Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Giambattista, Ch. 17 4, 10, 14, 17, 21, 28, 31, 33, 37, 40, 52, 57 4.Strategy The work done by the external agent is positive since the potential energy increases. Use Eq. (17-1). Solution Find the work done by the external agent. 10. Strategy Use the principle of superposition and Eq. (17-9). Solution Sum the electric fields at the center due to each charge. Do the same for the potential at the center. 14. Strategy Just outside the surface of the sphere, the electric potential is given by radius of the sphere. Solution Find the electric potential. 17. Strategy Use Eq. (17-9). Solution Find the electric potential at the third corner, B. 21. Strategy Rewrite each unit in terms of kg, m, s, and C. Solution Show that and therefore where r is the 28. (a)Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field is stronger. The electric field always points in the direction of maximum potential decrease. Solution The electric field lines are radial. They begin on the point charge and end on the inner surface of the shell. Then they begin again on the surface and extend to infinity. (b) Strategy Use Eqs. (16-5) and (17-8), and the principle of superposition. Solution For E is that due to the point charge, conductor. For For E once again is that due to the point charge, charge). For since this is inside a For (point since V is continuous, and it is constant in a conductor. For (to preserve continuity). 31.Strategy Use conservation of energy and Eq. (17-7). Solution Find the change in kinetic energy. 33.(a) Strategy The electric field always points in the direction of maximum potential decrease. Electrons, being negatively charged, move in the direction opposite the direction of the electric field; that is, in the direction of potential increase. Solution Since the speed of the electron decreased, it must have traveled in the direction of the electric field, so it moved in the direction of potential decrease; that is, (b) Strategy The kinetic energy of the electron decreased, so its potential energy increased. Use conservation of energy and Eq. (17-7). Solution Compute the potential difference the electron moved through. 37. Strategy and Solution (a) Electrons travel opposite the direction of the electric field, so (b) For a uniform electric field, so (c) Since the electron gains kinetic energy, its potential energy is directed Thus, 40.(a) Strategy Use Eq. (17-10). Solution Compute the maximum potential difference across the capacitor. (b) Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the magnitude of the greatest charge. 52.Strategy Use the definition of capacitance, Eq. (17-14). Solution Compute the capacitance of the capacitor. 57.(a) Strategy Use Eq. (17-18c). Solution Compute the capacitance. (b) Strategy Use Eq. (17-18a). Solution Compute the potential difference.