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Problem 54 Hard Difficulty

In Fig. $22-61,$ an electron is $\begin{array}{ll}{\text { shot }} & {\text { at } \quad \text { an initial speed of }} \\ {v_{0}=2.00 \times 10^{6} \mathrm{m} / \mathrm{s},} & {\text { at angle } \theta_{0}=}\end{array}$ $40.0^{\circ}$ from an $x$ axis. It moves
through a uniform electric field
$\vec{E}=(5.00 \mathrm{N} / \mathrm{C}) \hat{\mathrm{j}}$ . A screen for detecting electrons is positioned parallel to the $y$ axis, at distance $x=3.00 \mathrm{m} .$ In unit-vector notation,
what is the velocity of the electron when it hits the screen?

Answer

$$\vec{v}=\left(1.53 \times 10^{6} \mathrm{m} / \mathrm{s}\right) \hat{\mathrm{i}}-\left(4.34 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\mathrm{j}}$$

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Video Transcript

So did the fact that electron is negatively charged then, Ah, the field e is pointing in positive. Why direction Onda? We call that a poor direction leads to downward acceleration. So and also for calculating the ex elation will be using, um will be using a question come combination of the question 22 0.28 and Newton's second law. So we see that doing that. We have an exhibition off. Mmm Ah, a easy we'll do. Smalley Times biggie Over and M is the mass of the electron. So and also this quantity is 8.78 times 10 to the power 11 this 11 meters per second squared. And, ah, if you want detailed explanation off this quantity, how we got the expression, please look into the previous problems. It's been explained properly dead. But since it's been explained so many times, I just, uh, used, uh, I just mentioned that we'll be using this equation and Newton's second law toe. Find out the ex elation. All right, so now in, if we look at the projectile motion problem, that's in chapter for, uh, where we see that the G is just replaced with a And if we use a question 4.21 we see that time can be calculated with the following question. So time is equal to X over v o Co sign Beta zero. So that's the X component where we don't have any ex elation. So that's why we just have the X component of velocity and the displacement. So that's three meter on top. On in the denominator. We have two times 10 to the power, six meters per second and co Science Ada is co sign 40 in our gaze. This leads to a number 1.96 times 10 to the power negative six seconds. So then we can use a question for 0.23 to finally get the velocity in the airport. Directions of the wise defining velocity, which is the old sign, say that. Oh, so that's the velocity in the oh are the velocity component in wind direction. Then we have a minus sign because of the D solution. Any time's T is the ex elation here So we know all the values we can use it So two times 10 to the power six me never second Santa is sign 40 TV's than negative 8.78 times 10 to the bar, 11 meters per second squared and time is 1.96 times 10 to the bard. Negative six seconds. So combining this we see that the airport velocity is negative four point 34 times 10 to the bar fight meters per second since X component of velocity does not change, then the final velocity V becomes 1.53 times 10 to the power six meters per second. That's X component. So we use I had for that and then we have negative 4.34 times 10 to the five meters per second on then we have a J half because it's the wind direction. Thanks for watching as he

Kansas State University
Top Physics 102 Electricity and Magnetism Educators
Christina K.

Rutgers, The State University of New Jersey

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University of Washington

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University of Sheffield

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