the 3.0a magnitude of 600 N/C, is directed parallel to the positive x-aris, Ire
ID: 1868877 • Letter: T
Question
the 3.0a magnitude of 600 N/C, is directed parallel to the positive x-aris, Ire : what is the change in potential energy of a proton as it moves a. b. 8.0 x 10-17 1.9×10-16 J 0.30 10-21 J d. 500 J ANS: B TOP: 16.1 Potential Difference and Electric Potential PTS: 1 DIF: 2 ray tube is accelerated through a potential difference of 5.0 kV. What kinetic gain in the 6. An electron in a cathode energy does t a. 1.6×10-16J b. 8.0 x 10-16J c. 1.6 x 10-22J d, 8.0 × 10-2 J the e process? (e- 1.6 x 10-"C) ANS: B TOP: 16.1 Potential Difference and Electric Potential PTS: 1 DIF: 2 7. An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 5.0 mm and the potential difference across the plate is 5.0 V, with what velocity does the electron hit the positive plate? (m,-9.1 x 10-3 kg, e 1.6 x 10-1C) a. 2.6x 105 m/s b. 5.3 × 106 m/s c. 1.0x 106 m/s d. 1.3×106 m/s PTS: 1 DIF: 2 ANS: D TOP: 16.1 Potential Difference and Electric Potential 8. A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its h energy was delivered to the circuit? negative terminal. How muc a. 2.2 mJ b. 0.020J c. 0.18J d. 4.5 x 103 j PTS: 1 DIF: 2 ANS: C TOP: 16.1 Potential Difference and Electric Potential Find the electrical potential at 0.15 m from a point charge of 6.0m. (k, a. 5.4 × 104 V 9. 8.99 x 102 Nm?/C) b. 3.6x 103 v c. 2.4 x 106 v d. 1.2 x 10? v PTS: 1 DIF: 2 ANS: B TOP: 16.2 Electric Potential and Potential Energy Due to Point Charges ] 16.3 Potentials and Charged Conductors | 16.4 Equipotential SurfacesExplanation / Answer
Since electric field is negative gradient of potential and the electric field is directed in positive x direction, therefore, the potential is increasing in the negative x direction.
so,
the Potential Energy will be:
U = q[V - V'] = q[Ed]
=> U = 1.6 x 10-19 x [600][3 - 1] = 1.92 x 10-16 J. [option b].
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