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anonymous

  • one year ago

two horizontal plates erated by 4.00cm , potetinal differece 120V horizaonal beem of eletrons with speed 6.50*10^6 , the electron enter 1.00 above the negative plate , the horizontal distance tht the electrom travels before striking the postive plate?

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  1. IrishBoy123
    • one year ago
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    draw it first and you will get a lot of help

  2. anonymous
    • one year ago
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    |dw:1437369186841:dw|

  3. anonymous
    • one year ago
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    @dan815 @e.mccormick @Luigi0210

  4. anonymous
    • one year ago
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    What would be the electric field between the two parallel plates?

  5. anonymous
    • one year ago
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    i did fnet=Fe+Fg

  6. anonymous
    • one year ago
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    is that how u start?

  7. anonymous
    • one year ago
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    Yep, you can use the forces to do the problem. Then, F=ma. Then find where the electron intersects on of the plates.

  8. anonymous
    • one year ago
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    they didnt geive the accelration

  9. anonymous
    • one year ago
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    If you know the forces, you can find the acceleration.

  10. anonymous
    • one year ago
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    do we have to find it ?

  11. anonymous
    • one year ago
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    to find the distane?

  12. anonymous
    • one year ago
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    what formila are we goona conclude to ? in the end ?

  13. anonymous
    • one year ago
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    Yeah, we'll need to.

  14. anonymous
    • one year ago
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    Usually, in these types of problems, we'll ignore gravity. Find the electric field. F = qE=ma ==> a = qE/m We can then use: \[y(t) = \frac{1}{2}at^2 + v_ot + y_o\] to find when the electron meets one of the plates

  15. anonymous
    • one year ago
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    ok so for acc i get 2.85*10^17

  16. anonymous
    • one year ago
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    not sure if correct too mnay calcs maybe screwd

  17. anonymous
    • one year ago
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    Then, when you have t for that, you can find x There's no acceleration in the x direction, so you can just use: x(t) = v*t The v used here is the one given in the problem.

  18. anonymous
    • one year ago
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    not getting the answer

  19. anonymous
    • one year ago
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    first i found time , then found d but wrong ansswr1

  20. anonymous
    • one year ago
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    I get a different acceleration. Be careful with units.

  21. anonymous
    • one year ago
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    whatu get/

  22. anonymous
    • one year ago
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    what do u get as an answerr?

  23. anonymous
    • one year ago
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    5.276x10^14 is the acceleration I'm getting

  24. anonymous
    • one year ago
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    m/s^2

  25. anonymous
    • one year ago
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    even if i use ur acc i am getting wrong answr

  26. anonymous
    • one year ago
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    the answer is 6.94

  27. anonymous
    • one year ago
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    Show me what youre doing

  28. anonymous
    • one year ago
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    first speed didved by acc

  29. anonymous
    • one year ago
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    then i get the time , so i times by speed

  30. anonymous
    • one year ago
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    to find d

  31. anonymous
    • one year ago
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    No, can't do it quite that way. \[E = \frac{V}{d}\]gives the E field between the plates. V is the voltage between the plates, d is the separation between them (be careful of units! We want this in meters!). As shown above, we can find the acceleration using \[a = \frac{qE}{m} = \frac{V}{d}\frac{q}{m}\] You can look up the value for q/m online. It's very commonly used. After that, we'll find the time of flight. The electron starts 1cm above the bottom, negative plate. It will travel upwards to the positive plate, which sits at a height of 4cm. \[y(t) = \frac{1}{2}at^2 +v_ot + y_o\] From the information given, we see that vo is zero (no initial velocity in the y direction), and y_o = .01m (1cm). We want to know what t will be when y(t) = 4. That is: \[4 = \frac{1}{2}at^2+1\] Only one step after you find t. What do you get for t?

  32. anonymous
    • one year ago
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    Bleh. 4 should be .04, and 1 should be .01 in the last equation. Units.

  33. anonymous
    • one year ago
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    the answer is in cm

  34. anonymous
    • one year ago
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    You can convert to whatever units you'd like.

  35. anonymous
    • one year ago
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    ok yea got it thanks , could u help me with soe theory questions

  36. anonymous
    • one year ago
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    I'll try :)

  37. anonymous
    • one year ago
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    |dw:1437372625139:dw|

  38. anonymous
    • one year ago
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    which way will the electron accelerate?

  39. anonymous
    • one year ago
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    |dw:1437372687094:dw|

  40. anonymous
    • one year ago
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    You can think about this two ways. We can use F = qE = ma, or we can think about which way electric fields point between charges. Which way does the field point between a negative and positive charge?

  41. anonymous
    • one year ago
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    2?

  42. anonymous
    • one year ago
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    If we say that the direction the field is pointing in is the positive direction, then F = qE will be negative, because q is negative (charge of the electron). This means it would move in the direction opposite of the electric field.

  43. anonymous
    • one year ago
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    so 4

  44. anonymous
    • one year ago
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    ohh i was think why the answrr is 4

  45. anonymous
    • one year ago
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    beacuse i was 100% sure the direction would be 2 and then i looked at the answer it was 4 i ws soo confused

  46. anonymous
    • one year ago
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    its an electron so would make sence since they are nagative

  47. anonymous
    • one year ago
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    Exactly :)

  48. anonymous
    • one year ago
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    so if it was a proton the answr would be 2?

  49. anonymous
    • one year ago
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    Yep!

  50. anonymous
    • one year ago
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    their is one more

  51. anonymous
    • one year ago
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    their are 2 netual metal objcts x and y on a wooden table

  52. anonymous
    • one year ago
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    a negativly charged rod is brought near object y

  53. anonymous
    • one year ago
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    |dw:1437373183331:dw|

  54. anonymous
    • one year ago
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    so the answer is x received a negative charge by induction and y received a positive charge by induction

  55. anonymous
    • one year ago
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    i dont get what they mean

  56. anonymous
    • one year ago
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    Right. So, the negative charges in the rod will repel other negative charges. Since the objects are metal, and thus conductive, their electrons can move relatively freely. This means that the electrons in the metal will move away from the rod. Since the rod is near object y, they'll move away to object x. This will make object x have a net negative charge. The lack of electrons on object y will give it a net positive charge. This displacement of charge is called induction.

  57. anonymous
    • one year ago
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    i am reading from my notes about induction and it says it causes the chrages to become opposite

  58. anonymous
    • one year ago
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    so what ur saying is

  59. anonymous
    • one year ago
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    in the other way

  60. anonymous
    • one year ago
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    if the rod was postive then y would be negative and also x ?

  61. anonymous
    • one year ago
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    i mean x will be negative

  62. anonymous
    • one year ago
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    If the rod were positive, the y would become negative and x would be positive.

  63. anonymous
    • one year ago
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    oh yea

  64. anonymous
    • one year ago
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    some of the other answrrs are by conductions whats the differenr

  65. anonymous
    • one year ago
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    if the boxes were not neutral then it would be conductions right?

  66. anonymous
    • one year ago
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    I don't understand your questions. If the objects weren't conductors, then the charges wouldn't be able to move freely. If they were perfect insulators, they couldn't move at all, and thus no charge would be induced.

  67. anonymous
    • one year ago
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    like the correct answer is induction the wrong answer is conduction

  68. anonymous
    • one year ago
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    Oh, the phenomenon of the rod inducing a charge on the boxes is called induction.

  69. anonymous
    • one year ago
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    what would make it conduction

  70. anonymous
    • one year ago
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    if the boxes werent neutral?

  71. anonymous
    • one year ago
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    like lets say box x and y were postive then the answer would be conduction right

  72. anonymous
    • one year ago
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    Conduction is just the metal carrying the charge across a potential. Like a current moving down a wire.

  73. anonymous
    • one year ago
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    oh ok

  74. anonymous
    • one year ago
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    their is one more

  75. anonymous
    • one year ago
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    a posively charger sphere near a metal rod . both are insulated stands and the rod is grounded

  76. anonymous
    • one year ago
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    |dw:1437374305785:dw|

  77. anonymous
    • one year ago
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    the disturbution of the charge on the rod is ?

  78. anonymous
    • one year ago
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    This is somewhat similar to what we just did. What will happen when the sphere approaches the rod?

  79. anonymous
    • one year ago
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    umm x will become neagative?

  80. anonymous
    • one year ago
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    That's what I'd say. Charge would be drawn up through the ground wire toward the sphere.

  81. anonymous
    • one year ago
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    how do u knw tht?

  82. anonymous
    • one year ago
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    beacuse they are grounded ? like can't move right?

  83. anonymous
    • one year ago
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    Grounded just means that it can exchange charges with a very large, external source. So, the positive charge of the sphere will cause electrons to accumulate on the rod, mostly near the sphere.

  84. anonymous
    • one year ago
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    to show that an objct is negativly charged we need to ?

  85. anonymous
    • one year ago
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    it should be positve right?

  86. anonymous
    • one year ago
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    but the answer is negative?

  87. anonymous
    • one year ago
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    The positive sphere will cause the rod to become negatively charged.

  88. anonymous
    • one year ago
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    ok thats it for tonight

  89. anonymous
    • one year ago
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    do u come here often ?

  90. anonymous
    • one year ago
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    around this time ?

  91. anonymous
    • one year ago
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    occasionally. I'm kind of erratic about when I'm on

  92. anonymous
    • one year ago
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    ok i am going to sleep learned so much today need to settle in the brain

  93. anonymous
    • one year ago
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    thanks alot!

  94. anonymous
    • one year ago
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    hope to see u again (kinda hard to meet people here that are good at physics lol)

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