## JenniferSmart1 Group Title The capacitance of a parallel plate capacitor $C=\frac{Q}{V}$ $dV=-\int Edr$ $E=$ one year ago one year ago

1. JenniferSmart1 Group Title

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2. JenniferSmart1 Group Title

|dw:1362265934285:dw| should I use gauss's law to find the electric field?

3. experimentX Group Title

yes .. use the gauss law. consider a single plate first.

4. JenniferSmart1 Group Title

$\oint E dA=\frac{Q}{\epsilon_0}$ |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 Group Title

since we have two plates do we use $$2\sigma$$?

6. experimentX Group Title

yeah yeah ... that;s the way ... let me add few things ... |dw:1362266439292:dw|

7. experimentX Group Title

how should the flux lines go? fit it geometrically

8. JenniferSmart1 Group Title

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9. experimentX Group Title

yep!! Now what is flux??

10. JenniferSmart1 Group Title

$\sigma/\epsilon_0$? or do I need to consider the cylinder?

11. experimentX Group Title

|dw:1362266689639:dw| yes ... it's okay even if you consider square or rectangle.

12. JenniferSmart1 Group Title

I used L^2 for area

13. experimentX Group Title

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 Group Title

looks like you can use that on gravity too ...

15. JenniferSmart1 Group Title

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 Group Title

No ... calculate the charge density.

17. JenniferSmart1 Group Title

$\sigma=q/A= \frac{1.6\times10^{-19}}{.1^2}=1.6E-17 C/m^2$ $E=\frac {\sigma}{}$

18. experimentX Group Title

looks like you are going in wrong direction ... can you state the original question

19. JenniferSmart1 Group Title

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 Group Title

?

21. JenniferSmart1 Group Title

but what are we integration over....the distance between the plates?

22. experimentX Group Title

Q <-- this should the be the total charge on the surface ... and A <-- this is the total area of the surface.

23. JenniferSmart1 Group Title

is this different from the gaussian surface? do I need different variable to use gauss's law?

24. JenniferSmart1 Group Title

yeah it looks like I haven't quite grasped the concept of Gauss's law

25. JenniferSmart1 Group Title

should we use the cylinder?

26. experimentX Group Title

hmm ... what is your original question?

27. JenniferSmart1 Group Title

I have to find the capacitance of the parallel plate capacitor.

28. experimentX Group Title

what information are given initially?

29. JenniferSmart1 Group Title

the plate is square and has length 10cm....the plates are separated by 1mm

30. experimentX Group Title

ow ..

31. JenniferSmart1 Group Title

$dV=\int\frac{\sigma}{\epsilon_0}dA$

32. JenniferSmart1 Group Title

is that correct?

33. JenniferSmart1 Group Title

$V=\frac{2\sigma A}{\epsilon_0}$

34. experimentX Group Title

Woops!! error

35. experimentX Group Title

|dw:1362268232059:dw|

36. JenniferSmart1 Group Title

but would that still be $V=\frac{\sigma A}{\epsilon_0}$

37. experimentX Group Title

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38. JenniferSmart1 Group Title

Let me try ...One moment....

39. experimentX Group Title

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40. JenniferSmart1 Group Title

$\oint EdA=\frac Q{\epsilon_0}$ $E\pi r^2=\frac{Q}{\epsilon_0}$ $\rho=\frac Q {\pi r^2L}$ $E\pi r^2=\frac{\rho \pi r^2 L}{\epsilon_0}$

41. JenniferSmart1 Group Title

so the charge density is that of the cylinder?

42. JenniferSmart1 Group Title

$E=\frac{\rho L}{\epsilon_0}$

43. JenniferSmart1 Group Title

I completely missed the point didn't I?

44. experimentX Group Title

use sigma for surface charge density.

45. experimentX Group Title

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46. JenniferSmart1 Group Title

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 Group Title

yes?

48. JenniferSmart1 Group Title

$\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}$

49. experimentX Group Title

No ... we are not integrating over the circle.

50. JenniferSmart1 Group Title

closed integral from a to b?

51. experimentX Group Title

|dw:1362269839865:dw|

52. experimentX Group Title

to understand Gauss law ... you should understand flux. Find the total flux .. this case |dw:1362269948919:dw|

53. JenniferSmart1 Group Title

is this a circular surface of charge +q ?

54. experimentX Group Title

no ... not circular ... spherical!!

55. JenniferSmart1 Group Title

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

56. experimentX Group Title

consider ... the charge in enclosed by sphere.

57. JenniferSmart1 Group Title

|dw:1362270254679:dw| in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere

58. JenniferSmart1 Group Title

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59. experimentX Group Title

yes ...

60. JenniferSmart1 Group Title

Yay!!! sigh...

61. experimentX Group Title

what is the total flux?

62. JenniferSmart1 Group Title

zero?

63. JenniferSmart1 Group Title

the sum of all of the electric field lines?

64. JenniferSmart1 Group Title

zero

65. JenniferSmart1 Group Title

66. experimentX Group Title

No ...

67. JenniferSmart1 Group Title

sum of all the Electric field lines $\phi=\oint E dr=\frac{Q}{\epsilon_0}$

68. JenniferSmart1 Group Title

volume?

69. experimentX Group Title

You should understand what is flux first ... Flux is like ... |dw:1362270773063:dw|

70. JenniferSmart1 Group Title

the amount of water that flows through it is A times $$v$$

71. JenniferSmart1 Group Title

$\frac{meter^2}{1}\times\frac{meters}{second}$ that doesn't seem right

72. experimentX Group Title

yeah .. that's correct. It's easy to visualize water ... |dw:1362271042142:dw|

73. JenniferSmart1 Group Title

yep

74. JenniferSmart1 Group Title

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75. JenniferSmart1 Group Title

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

76. JenniferSmart1 Group Title

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

77. experimentX Group Title

78. JenniferSmart1 Group Title

oh surface area of the sphere!

79. experimentX Group Title

yeah ... what is the total flux?

80. JenniferSmart1 Group Title

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 Group Title

i meant surface area...not areas

82. experimentX Group Title

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83. JenniferSmart1 Group Title

what's s? surface area?

84. JenniferSmart1 Group Title

ds...?

85. experimentX Group Title

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86. JenniferSmart1 Group Title

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87. JenniferSmart1 Group Title

what's that?

88. experimentX Group Title

sphere.

89. JenniferSmart1 Group Title

oh are you saying that $\int ds=s$ yeah I believe that but is s ...oh i see sphere

90. JenniferSmart1 Group Title

we're integrating with respect to the sphere, correct?

91. JenniferSmart1 Group Title

the surface area of the sphere

92. experimentX Group Title

yes ... but electric Field is constant .. since it's sphere.

93. JenniferSmart1 Group Title

yep

94. JenniferSmart1 Group Title

what would you limits be?

95. experimentX Group Title

and we are integrating surface ...not distance. lol ... the limits ... don't worry

96. JenniferSmart1 Group Title

so $\oint \textrm{never has limits?}$

97. experimentX Group Title

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98. JenniferSmart1 Group Title

yeah I get that E is constant everywhere over the sphere

99. JenniferSmart1 Group Title

surface*

100. experimentX Group Title

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101. JenniferSmart1 Group Title

oh because were integrating $\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}$

102. JenniferSmart1 Group Title

since the answer is just s

103. experimentX Group Title

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104. JenniferSmart1 Group Title

yep , makes sense....

105. JenniferSmart1 Group Title

I finally understand flux.... :)

106. experimentX Group Title

Not complete yet!!

107. JenniferSmart1 Group Title

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108. experimentX Group Title

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109. JenniferSmart1 Group Title

i'm not gonna look

110. JenniferSmart1 Group Title

not gonna look

111. JenniferSmart1 Group Title

not gonna look

112. JenniferSmart1 Group Title

ok fine lol

113. JenniferSmart1 Group Title

keep going...i'm sorry

114. experimentX Group Title

Now ... |dw:1362272365512:dw|

115. experimentX Group Title

this is called conservation of flux.

116. experimentX Group Title

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117. experimentX Group Title

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118. experimentX Group Title

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119. JenniferSmart1 Group Title

through both spheres? yes

120. experimentX Group Title

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 Group Title

interesting! yeah $$E\propto \frac 1{r^2}$$ $F_g\propto \frac 1{r^2}$

122. JenniferSmart1 Group Title

or is it $F_e\propto \frac 1{r^2}$ and $F_g\propto \frac 1{r^2}$

123. experimentX Group Title

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 Group Title

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125. experimentX Group Title

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126. JenniferSmart1 Group Title

oh|dw:1362273163066:dw| flux is still the same