Answer:
Explanation:
- given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
- acceleration = 0.032 X 2 /(1.30×10−8)^2
a = 3.79 x 10^14m/s^2
E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19
E = magnitude of this electric field. = 2156.3N/C
b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as
= 2 X 3.79 x 10^14 X 0.032
= 4.92 X 10^6m/s
Answer:
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Explanation:
