Two parallel plates (as in procedure 2) are parallel to the floor, distance 4.17

Question

Two parallel plates (as in procedure 2) are parallel to the floor, distance 4.17 cm. The difference in potential between the plates is 435 V, with the upper plate at the higher potential. Assume the plates are very large (so the electric field is approximately straight lines), and no other charges are nearby

c) An electron (mass 9.11x10-31 kg, charge 1.60x10-19 C) is placed near the negative plate. The force felt by the electron at this point is
N

d) Suppose the electron starts at rest on the negative plate. When it arrives at the positive plate, its speed will be
m/s

Explanation / Answer

V= 435
D=4.17*10-2
E= 435/ 4.17*10-2 = 1.0432 * 104 V/m

F= E*q = 1.0432*104 *1.60x10-19 = 1.669*10-15 N

F = ma

a = F/m = 1.832*1015 m/s2

V= 2aD= 1.236*107 m/s

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