Two parallel plates (as in procedure 2) are parallel to the floor, distance 1.24

Question

Two parallel plates (as in procedure 2) are parallel to the floor, distance 1.24 cm. The difference in potential between the plates is 146 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.

1) The direction of the field is:
a. upward above the plates, downward below the plates, upward between the plates
b. downward above the plates, upward below the plates, downward between the plates
c. upward above the plates, downward below the plates, downward between the plates
d. downward above the plates, upward below the plates, zero between the plates
e. zero outside of the plates, upward between the plates
f. downward above and below the plates, zero between the plates
g. upward above and below the plates, zero between the plates
h. downward above the plates, downward below the plates, downward between the plates
i. downward above the plates, upward below the plates, upward between the plates
j. zero outside of the plates, downward between the plates
k. upward above and below the plates, downward between the plates
l. upward above the plates, downward below the plates, zero between the plates
m. downward above and below the plates, upward between the plates
n. upward above and below the plates, upward between the plates

2) As you move upward, from the negative plate to the positive plate, the electric field will
a. have the same magnitude and remain in the same direction
b.increase in magnitude, remain in the same direction
c.decrease in magnitude, remain the the same direction
d.decrease in magnitude, reverse direction
e.have the same magnitude, reverse direction
f.increase in magnitude, reverse 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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