Hi, cab you please help me to understand this example •54GOIn Fig. 22-61, an electron is shot at an initial speed of vo = 2.00 x100 m/s, at angle 0o = 40.0° from an x axis. It moves through a uniform(5.00 N/C) j. A screen for detecting electrons is3.00 m. In unit-vector%3Delectric field Epositioned parallel to the y axis, at distance xnotation, what is the velocity of the electron when it hits the screen?%3DDetectingvoscreenX-Figure 22-61 Problem 54.

given V = 2.00 x 106 m/s θ=400 Using E→ = (5.00N/C)j^ ↑ Force on electron due to electric field =…

Answered: GO •54 In Fig. 22-61, an electron is…

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Hi, cab you please help me to understand this example

•54
GO
In Fig. 22-61, an electron is shot at an initial speed of vo = 2.00 x
100 m/s, at angle 0o = 40.0° from an x axis. It moves through a uniform
(5.00 N/C) j. A screen for detecting electrons is
3.00 m. In unit-vector
%3D
electric field E
positioned parallel to the y axis, at distance x
notation, what is the velocity of the electron when it hits the screen?
%3D
Detecting
vo
screen
X-
Figure 22-61 Problem 54.

Expert Solution

Step 1

given

V = 2.00 x 10⁶ m/s

$θ = 40^{0}$

Using

$\vec{E}$ = (5.00N/C) $\hat{j}$ ↑

Force on electron due to electric field = q $\vec{E}$

= $(1.6 \times 10^{- 19} \times 5) = (8 \times 10^{- 19} N)$

The force will be in downward direction.

$A c c e l e r a t i o n = \frac{F o r c e}{m a s s o f e l e c t r o n} = \frac{q \vec{E}}{m_{e}} = (\frac{1.6 \times 10^{- 19} \times 5}{9.109 \times 10^{- 31}} \frac{m}{s^{2}}) = 0.879 \times 10^{12} \frac{m}{s^{2}} (a c t i n g d o w n w a r d) = 8.79 \times 10^{11} \frac{m}{s^{2}}$

Initial Velocity of electron

$\vec{v} = {\vec{v}}_{x} \hat{i} + {\vec{v}}_{y} \hat{j} = v_{0} \cos θ \hat{i} + v_{0} \sin θ \hat{j}$

there is no acceleration in horizontal direction .

So, electrons will hit the screen after time

(t)= $\frac{x_{0}}{v_{0} \cos θ} = \frac{3.00}{2 \times 10^{6} \times \cos 40^{0}} s e c = 1.958 \times 10^{- 6} s e c$

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