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JenniferSmart1

  • 2 years ago

The capacitance of a parallel plate capacitor \[C=\frac{Q}{V}\] \[dV=-\int Edr\] \[E=\]

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  1. JenniferSmart1
    • 2 years ago
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    |dw:1362265815092:dw|

  2. JenniferSmart1
    • 2 years ago
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    |dw:1362265934285:dw| should I use gauss's law to find the electric field?

  3. experimentX
    • 2 years ago
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    yes .. use the gauss law. consider a single plate first.

  4. JenniferSmart1
    • 2 years ago
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    \[\oint E dA=\frac{Q}{\epsilon_0}\] |dw:1362266179361:dw| \[Q=\sigma A\] \[EL^2=\frac{Q}{\epsilon_0}=\frac{\sigma L^2}{\epsilon_0}\] \[E=\frac{\sigma}{\epsilon_0}\]

  5. JenniferSmart1
    • 2 years ago
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    since we have two plates do we use \(2\sigma\)?

  6. experimentX
    • 2 years ago
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    yeah yeah ... that;s the way ... let me add few things ... |dw:1362266439292:dw|

  7. experimentX
    • 2 years ago
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    how should the flux lines go? fit it geometrically

  8. JenniferSmart1
    • 2 years ago
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    |dw:1362266583983:dw|

  9. experimentX
    • 2 years ago
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    yep!! Now what is flux??

  10. JenniferSmart1
    • 2 years ago
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    \[\sigma/\epsilon_0\]? or do I need to consider the cylinder?

  11. experimentX
    • 2 years ago
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    |dw:1362266689639:dw| yes ... it's okay even if you consider square or rectangle.

  12. JenniferSmart1
    • 2 years ago
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    I used L^2 for area

  13. experimentX
    • 2 years ago
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    everything is okay ... the answer is interesting ... If you have infinite charged sheet ... not matter how far you fly, the electric field will never decrease.

  14. experimentX
    • 2 years ago
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    looks like you can use that on gravity too ...

  15. JenniferSmart1
    • 2 years ago
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    should I try putting in values? Let's say 10cm=L and the plate seperation is 1mm. wait, so the electric field does not depend on the dimensions? would I need to substitute for sigma and epsilon to find the electric field if I'm only given geometric dimensions?

  16. experimentX
    • 2 years ago
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    No ... calculate the charge density.

  17. JenniferSmart1
    • 2 years ago
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    \[\sigma=q/A= \frac{1.6\times10^{-19}}{.1^2}=1.6E-17 C/m^2\] \[E=\frac {\sigma}{}\]

  18. experimentX
    • 2 years ago
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    looks like you are going in wrong direction ... can you state the original question

  19. JenniferSmart1
    • 2 years ago
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    I think we're on the right track...once we have E we can solve for V and so on....should I leave E in variable form for now.

  20. JenniferSmart1
    • 2 years ago
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    ?

  21. JenniferSmart1
    • 2 years ago
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    but what are we integration over....the distance between the plates?

  22. experimentX
    • 2 years ago
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    Q <-- this should the be the total charge on the surface ... and A <-- this is the total area of the surface.

  23. JenniferSmart1
    • 2 years ago
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    is this different from the gaussian surface? do I need different variable to use gauss's law?

  24. JenniferSmart1
    • 2 years ago
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    yeah it looks like I haven't quite grasped the concept of Gauss's law

  25. JenniferSmart1
    • 2 years ago
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    should we use the cylinder?

  26. experimentX
    • 2 years ago
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    hmm ... what is your original question?

  27. JenniferSmart1
    • 2 years ago
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    I have to find the capacitance of the parallel plate capacitor.

  28. experimentX
    • 2 years ago
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    what information are given initially?

  29. JenniferSmart1
    • 2 years ago
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    the plate is square and has length 10cm....the plates are separated by 1mm

  30. experimentX
    • 2 years ago
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    ow ..

  31. JenniferSmart1
    • 2 years ago
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    \[dV=\int\frac{\sigma}{\epsilon_0}dA\]

  32. JenniferSmart1
    • 2 years ago
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    is that correct?

  33. JenniferSmart1
    • 2 years ago
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    \[V=\frac{2\sigma A}{\epsilon_0}\]

  34. experimentX
    • 2 years ago
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    Woops!! error

  35. experimentX
    • 2 years ago
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    |dw:1362268232059:dw|

  36. JenniferSmart1
    • 2 years ago
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    but would that still be \[V=\frac{\sigma A}{\epsilon_0}\]

  37. experimentX
    • 2 years ago
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    |dw:1362268629160:dw|

  38. JenniferSmart1
    • 2 years ago
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    Let me try ...One moment....

  39. experimentX
    • 2 years ago
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    |dw:1362268884843:dw|

  40. JenniferSmart1
    • 2 years ago
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    \[\oint EdA=\frac Q{\epsilon_0}\] \[E\pi r^2=\frac{Q}{\epsilon_0}\] \[\rho=\frac Q {\pi r^2L}\] \[E\pi r^2=\frac{\rho \pi r^2 L}{\epsilon_0}\]

  41. JenniferSmart1
    • 2 years ago
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    so the charge density is that of the cylinder?

  42. JenniferSmart1
    • 2 years ago
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    \[E=\frac{\rho L}{\epsilon_0}\]

  43. JenniferSmart1
    • 2 years ago
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    I completely missed the point didn't I?

  44. experimentX
    • 2 years ago
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    use sigma for surface charge density.

  45. experimentX
    • 2 years ago
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    |dw:1362269453040:dw|

  46. JenniferSmart1
    • 2 years ago
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    oh so the flux is only though the circle... \(\phi\) is of that surface....because we're looking for the flux through the plate, and we're using a cylinder to model that....I think I got it now

  47. JenniferSmart1
    • 2 years ago
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    yes?

  48. JenniferSmart1
    • 2 years ago
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    \[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]

  49. experimentX
    • 2 years ago
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    No ... we are not integrating over the circle.

  50. JenniferSmart1
    • 2 years ago
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    closed integral from a to b?

  51. experimentX
    • 2 years ago
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    |dw:1362269839865:dw|

  52. experimentX
    • 2 years ago
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    to understand Gauss law ... you should understand flux. Find the total flux .. this case |dw:1362269948919:dw|

  53. JenniferSmart1
    • 2 years ago
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    is this a circular surface of charge +q ?

  54. experimentX
    • 2 years ago
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    no ... not circular ... spherical!!

  55. JenniferSmart1
    • 2 years ago
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    the flux would be |dw:1362270059079:dw| oh ok...I guess we'll work with a sphere instead...Let's see...

  56. experimentX
    • 2 years ago
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    consider ... the charge in enclosed by sphere.

  57. JenniferSmart1
    • 2 years ago
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    |dw:1362270254679:dw| in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere

  58. JenniferSmart1
    • 2 years ago
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    |dw:1362270435029:dw|

  59. experimentX
    • 2 years ago
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    yes ...

  60. JenniferSmart1
    • 2 years ago
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    Yay!!! sigh...

  61. experimentX
    • 2 years ago
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    what is the total flux?

  62. JenniferSmart1
    • 2 years ago
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    zero?

  63. JenniferSmart1
    • 2 years ago
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    the sum of all of the electric field lines?

  64. JenniferSmart1
    • 2 years ago
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    zero

  65. JenniferSmart1
    • 2 years ago
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    final answer

  66. experimentX
    • 2 years ago
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    No ...

  67. JenniferSmart1
    • 2 years ago
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    sum of all the Electric field lines \[\phi=\oint E dr=\frac{Q}{\epsilon_0}\]

  68. JenniferSmart1
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    volume?

  69. experimentX
    • 2 years ago
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    You should understand what is flux first ... Flux is like ... |dw:1362270773063:dw|

  70. JenniferSmart1
    • 2 years ago
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    the amount of water that flows through it is A times \(v\)

  71. JenniferSmart1
    • 2 years ago
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    \[\frac{meter^2}{1}\times\frac{meters}{second}\] that doesn't seem right

  72. experimentX
    • 2 years ago
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    yeah .. that's correct. It's easy to visualize water ... |dw:1362271042142:dw|

  73. JenniferSmart1
    • 2 years ago
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    yep

  74. JenniferSmart1
    • 2 years ago
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    |dw:1362271173946:dw|

  75. JenniferSmart1
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    so the flux through the area would be the volume of the sphere times the electric field

  76. JenniferSmart1
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    I tried to make your circle look more like a sphere :P

  77. experimentX
    • 2 years ago
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    check your answer again!!

  78. JenniferSmart1
    • 2 years ago
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    oh surface area of the sphere!

  79. experimentX
    • 2 years ago
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    yeah ... what is the total flux?

  80. JenniferSmart1
    • 2 years ago
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    because the flux is through the surface of the sphere :D The total flux is the sum of all of (the surface areas multiplied by the electric field)

  81. JenniferSmart1
    • 2 years ago
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    i meant surface area...not areas

  82. experimentX
    • 2 years ago
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    |dw:1362271421314:dw|

  83. JenniferSmart1
    • 2 years ago
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    what's s? surface area?

  84. JenniferSmart1
    • 2 years ago
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    ds...?

  85. experimentX
    • 2 years ago
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    |dw:1362271575577:dw|

  86. JenniferSmart1
    • 2 years ago
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    |dw:1362271615713:dw|

  87. JenniferSmart1
    • 2 years ago
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    what's that?

  88. experimentX
    • 2 years ago
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    sphere.

  89. JenniferSmart1
    • 2 years ago
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    oh are you saying that \[\int ds=s\] yeah I believe that but is s ...oh i see sphere

  90. JenniferSmart1
    • 2 years ago
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    we're integrating with respect to the sphere, correct?

  91. JenniferSmart1
    • 2 years ago
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    the surface area of the sphere

  92. experimentX
    • 2 years ago
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    yes ... but electric Field is constant .. since it's sphere.

  93. JenniferSmart1
    • 2 years ago
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    yep

  94. JenniferSmart1
    • 2 years ago
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    what would you limits be?

  95. experimentX
    • 2 years ago
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    and we are integrating surface ...not distance. lol ... the limits ... don't worry

  96. JenniferSmart1
    • 2 years ago
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    so \[\oint \textrm{never has limits?}\]

  97. experimentX
    • 2 years ago
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    |dw:1362271874243:dw|

  98. JenniferSmart1
    • 2 years ago
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    yeah I get that E is constant everywhere over the sphere

  99. JenniferSmart1
    • 2 years ago
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    surface*

  100. experimentX
    • 2 years ago
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    |dw:1362271927389:dw|

  101. JenniferSmart1
    • 2 years ago
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    oh because were integrating \[\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}\]

  102. JenniferSmart1
    • 2 years ago
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    since the answer is just s

  103. experimentX
    • 2 years ago
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    |dw:1362272111497:dw|

  104. JenniferSmart1
    • 2 years ago
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    yep , makes sense....

  105. JenniferSmart1
    • 2 years ago
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    I finally understand flux.... :)

  106. experimentX
    • 2 years ago
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    Not complete yet!!

  107. JenniferSmart1
    • 2 years ago
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    |dw:1362272196631:dw|

  108. experimentX
    • 2 years ago
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    |dw:1362272206784:dw|

  109. JenniferSmart1
    • 2 years ago
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    i'm not gonna look

  110. JenniferSmart1
    • 2 years ago
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    not gonna look

  111. JenniferSmart1
    • 2 years ago
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    not gonna look

  112. JenniferSmart1
    • 2 years ago
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    ok fine lol

  113. JenniferSmart1
    • 2 years ago
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    keep going...i'm sorry

  114. experimentX
    • 2 years ago
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    Now ... |dw:1362272365512:dw|

  115. experimentX
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    this is called conservation of flux.

  116. experimentX
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    |dw:1362272482209:dw|

  117. experimentX
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    |dw:1362272583908:dw|

  118. experimentX
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    |dw:1362272690179:dw|

  119. JenniferSmart1
    • 2 years ago
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    through both spheres? yes

  120. experimentX
    • 2 years ago
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    This is why Gravity ... and Electrostatic force follow inverse square laws. Any shape ... but must be closed. and charge must be inside it.

  121. JenniferSmart1
    • 2 years ago
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    interesting! yeah \(E\propto \frac 1{r^2}\) \[F_g\propto \frac 1{r^2}\]

  122. JenniferSmart1
    • 2 years ago
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    or is it \[F_e\propto \frac 1{r^2}\] and \[F_g\propto \frac 1{r^2}\]

  123. experimentX
    • 2 years ago
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    just think that ... charge is mouth of that pipe on that water example ... and outer envelopes is the other hose water comes out thorugh. It is independent of spape of pipe.

  124. experimentX
    • 2 years ago
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    |dw:1362273084528:dw|

  125. experimentX
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    |dw:1362273139951:dw|

  126. JenniferSmart1
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    oh|dw:1362273163066:dw| flux is still the same

  127. experimentX
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    yes ... the flux is still the same.

  128. experimentX
    • 2 years ago
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    |dw:1362273240830:dw|