Two parallel plates (as in procedure 2) are parallel to the floor, distance 3.6

Question

Two parallel plates (as in procedure 2) are parallel to the floor, distance 3.6 cm. The difference in potential between the plates is 467 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.

a) The direction of the field is
downward above the plates, downward below the plates, downward between the plates
upward above the plates, downward below the plates, downward between the plates
downward above and below the plates, zero between the plates
downward above the plates, upward below the plates, downward between the plates
downward above and below the plates, upward between the plates
upward above the plates, downward below the plates, zero between the plates
zero outside of the plates, downward between the plates
upward above and below the plates, downward between the plates
upward above and below the plates, upward between the plates
downward above the plates, upward below the plates, zero between the plates
downward above the plates, upward below the plates, upward between the plates
upward above the plates, downward below the plates, upward between the plates
zero outside of the plates, upward between the plates
upward above and below the plates, zero between the plates

b) As you move upward, from the negative plate to the positive plate, the electric field will
decrease in magnitude, remain the the same direction
increase in magnitude, reverse direction
have the same magnitude, reverse direction
increase in magnitude, remain in the same direction
decrease in magnitude, reverse direction
have the same magnitude and remain in the same direction


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

Given that the potential difference between the plates is V = 467 V distance d = 3.6 cm a ) upward above the plates, downward below the plates, downward between the plates b ) As you move upward, from the negative plate to the positive plate, the electric field will have the same magnitude and remain in the same direction
c ) Given that charge q = 1.6*10^-19 C          Electric field E = V / d                                    = 467 V / 3.6 *10^-2 m                                    = 12.9*10^3 V /m                       Force F = q E                                       = 1.6*10^-19 C * 12.9*10^3 V /m                                       = 20.7 *10^-16 N d ) the electron is rest when it arrives to positive plate the speed is          1/2 m v^2 = q V          1/2 *9.31* 10^-31 kg * v^2 = 1.6*10^-19 * 467 V                           v = 4.0*10^5 m/s                                    

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