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Unformatted text preview: Physics 1B(a): Electricity and Magnetism Chapter 20 Homework Solutions 20 . 1 (a) Conservation of energy is the best way to solve this problem. Initially the proton has zero energy (kinetic or potential) and picks up kinetic energy as it moves through the potential V = V f- V i =- 120 V . E i = 0 = E f = 1 2 m p v 2 p + q V = 1 2 m p v 2 p + ( e )(- 120 V ) or equivalently K i + U i = 0 + qV i = 0 + ( e )(120 V ) = K f + U f = 1 2 m p v 2 p + ( e )(0 V ) since the proton begins at the higher potential (which is the higher volage for positive charges), yielding v p = s (- 2 e V ) m p = s- 2(1 . 6x10- 19 C )(- 120 V ) 1 . 67x10- 27 kg 1 J 1 V C = 1 . 52x10 5 m/s Note that whatever the answer is, when we determine the velocity we should always be taking the square root of a positive quantity. This way, if you incorrectly assign minus signs for positively or negatively charged particles, you can always check yourself when computing energies or velocities. (b) The computation is the same for the electron that is accelerated through the same potential, except we change the mass to that of the electrons. Note the higher starting potential for the electron will be 0 V, since electrons complete motion the opposite way a proton would (by our conventions). The electron ends at the lower voltage of 120 V, because it has the opposite sign of the proton: K i + U i = 0 + (- e )(0 V ) = K f + U f = 1 2 m e v 2 e + (- e )(120 V ) v e = sqrt +2 e (120 V ) m e = 6 . 49x10 6 m/s Although we are not asked to calculate the velocity, the electron moves in the opposite direction from that of the proton. 20 . 3 (a) Recall that the change in the potential energy of a charge is U =- ( qE ) x where x is directed along the electric field lines. Calculating the change in potential energy is the same as calculating -(work done along path taken), so we choose the easiest path for our calculations, going from (0 cm, 0 cm) to (20 cm, 0 cm), then from (20 cm, 0 cm) to (20 cm, 50 1 cm). The first path segment is along the direction of the electric field (pointed in the positive x direction) and so the change in potential energy along that path is U 1 =- (12 . 0x10- 6 )(250 V/m )(0 . 200 m ) =- 6 . 00x10- 4 J while along the second path the change in potential energy is zero since we move perpendicular to the direction of the electric field. So the total change in potential energy is given in the above answer. (b) The change in the electric potential is the change in electric potential energy per unit charge and so is V = U q =- 6 . 00x10- 4 J 12 . 0x10- 6 C =- 50 . V 20 . 5 We use conservation of energy, which leads to the change in the potential energy being equal to-(work done on system): U = q V =- 1 2 m e ( v 2 f- v 2 i ) = . 5(9 . 11x10- 31 kg )((1 . 40x10 5 m/s ) 2- (3 . 70x10 6 m/s ) 2 ) = +6 . 23x10- 18 J q V = (- 1 . 60x10- 19 C ) V = +6 . 23x10- 18 J V =- 38 . 9 V so the final voltage is lower than the initial voltage since...
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