38 1 Model Current is defined as the rate at which charge flows across an area of cross section Solve Since the current is the number of electrons per second is Q N e Q t and N t 10 nA 1 0 10 C s C 1 60 10 e 19 8 6 25 10 s 10 1 6 3 10 s 10 1 38 2 Model Assume the fields between the electrodes are uniform and that they are zero outside the electrodes Visualize Please refer to Figures 38 7 and 38 8 Solve a The speed with which a particle can pass without deflection is b The radius of cyclotron motion in a magnetic field is v E B V d B 3 600 V 5 0 10 m 2 0 10 T 3 6 0 10 m s 7 r m v e B 9 11 10 1 60 10 31 19 kg C 7 6 0 10 m 2 0 10 T 3 0 17 m 17 cm 38 3 Model Assume the fields between the electrodes are uniform and that they are zero outside the electrodes Visualize Without the external magnetic field B the electrons will be deflected up toward the positive electrode The magnetic field must therefore be directed out of the page to exert a balancing downward force on the negative electron Solve given by Equation 38 4 The magnitude of the magnetic field is In a crossed field experiment the magnitudes of the electric and magnetic forces on the electron are E v cid 71 B 5 0 10 3 T out of page Thus B V d v 3 200 V 8 0 10 m 5 0 10 m 6 5 0 10 T 3 38 4 Model Assume the electric field E V d between the plates is uniform Visualize Please refer to Figure 38 9 Solve a The mass of the droplet is m drop V 4 3 3 R 885 kg m 3 4 3 0 4 10 m 6 2 37 10 16 kg 2 4 10 16 kg 3 b In order for the upward electric force to balance the gravitational force the charge on the droplet must be q drop m g drop E 2 37 10 16 kg 9 8 m s 2 20 V 11 10 m 3 1 28 10 18 C 1 3 10 18 C c Because the electric force is directed toward the electrode at the higher potential or more positive plate the charge on the droplet is negative The number of surplus electrons is N q droplet e 1 28 10 1 60 10 18 19 C C 8 38 5 Model Assume the electric field E V d between the plates is uniform Visualize Please refer to Figure 38 9 To balance the weight the electric force must be directed toward the upper electrode which is more positive than the lower electrode 4 R q E Solve Since drop 3 is m drop 3 3 R 15 e V d 4 g 3 V the equation 25 V 0 012 m 15 1 60 10 3 860 kg m 9 8 m s 4 3 2 m g drop C 19 1 4163 10 19 3 m R 0 52 m 38 6 Model The charge on an object is an integral multiple of a certain minimum charge value Solve From smallest charge to the largest charge the measured charges on the drops are 2 66 10 19 C 3 99 10 19 C 6 65 10 19 C 9 31 10 19 C and 10 64 10 19 C The differences between these values are 1 33 10 19 C 2 66 10 19 C 2 66 10 19 C and 1 33 10 19 C These differences are a multiple of 1 33 10 19 C Thus the largest value of the fundamental unit of charge that is consistent with the charge measurements is 1 33 10 19 C 38 7 Model The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt Solve a Converting electron volts to joules Using the definition of kinetic energy 1 2K 2 mv b Likewise the speed of the neutron is 100 eV 100 eV 1 60 10 17 J 19 J 1 60 10 1 eV v K 2 m 2 1 60 10 9 11 10 17 J 31 kg 5 9 10 m s 6 v 2 5 0 MeV 1 67 10 kg 27 13 2 8 0 10 1 67 10 27 J kg 3 1 10 m s 7 c The mass of the particle is m K 2 2 v 2 3 34 10 J 1 0 10 m s The mass of the particle is the same as the mass of four protons or two protons and two neutrons It is an alpha particle 2 2 09 MeV 1 0 10 m s 6 69 10 kg 27 13 7 7 2 2 38 8 Model The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt Solve a Converting electron volts to joules Using the definition of kinetic energy 1 2K 2 mv 6 MeV 6 6 10 eV 9 6 10 13 J 19 J 1 6 10 1 eV v K 2 m 13 2 9 6 10 1 67 10 27 J kg 3 4 10 m s 7 b Likewise the speed of the helium atom is v 2 20 MeV 27 4 1 67 10 kg 2 3 20 10 4 1 67 10 12 27 J kg 3 1 10 m s 7 m K 2 2 v 2 1 14 keV 2 0 10 m s 7 2 16 2 1 82 10 2 0 10 m s J 2 7 9 12 10 kg 31 c The mass of the particle is The particle is an electron 38 9 Model The electron volt is a unit of energy It is defined as the energy gained by an electron if it accelerates through a potential difference of 1 volt Solve a The kinetic energy is K 1 2 2 mv 1 2 9 11 10 31 kg 5 0 10 m s 6 1 139 10 17 J 2 1 eV 1 60 10 19 J 71 eV b The potential energy is U 1 4 0 e e 0 10 nm 2 9 0 10 N m C 1 60 10 9 19 2 2 0 10 10 m 9 2 30 10 18 J 14 eV 1 eV 1 60 10 19 J c The figure shows a proton accelerating from rest across a parallel plate capacitor with a potential difference of V 5000 V The energy conservation equation Kf qVf Ki qVi is Kf Ki q Vi Vf 0 J e V e 5000 V 5000 eV 5 0 …
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