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JenniferSmart1 Group Title

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 Group Title
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    |dw:1362265815092:dw|

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

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

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

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

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

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

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

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

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

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

    • one year ago
  13. experimentX Group Title
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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 Group Title
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    looks like you can use that on gravity too ...

    • one year ago
  15. JenniferSmart1 Group Title
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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 Group Title
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    No ... calculate the charge density.

    • one year ago
  17. JenniferSmart1 Group Title
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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 Group Title
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    looks like you are going in wrong direction ... can you state the original question

    • one year ago
  19. JenniferSmart1 Group Title
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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 Group Title
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    ?

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

    • one year ago
  22. experimentX Group Title
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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 Group Title
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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 Group Title
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    yeah it looks like I haven't quite grasped the concept of Gauss's law

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

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

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

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

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

    • one year ago
  30. experimentX Group Title
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    ow ..

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

    • one year ago
  32. JenniferSmart1 Group Title
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    is that correct?

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

    • one year ago
  34. experimentX Group Title
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    Woops!! error

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

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

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

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

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

    • one year ago
  40. JenniferSmart1 Group Title
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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 Group Title
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    so the charge density is that of the cylinder?

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

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

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

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

    • one year ago
  46. JenniferSmart1 Group Title
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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 Group Title
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    yes?

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

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

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

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

    • one year ago
  52. experimentX Group Title
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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 Group Title
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    is this a circular surface of charge +q ?

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

    • one year ago
  55. JenniferSmart1 Group Title
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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 Group Title
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    consider ... the charge in enclosed by sphere.

    • one year ago
  57. JenniferSmart1 Group Title
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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 Group Title
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    |dw:1362270435029:dw|

    • one year ago
  59. experimentX Group Title
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    yes ...

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

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

    • one year ago
  62. JenniferSmart1 Group Title
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    zero?

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

    • one year ago
  64. JenniferSmart1 Group Title
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    zero

    • one year ago
  65. JenniferSmart1 Group Title
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    final answer

    • one year ago
  66. experimentX Group Title
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    No ...

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

    • one year ago
  68. JenniferSmart1 Group Title
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    volume?

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

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

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

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

    • one year ago
  73. JenniferSmart1 Group Title
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    yep

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

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

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

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

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

    • one year ago
  80. JenniferSmart1 Group Title
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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 Group Title
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    i meant surface area...not areas

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

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

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

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

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

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

    • one year ago
  88. experimentX Group Title
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    sphere.

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

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

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

    • one year ago
  93. JenniferSmart1 Group Title
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    yep

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

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

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

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

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

    • one year ago
  99. JenniferSmart1 Group Title
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    surface*

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

    • one year ago
  101. JenniferSmart1 Group Title
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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 Group Title
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    since the answer is just s

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

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

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

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

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

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

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

    • one year ago
  110. JenniferSmart1 Group Title
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    not gonna look

    • one year ago
  111. JenniferSmart1 Group Title
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    not gonna look

    • one year ago
  112. JenniferSmart1 Group Title
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    ok fine lol

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

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

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

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

    • one year ago
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    |dw:1362272583908:dw|

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

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

    • one year ago
  120. experimentX Group Title
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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 Group Title
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    interesting! yeah \(E\propto \frac 1{r^2}\) \[F_g\propto \frac 1{r^2}\]

    • one year ago
  122. JenniferSmart1 Group Title
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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 Group Title
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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
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    |dw:1362273084528:dw|

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

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

    • one year ago