An electron (mass m = 9.11 ? 10 -31 kg) is accelerated in the uniform field E (
ID: 1375936 • Letter: A
Question
An electron (mass m = 9.11 ? 10-31 kg) is accelerated in the uniform field E (E = 2.00 ? 104 N/C) between two parallel charged plates. The separation of the plates is 1.05 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, Fig. 16-60.
(a) With what speed does it leave the hole?
m/s
(b) Explain why the gravitational force can be ignored.
mg/qE is on the order of 1015 N.The gravitational force is perpendicular to the electric field.
mg/qE is on the order of 10-15 N.
mg/qE is on the order of 107 N.
mg/qE is on the order of 10-7 N.
An electron (mass m = 9.11 ? 10-31 kg) is accelerated in the uniform field E (E = 2.00 ? 104 N/C) between two parallel charged plates. The separation of the plates is 1.05 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, Fig. 16-60. (a) With what speed does it leave the hole? m/s (b) Explain why the gravitational force can be ignored. mg/qE is on the order of 1015 N. The gravitational force is perpendicular to the electric field. mg/qE is on the order of 10-15 N. mg/qE is on the order of 107 N. mg/qE is on the order of 10-7 N.Explanation / Answer
(a)ma = F = e E so the acceleration a = (1.6 x 10-19) (2.0 x 10 4) / (9.1 x 10 -31) (m/s2).
a = 3.516 x 10 15 m/s2
v = (a)(t), x = 1/2 a t2, so t = [(2)(x) / (a)] 1/2 and v = [(2)(x)(a)]1/2
v = [2(0.0105)(3.516 x 10 15]1/2 (m/s) = 8.5933 x 106 m/s Ans.
(b)gravitational force magnitude = mg = (9.1 x 10 -31 kg) (9.8 m/s2) = 8.9 x 10 -30 N
electric force magnitude = e E = 3.2 x 10 -15 N , so mg << e E and the effects due to gravity are extremely
small.
mg/qE = 8.9 x 10 -30 N / 3.2 x 10 -15 N = 2.78 x 10-15
hence mg/qE is on the order of 10-15 N.
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