## JenniferSmart1 Group Title Question one year ago one year ago

1. sami-21 Group Title

wow so many people here to help !!

2. JenniferSmart1 Group Title

@zepdrix

3. JenniferSmart1 Group Title

First there is no force acting on the particle, just the velocity in the positive x direction. Then there is a force acting in the y direction forcing the electron upwards, Then in the last scenario there is no force acting on the particle except for the velocity... THere should be a y displacement, because of the vertical acceleration

4. JenniferSmart1 Group Title

|dw:1361683871212:dw|

5. JenniferSmart1 Group Title

|dw:1361684180784:dw|

6. JenniferSmart1 Group Title

A B and C are length...distances. So the electron travels along "c" and feels and acceleration upwards, but when it starts traveling in the distance B it doesn't feel any acceleration but it will continue moving to the right but at an angle upwards. I can't type sorry lol

7. JenniferSmart1 Group Title

oh so after it leaves c , it continues like this... |dw:1361684734571:dw| with a velocity that has an x and y component...oh that's why it's going at an angle =)

8. JenniferSmart1 Group Title

so the v_1 is the original velocity in the x direction, which has been consist throughout this path. Hhhhhmmm how do we explain the y component of the velocity?

9. zepdrix Group Title

Yah why does the wind stop blowing when it reaches B? :c hmm that's confusing me...

10. zepdrix Group Title

Because it's making me think that it would do this,|dw:1361685064192:dw| I'm probably not interpreting that correctly though :\

11. JenniferSmart1 Group Title

it's actually going through an electric field for a distance c which is causing it to move upwards. the particle is an electron (negative) and it wants to accelerate towards the positive plate. (opposites attract remember?) but you can think of it as wind blowing and then it stops blowing miraculously :) |dw:1361685140543:dw|

12. zepdrix Group Title

yah i was just being silly by saying wind :) lol

13. JenniferSmart1 Group Title

since it acquired a $$v_{net}$$ at an angle $$\theta$$ while traveling in c, it will continue in that manner after it leaves c because there are no forces acting upon it, and it always had this horizontal velocity.

14. zepdrix Group Title

Ok fair enough c:

15. JenniferSmart1 Group Title

LOL! This is how printers work apparently. The ink particle acquires a negative charge and it's path towards the paper is manipulated by an electric field that it travels through. So the stronger the electric field that it travels through (positive and negative plates), the more it deflects

16. zepdrix Group Title

Oooo interesting! c:

17. zepdrix Group Title

Hmm your class seems hard :\ I think JenJen has gotten a lil too smart1 for me to help.

18. JenniferSmart1 Group Title

*you're

19. JenniferSmart1 Group Title

displacement is $$\Delta y= vt^2+\frac 12 at^2$$ we probably have two y displacements.

20. JenniferSmart1 Group Title

One displacement for when the particle is within that electric field and one for when it exits. Because the first displacement has an acceleration $$\uparrow$$ factor and the second displacement doesn't

21. zepdrix Group Title

$\large \Delta y= vt+\frac 12 at^2$ I don't think it's a t^2 on the v is it? :O

22. JenniferSmart1 Group Title

no i don't think so, because velocity is m/s and delta y is in m, so just one s would cancel it out. $\frac m s\cdot s$

23. JenniferSmart1 Group Title

so $m=\frac ms\cdot s+\frac 12 \frac m{s^2}s^2$ and acceleration is in $$\frac m {s^2}$$

24. zepdrix Group Title

$\large \Delta y= vt+\frac 12 at^2 \qquad \rightarrow \qquad \Delta y= \frac{m}{s}(s)+\frac 12 \left(\frac{m}{s^2}\right)s^2$ So $$\Delta y$$ equals some distance in meters. Bah you wrote it out already, yer too fast for me! lol

25. JenniferSmart1 Group Title

lol, yep you got it!

26. zepdrix Group Title

So doesn't that help show that we dont' want the square on the first t? D:

27. JenniferSmart1 Group Title

what do you mean?

28. zepdrix Group Title

Maybe you just made a typo :O$\huge \Delta y= vt^{\color{orangered}{2}}+\frac 12 at^2$Why is the orange part there?

29. JenniferSmart1 Group Title

ooooops!!!! yeah that's a BIG TYPO LOL sorry...yeah that makes it more confusing haha

30. JenniferSmart1 Group Title

so we have to $$\Delta y$$ $\Delta y_1$ and $\Delta y_2$ the second $$\Delta y$$ doesn't have an acceleration...so we can cancel out the $$\frac 12 at^2$$ $\large \Delta y_2= vt+\cancel{\frac 12 at^2}$

31. JenniferSmart1 Group Title

and $$\Delta y_{total}=\Delta y_1+\Delta y_2$$

32. JenniferSmart1 Group Title

$\Delta y_{total}=\Delta y_1+\Delta y_2$ $=vt^2+\frac 12at^2+vt^2$

33. zepdrix Group Title

There you go doing it again >:O

34. JenniferSmart1 Group Title

oh my!!!! LOL!!!! I like squaring my velocities LOL!!!! $=vt+\frac 12at^2+vt$

35. JenniferSmart1 Group Title

THis is why I need you here =D

36. JenniferSmart1 Group Title

now we need to substitute for the t's because we're not given the time that it takes the particle to travel along that path.

37. JenniferSmart1 Group Title

Let's see....$$\Delta x=vt+\frac 12at^2$$ inside and outside of the electric field? or just one x displacement?|dw:1361686876935:dw|

38. JenniferSmart1 Group Title

or maybe to 2 $$\Delta x's$$ because the speed at which the particle goes within in the field is different form the speed outside of that field? What do you think?

39. JenniferSmart1 Group Title

i mean the time it takes to travel inside of c could be affected by that vertical acceleration.

40. JenniferSmart1 Group Title

*two not "to 2"

41. zepdrix Group Title

I dunno :\ I give up. Pretend I'm still here and keep talking XD lol

42. JenniferSmart1 Group Title

LOL sure. Thanks a lot btw, it honestly helps a lot. I would have cried myself to sleep already :'(

43. JenniferSmart1 Group Title

Let's see. $\Delta x_1=v_xt+\frac 12 a_xt^2$ $\Delta x_2=v_xt+\frac 12 a_xt^2$ there isn't an acceleration in the x direction at any point through this journey... $\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}$ $\Delta x_2=v_xt+\cancel{\frac 12 a_xt^2}$ and the velocity the x direction isn't changing either. That leaves us with: $t=\frac{\Delta x}v$ cool now I can just substitute.

44. JenniferSmart1 Group Title

$\Delta y_{total}=vt+\frac 12at^2+vt$ $\Delta y_{total}=2vt+\frac 12at^2$ $\Delta y_{total}=2vt+\frac 12at^2$ and $t=\frac{\Delta x}v$ $\Delta y_{total}=2v\frac{\Delta x}{v}+\frac 12a\frac{\Delta x}{v}^2$

45. JenniferSmart1 Group Title

OOopppps!!!!!! $\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2$

46. JenniferSmart1 Group Title

right zep?

47. zepdrix Group Title

Why did the acceleration get crossed out during the first displacement?$\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}$

48. zepdrix Group Title

Oh because this is displacement in the x direction? The acceleration is 0, and the velocity is constant?

49. JenniferSmart1 Group Title

yes sir! you are correct!

50. JenniferSmart1 Group Title

$\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2$ Let's see, now I have to sub for the acceleration. $F=ma$ $F=qE$ q is the charge of one electron? I think? LOL i should know this haha. E is the Electric field

51. JenniferSmart1 Group Title

no that's wrong!

52. JenniferSmart1 Group Title

if the electric field is the force per unit charge....then.... that is correct....

53. JenniferSmart1 Group Title

$E=\frac F q$

54. JenniferSmart1 Group Title

but i need to substitute for the acceleration, because that's not something I can measure. I can only measure the deflection (in mm)

55. zepdrix Group Title

hmmm true

56. JenniferSmart1 Group Title

$F=ma$ $\frac E q=ma$ $a=\frac E{mq}$ but there is more....I can't just make that substitution apparently. I need to use potential and kinetic energies instead :-(

57. zepdrix Group Title

ikr!

58. JenniferSmart1 Group Title

Potential energy equals kinetic energy $UE=KE$

59. JenniferSmart1 Group Title

Kinetic energy $$KE=\frac 12 mv^2$$

60. JenniferSmart1 Group Title

and for reasons I don't know $UE=V\cdot q$ Potential energy $$U$$ is supposedly different from electric potential $$V$$ Electric potential is the Potential energy per unit charge. $\frac U q=V$

61. JenniferSmart1 Group Title

this allows be to sub for velocity, but not for the acceleration :'-(

62. JenniferSmart1 Group Title

I'd have to use $a=\frac E{mq}$ Let's see what happens.... $\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2$ $\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)^2$ ugh!!!!!!!!!!!!! $\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)$ $\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)$

63. JenniferSmart1 Group Title

we made a mistake....there was no velocity in the y at any point, huh? I don't know anymore....time to sleep haha.

64. JenniferSmart1 Group Title

@JamesJ

65. TuringTest Group Title

What exactly do you want to know? The vertical displacement over A? in terms of which variables? and do you *have* to use energy formulas to get the answer?

66. JenniferSmart1 Group Title

@TuringTest My goal is to get the displacement as a function of the electric field It should look like this $\Delta y=\Delta y_1+\Delta y_2 =\frac{Ec}{2v}\left(B+\frac c 2\right)$