Question

Each of the electrons in a particle beam has a kinetic energy of 1.58 ✕ 10⁻¹⁷ J.

(a) What is the magnitude of the uniform electric field (pointing in the direction of the electrons' movement) that will stop these electrons in a distance of 10.5 cm?
N/C

(b) How long will it take to stop the electrons?
ns

(c) After the electrons stop, the electric field will continue to act on them, causing the electrons to accelerate in a direction

at what rate?

Answer 1

a)

as we know

kE = q E d

1.58* 10^-17 = 1.6* 10^-19* E * 0.105

E = 940.48 N/C

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b)

acceleration acts on electrons

a = qE / m

a = 1.6* 10^-19 * 940.48 / (9.11* 10^-31)

a = 1.652* 10^14 m/s^2

using 1st equation of motion

t = v/a

t = sqrt (2ad) / a

t = sqrt (2 * 0.105 / 1.652* 10^14)

t = 3.566* 10^-8 s

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c)

a = 1.652* 10^14 m/s^2

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