Each of the electrons in a particle beam has a kinetic energy of 1.54 10 17 J. (

Question

Each of the electrons in a particle beam has a kinetic energy of 1.54 1017 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 18.0 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 opposite towards the field perpendicular to the field opposite to the field at what rate?

Explanation / Answer

V2^2 - V1^2 = 2*a*s

where V2 = final velocity = 0

V1 = calculated below

s = 0.18 m

now,

Kinetic Energy = 1.54 1017 J = KE = (1/2)mv^2

where, m = mass  of the electron

v = velocity

so, v = 5814748 m/s = V1 used above

now, from

V2^2 - V1^2 = 2*a*s

a = 9.39*10^13 m/s^2

from Eq = F = ma,

(a) E = m*a/q = 533.99

for stopping the electrons, V2 = V1 + a*t

or, (b) t = 5814748/9.30*10^13 = 6.25*10^-8 s = 62.5 ns

F = m*a = E*q

where q = charge of electron

E = 533.99

so, (c) a = E*q/m = 9.39*10^13 m/s^2

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