Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 17, Problem 109P

(a)

To determine

What is the direction and magnitude of the change in velocity of electrons between plates.

(a)

Expert Solution
Check Mark

Answer to Problem 109P

The magnitude of the change in velocity is 7.0×106m/s_ and is directed upward.

Explanation of Solution

Write newton’s equation to find the force acting on the electrons.

    Fy=may                                                                                                              (I)

Here, Fy is the force, m is the mass and ay is the acceleration

Write the equation to find the force due to electric field.

    Fy=eE                                                                                                                   (II0

Here, e is charge of electron and E is the electric field

Equate equations (I) and (II) and get an equation for ay.

    eE=mayay=eEm                                                                                                                 (III)

Write the equation to find the time of travel of charge.

    Δt=Δxvx                                                                                                             (IV)

Here, Δt is the time of travel, Δx is change in distance and vx is the velocity

Write the equation to find the change in velocity.

  Δvy=ayΔt=(eEm)(Δxvx)                                                                                                  (V)

Conclusion

Substitute 1.602×1019C for e , 2.0×104N/C for E , 6.0cm for Δx , 9.109×1031kg for m and 3.0×107m/s for vx in equation (V) to get Δvy

    Δvy=(1.602×1019C)(2.0×104N/C)(6.0cm(1m102cm))(9.109×1031kg)(3.0×107m/s)=7.0×106m/s

Therefore, the magnitude of the change in velocity is 7.0×106m/s_ and is directed upward. Since the change is positive it is upward in direction.

(b)

To determine

How far are the electrons deflected in the ±y direction between the plates.

(b)

Expert Solution
Check Mark

Answer to Problem 109P

Electrons get deflected to a distance of 7.0mm_.

Explanation of Solution

Write the equation to find the displacement of electrons.

    Δy=12ay(Δt)2                                                                                                    (VI)

Substitute equation (III) and (IV) in equation (VI) to get Δy

    Δy=12(eEm)(Δxvx)2                                                                                                (VII)

Conclusion:

Substitute 1.602×1019C for e , 2.0×104N/C for E , 6.0cm for Δx , 9.109×1031kg for m and 3.0×107m/s for vx in equation (VII) to get Δy

    Δy=(1.602×1019C)(2.0×104N/C)(6.0cm(1m102cm))2(9.109×1031kg)(3.0×107m/s)=7.0mm

Therefore, electrons get deflected to a distance of 7.0mm_.

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Chapter 17 Solutions

Physics

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Charges of + 2.0 nC and − 1.0 nC are located...Ch. 17 - Prob. 27PCh. 17 - 28. (a) Find the potential at points a and b in...Ch. 17 - 29. (a) In the diagram, what are the potentials at...Ch. 17 - 30. (a) In the diagram, what are the potentials at...Ch. 17 - Prob. 31PCh. 17 - 32. By rewriting each unit in terms of kilograms,...Ch. 17 - 33. Rank points A–E in order of the potential,...Ch. 17 - Prob. 34PCh. 17 - Prob. 35PCh. 17 - Prob. 36PCh. 17 - Prob. 37PCh. 17 - Prob. 38PCh. 17 - Prob. 39PCh. 17 - Prob. 40PCh. 17 - Prob. 41PCh. 17 - Prob. 43PCh. 17 - 43. A positive point charge is located at the...Ch. 17 - Prob. 44PCh. 17 - Prob. 45PCh. 17 - 46. Point P is at a potential of 500.0 kV, and...Ch. 17 - 47. An electron is accelerated from rest through a...Ch. 17 - 48. As an electron moves through a region of...Ch. 17 - Prob. 49PCh. 17 - 50. An electron beam is deflected upward through...Ch. 17 - 51. In the electron gun of Example 17.8, if the...Ch. 17 - 52. 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