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3 years ago

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A uniform electric ﬁeld exists in the region between two oppositely charged plane-parallel plates. An electron is released from

rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.20 cm distant from the ﬁrst, in a time interval of
a. Find the magnitude of this electric ﬁeld.
b. Find the speed of the electron when it strikes the second plate

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

Explanation:

given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s

q = 1.6 x 10_19C

using S = at^2/2

acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

From F = ma

F = qE

ma = qE

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

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