a) Points A and B have electric potentials of 255 V and 129 V, respectively. Whe
ID: 1706688 • Letter: A
Question
a) Points A and B have electric potentials of 255 V and 129 V, respectively. When an electron released from rest at point A arrives at point C its kinetic energy is KEA. When the electron is released from rest at point B, however, its kinetic energy when it reaches point C is KB = 4KEA. What is the electric potential at point C?answer is in V
b) A charge of 2.87 µC is held fixed at the origin. A second charge of 2.87 µC is released from rest at the position (1.15 m, 0.350 m) . If the mass of the second charge is 3.10 g, what is its speed when it moves infinitely far from the origin?
answer is in m/s
Explanation / Answer
Use energy principle. energy of charge 1.6 x 10^-19 at point a where potential is 255V is eV where V = 255 and e is the charge on the electron (i.e.) 1.6x10^-19. When it arrives at C, the potential is say Vc so energy is eVc. Thus, change in energy is eVc - 255e. This loss in electric potential energy corresponds to increase in kinetic energy which is KEA. Similarly, for the second scenario where it is released from point B, eVc - 129e = 4KEA. Solve these equations simultaneously to get the value of Vc. Let me know if you need further guidance. (b) If you recall, potential at infinity is 0. Also, in a field, the potential is (1/4piE)(Q1Q2/r) where E is epsilon and Q1 and Q2 are the value of the charges. r is the distance between them. the distance between them from the graph will be sqrt(1.15^2 + 0.35^2). The value you get minus 0 is the change in energy of the charge. This change in energy is equal to increase in kinetic energy 0.5 m v^2 where v is velocity. Solve for v. Again, let me know if you need help witht he algebra:)
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