18.77. Set Up: With air between the layers, and The energy density in the electric field is The volume of a shell of thickness t and average radius R is The volume of a solid sphere of radius R is With the dielectric present, and Solve: (a) (b) The outer wall of the cell is at higher potential, since it has positive charge. (c) For the cell, and The volume of the cell wall is The total electric field in the cell wall is (d) and 18.78. Set Up: Let a be the initial situation, where the alpha particle is very far from the gold nucleus and has kinetic energy At a the gold nucleus has zero kinetic energy. Let b be at the distance of closest approach, when the distance between the two particles is Conservation of energy says The alpha particle has charge and the gold nucleus has charge Solve: and since at the distance of closest approach the alpha particle has momentar-ily come to rest. since is very large. Conservation of energy gives
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