field of strength 3.0 N/C. The electrical potential difference between the proton’s initial and ending points is: a. 4.8  1019 V. b. 0.30 V. c. 0.033 V. d. 30 V. ANS: B 2. The unit of electrical potential, the volt, is dimensionally equivalent to: a. JC. b. J/C. c. C/J. d. FC. ANS: B 3. A free electron is in an electric field. With respect to the field, it experiences a force acting: a. parallel. b. anti-parallel (opposite in direction). c. perpendicular. d. along a constant potential line. ANS: B 4. If the distance between two negative point charges is increased by a factor of three, the resultant potential energy is what factor times the initial potential energy? a. 3.0 b. 9.0 c. 1/3 d. 1/9 ANS: C 4 5. Which of the following characteristics are held in common by both gravitational and electrostatic forces when dealing with either point masses or charges? a. inverse square distance law applies b. forces are conservative c. potential energy is a function of distance of separation d. all of the above choices are valid ANS: D 6. Two point charges of values +3.4 and +6.6 C are separated by 0.10 m. What is the electrical potential at the point midway between the two point charges? (ke = 8.99  109 Nm2/C2) a. +1.8  106 V b. 0.90  106 V c. +0.90  106 V d. +3.6  106 V ANS: A 7. Increasing the voltage across the two plates of a capacitor will produce what effect on the capacitor? a. increase charge b. decrease charge c. increase capacitance d. decrease capacitance ANS: A 8. Three capacitors of 1.0, 1.5, and 2.0 F are connected in series. Find the combined capacitance. a. 4.5 F b. 4.0 F c. 2.2 F d. 0.46 F ANS: D 9. Two capacitors with capacitances of 1.5 and 0.25 F, respectively, are connected in parallel. The system is connected to a 50-V battery. What charge accumulates on the 1.5-F capacitor? a. 100 C b. 75 C c. 50 C d. 33 C ANS: B 10. What is the equivalent capacitance between points a and b? All capacitors are 1.0 F. a. 4.0 µF b. 1.7 µF c. 0.60 µF d. 0.25 µF ANS: C 11. If C = 36 µF, determine the equivalent capacitance for the combination shown. a. 36 µF b. 32 µF c. 28 µF d. 24 µF ANS: D 5 12. If C = 10 µF, what is the equivalent capacitance for the combination shown? a. 7.5 µF b. 6.5 µF c. 7.0 µF d. 5.8 µF ANS: D 13. What is the equivalent capacitance of the combination shown? a. 29 µF b. 10 µF c. 40 µF d. 25 µF ANS: B 14. What is the equivalent capacitance of the combination shown? a. 24 µF b. 100 µF c. 12 µF d. 4.6 µF ANS: A 15. By what factor is the total pressure greater at a depth of 850 m in water than at the surface where pressure is one atmosphere? (water density = 1.0  103 kg/m3, 1 atmosphere pressure = 1.01  105 N/m2, and g = 9.8 m/s2) a. 100 b. 83 c. 74 d. 19 ANS: B 16. In a large tank of liquid, the hydrostatic pressure at a given depth is a function of: a. depth. b. surface area. c. liquid density. d. Choices a and c are both valid. ANS: D 6