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JenniferSmart1

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

  • one year ago
  • one year ago

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  1. JenniferSmart1
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    |dw:1362265815092:dw|

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

    • one year ago
  3. experimentX
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    yes .. use the gauss law. consider a single plate first.

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

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

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

    • one year ago
  7. experimentX
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    how should the flux lines go? fit it geometrically

    • one year ago
  8. JenniferSmart1
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    |dw:1362266583983:dw|

    • one year ago
  9. experimentX
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    yep!! Now what is flux??

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

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

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

    • one year ago
  13. experimentX
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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.

    • one year ago
  14. experimentX
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    looks like you can use that on gravity too ...

    • one year ago
  15. JenniferSmart1
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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?

    • one year ago
  16. experimentX
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    No ... calculate the charge density.

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

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

    • one year ago
  19. JenniferSmart1
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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.

    • one year ago
  20. JenniferSmart1
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    ?

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

    • one year ago
  22. experimentX
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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.

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

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

    • one year ago
  25. JenniferSmart1
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    should we use the cylinder?

    • one year ago
  26. experimentX
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    hmm ... what is your original question?

    • one year ago
  27. JenniferSmart1
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    I have to find the capacitance of the parallel plate capacitor.

    • one year ago
  28. experimentX
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    what information are given initially?

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

    • one year ago
  30. experimentX
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    ow ..

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

    • one year ago
  32. JenniferSmart1
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    is that correct?

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

    • one year ago
  34. experimentX
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    Woops!! error

    • one year ago
  35. experimentX
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    |dw:1362268232059:dw|

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

    • one year ago
  37. experimentX
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    |dw:1362268629160:dw|

    • one year ago
  38. JenniferSmart1
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    Let me try ...One moment....

    • one year ago
  39. experimentX
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    |dw:1362268884843:dw|

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

    • one year ago
  41. JenniferSmart1
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    so the charge density is that of the cylinder?

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

    • one year ago
  43. JenniferSmart1
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    I completely missed the point didn't I?

    • one year ago
  44. experimentX
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    use sigma for surface charge density.

    • one year ago
  45. experimentX
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    |dw:1362269453040:dw|

    • one year ago
  46. JenniferSmart1
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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

    • one year ago
  47. JenniferSmart1
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    yes?

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

    • one year ago
  49. experimentX
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    No ... we are not integrating over the circle.

    • one year ago
  50. JenniferSmart1
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    closed integral from a to b?

    • one year ago
  51. experimentX
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    |dw:1362269839865:dw|

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

    • one year ago
  53. JenniferSmart1
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    is this a circular surface of charge +q ?

    • one year ago
  54. experimentX
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    no ... not circular ... spherical!!

    • one year ago
  55. JenniferSmart1
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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...

    • one year ago
  56. experimentX
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    consider ... the charge in enclosed by sphere.

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

    • one year ago
  58. JenniferSmart1
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    |dw:1362270435029:dw|

    • one year ago
  59. experimentX
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    yes ...

    • one year ago
  60. JenniferSmart1
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    Yay!!! sigh...

    • one year ago
  61. experimentX
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    what is the total flux?

    • one year ago
  62. JenniferSmart1
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    zero?

    • one year ago
  63. JenniferSmart1
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    the sum of all of the electric field lines?

    • one year ago
  64. JenniferSmart1
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    zero

    • one year ago
  65. JenniferSmart1
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    final answer

    • one year ago
  66. experimentX
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    No ...

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

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  68. JenniferSmart1
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    volume?

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

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

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

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

    • one year ago
  73. JenniferSmart1
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    yep

    • one year ago
  74. JenniferSmart1
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    |dw:1362271173946:dw|

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

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

    • one year ago
  77. experimentX
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    check your answer again!!

    • one year ago
  78. JenniferSmart1
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    oh surface area of the sphere!

    • one year ago
  79. experimentX
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    yeah ... what is the total flux?

    • one year ago
  80. JenniferSmart1
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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)

    • one year ago
  81. JenniferSmart1
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    i meant surface area...not areas

    • one year ago
  82. experimentX
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    |dw:1362271421314:dw|

    • one year ago
  83. JenniferSmart1
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    what's s? surface area?

    • one year ago
  84. JenniferSmart1
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    ds...?

    • one year ago
  85. experimentX
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    |dw:1362271575577:dw|

    • one year ago
  86. JenniferSmart1
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    |dw:1362271615713:dw|

    • one year ago
  87. JenniferSmart1
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    what's that?

    • one year ago
  88. experimentX
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    sphere.

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

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

    • one year ago
  91. JenniferSmart1
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    the surface area of the sphere

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

    • one year ago
  93. JenniferSmart1
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    yep

    • one year ago
  94. JenniferSmart1
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    what would you limits be?

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

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

    • one year ago
  97. experimentX
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    |dw:1362271874243:dw|

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

    • one year ago
  99. JenniferSmart1
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    surface*

    • one year ago
  100. experimentX
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    |dw:1362271927389:dw|

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

    • one year ago
  102. JenniferSmart1
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    since the answer is just s

    • one year ago
  103. experimentX
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    |dw:1362272111497:dw|

    • one year ago
  104. JenniferSmart1
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    yep , makes sense....

    • one year ago
  105. JenniferSmart1
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    I finally understand flux.... :)

    • one year ago
  106. experimentX
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    Not complete yet!!

    • one year ago
  107. JenniferSmart1
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    |dw:1362272196631:dw|

    • one year ago
  108. experimentX
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    |dw:1362272206784:dw|

    • one year ago
  109. JenniferSmart1
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    i'm not gonna look

    • one year ago
  110. JenniferSmart1
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    not gonna look

    • one year ago
  111. JenniferSmart1
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    not gonna look

    • one year ago
  112. JenniferSmart1
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    ok fine lol

    • one year ago
  113. JenniferSmart1
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    keep going...i'm sorry

    • one year ago
  114. experimentX
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    Now ... |dw:1362272365512:dw|

    • one year ago
  115. experimentX
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    this is called conservation of flux.

    • one year ago
  116. experimentX
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    |dw:1362272482209:dw|

    • one year ago
  117. experimentX
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    |dw:1362272583908:dw|

    • one year ago
  118. experimentX
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    |dw:1362272690179:dw|

    • one year ago
  119. JenniferSmart1
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    through both spheres? yes

    • one year ago
  120. experimentX
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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.

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

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

    • one year ago
  123. experimentX
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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.

    • one year ago
  124. experimentX
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    |dw:1362273084528:dw|

    • one year ago
  125. experimentX
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    |dw:1362273139951:dw|

    • one year ago
  126. JenniferSmart1
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    oh|dw:1362273163066:dw| flux is still the same

    • one year ago
  127. experimentX
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    yes ... the flux is still the same.

    • one year ago
  128. experimentX
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    |dw:1362273240830:dw|

    • one year ago