Electric Flux - Revision history

Ericabush at 00:39, 2 December 2015

2015-12-02T00:39:33Z

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	~~== Connectedness ==~~

	~~Is there an interesting industrial application?~~

	== History ==		== History ==

	Carl Gauss discovered this relation in 1835 and the equation was published in 1867. It is considered to be one of the four equations that are the basis for electrodynamics.		Carl Gauss discovered this relation in 1835 and the equation was published in 1867. It is considered to be one of the four equations that are the basis for electrodynamics.

	~~See also~~
	~~Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?~~

	~~Further reading~~
	~~Books, Articles or other print media on this topic~~

	~~External links~~
	~~Internet resources on this topic~~

Ericabush at 00:30, 2 December 2015

2015-12-02T00:30:34Z

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	2 m away from an unknown particle there is an electric field of magnitude 700 N pointing outward in all directions. What is the charge of the particle?		2 m away from an unknown particle there is an electric field of magnitude 700 N pointing outward in all directions. What is the charge of the particle?


	Because the field is radiating outward and is constant at a given location from the particle, the closed surface used for Gauss's law should be a spherical shell of radius 2 m with the particle at the center. This allows the field to be perpendicular to the surface at any given location and constant over the closed surface.		Because the field is radiating outward and is constant at a given location from the particle, the closed surface used for Gauss's law should be a spherical shell of radius 2 m with the particle at the center. This allows the field to be perpendicular to the surface at any given location and constant over the closed surface.


	<math display="block"> ~~SA sphere~~ = 4\pi r^2</math>		<math display="block"> SA_sphere = 4\pi r^2</math>

Ericabush at 00:28, 2 December 2015

2015-12-02T00:28:10Z

@@ Line 51: / Line 51: @@
 <math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
 <math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>
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 '''Difficult:'''

Ericabush at 00:07, 2 December 2015

2015-12-02T00:07:20Z

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	'''Simple:'''		'''Simple:'''


	A box with a height of 2m, a width of 3m, and a length of 4m has an electric field of (0, -1400, 0) N uniformly covering all sides. What is the total electric flux of the box?		A box with a height of 2m, a width of 3m, and a length of 4m has an electric field of (0, -1400, 0) N uniformly covering all sides. What is the total electric flux of the box?

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	'''Middle:'''		'''Middle:'''


	A box with a height of 2 m, a width of 3 m, and a length of 4 m has an electric field, E1, of magnitude 900 N that points in the positive x and y with a 30 degree angle to the top surface and electrics field, E2, of magnitude 900 N in the negative x and y with a 30 degree angle to the bottom surface. How much charge is in the box?		A box with a height of 2 m, a width of 3 m, and a length of 4 m has an electric field, E1, of magnitude 900 N that points in the positive x and y with a 30 degree angle to the top surface and electrics field, E2, of magnitude 900 N in the negative x and y with a 30 degree angle to the bottom surface. How much charge is in the box?

	<math display="block"> Φelectric= \int \ {\vec{E}cosθdA} </math>		<math display="block"> Φelectric= \int \ {\vec{E}cosθdA} </math>

	Pulling out the constant E and theta, and integrating dA give: ~~electric flux~~= EAcos(θ)		Pulling out the constant E and theta, and integrating dA give:

			<math display="block">Φelectric= EAcos(θ) </math>

	The angle between the surface normal and the electric field is not given.		The angle between the surface normal and the electric field is not given.
	Making a right triangle using the normal and the electric field. The angles in a triangle must add up to 180 and two of them, 30 and 90, are known. The angle between the field and the normal must be 180- 90-30= 60.		Making a right triangle using the normal and the electric field. The angles in a triangle must add up to 180 and two of them, 30 and 90, are known. The angle between the field and the normal must be 180- 90-30= 60.
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	<math display="block">Φtop= 900cos(60)3*4= 5400 </math>		<math display="block">Φtop= 900cos(60)3*4= 5400 </math>

	The bottom surface is ~~done~~ the same way, and the angle between the electric field and the normal is found to be 60 degrees, so:		The bottom surface is calculated the same way, and the angle between the electric field and the normal is found to be 60 degrees, so:

	<math display="block"> Φbottom= 900cos(60)3*4= 5400 </math>		<math display="block"> Φbottom= 900cos(60)3*4= 5400 </math>
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	<math display="block"> 10800 = {Q\over ε_0} </math>		<math display="block"> 10800 = {Q\over ε_0} </math>
	ε_0 = 8.84e-12

			<math display="block"> ε_0 = 8.84e-12 </math>

	Solving for Q gives:		Solving for Q gives:
	math display="block"> Q= {10800*8.84e-12}=9.56e-8 C </math>		<math display="block"> Q= {10800*8.84e-12}= 9.56e-8 </math> C

Ericabush at 00:02, 2 December 2015

2015-12-02T00:02:45Z

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 '''Simple:'''
-A box with a height of 2m, a width of 3m, and a length of 4m has an Electric field of (0, -1400, 0) N uniformly covering it. What is the total electric flux of the box?
+A box with a height of 2m, a width of 3m, and a length of 4m has an electric field of (0, -1400, 0) N uniformly covering all sides. What is the total electric flux of the box?
 Electric flux= <math display="block"> Φelectric= \int \ {\vec{E}cosθdA} </math>
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 '''Middle:'''

Ericabush at 21:25, 1 December 2015

2015-12-01T21:25:42Z

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	<math display="block">\text{Electric Flux:~~} Φelectric ={ Q \over ε_0~~} </math>		<math display="block">\text{Electric Flux:}</math>


	<math display="block">\~~text {Electric Flux:~~} Φelectric= \int \ {\vec{E}cosθdA} </math>		<math display="block"> Φelectric ={ Q \over ε_0} </math>


			<math display="block"> Φelectric= \int \ {\vec{E}cosθdA} </math>

	Where theta is the angle between the electric field vector and the surface normal.		Where theta is the angle between the electric field vector and the surface normal.

Ericabush at 02:01, 1 December 2015

2015-12-01T02:01:59Z

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	'''Middle:'''		'''Middle:'''


	'''Difficult:'''		'''Difficult:'''
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	Internet resources on this topic		Internet resources on this topic

	References
			== References ==

	[http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1]		[http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1]
	[https://en.wikipedia.org/wiki/Gauss%27s_law]		[https://en.wikipedia.org/wiki/Gauss%27s_law]

Ericabush at 02:00, 1 December 2015

2015-12-01T02:00:27Z

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	'''A Mathematical Model'''		'''A Mathematical Model:'''


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	'''~~A Computational Model~~ '''		== Examples ==

			'''Simple:'''
			A box with a height of 2m, a width of 3m, and a length of 4m has an Electric field of (0, -1400, 0) N uniformly covering it. What is the total electric flux of the box?

			Electric flux= <math display="block"> Φelectric= \int \ {\vec{E}cosθdA} </math>

	~~How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript~~		The electric field and area of each side are constant, so they can be pulled out of the integral to give:
			<math display="block">Φelectric= EcosθA </math>

	== ~~Examples~~ ==		For the front, back, left and right sides the angle between the normal and the electric field is 90 degrees, and cos(90)=0. For each of those sides, <math display="block"> Φelectric= E0A</math> , so there is no electric flux through those sides.

			On the top, the normal points up and the electric field points down, so the angle between them is 180. cos(180)= -1. Electric field and the dimensions are plugged in to give:
			<math display="block">Φtop= 1400-13*4 = -16,800</math>

			On the bottom face, the normal points down and the electric field vector also points dowm, so the angles between them is 0. cos(0)=1. When the electric field and dimensions are plugged in the electric flux equals:
			<math display="block">Φbottom= 140013*4 = 16,800</math>

			The total electric flux is the sum of all of the fluxes.

	~~Be sure to show all steps in your solution and include diagrams whenever possible~~		<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
			<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

			'''Middle:'''

	~~== Simple~~		'''Difficult:'''
	~~Middle~~
	Difficult ==

Ericabush at 01:16, 1 December 2015

2015-12-01T01:16:17Z

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	References		References
	http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1		[http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1]
			[https://en.wikipedia.org/wiki/Gauss%27s_law]

Ericabush at 01:14, 1 December 2015

2015-12-01T01:14:05Z

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	Electric flux through an area is the electric field multiplied by the area of a plane that is perpendicular to the field. Gauss's Law relates the electric flux through an area to the amount of charge enclosed in that area. Gauss's law ~~must~~ be ~~used along a~~ closed surface~~, but any chosen surface that contains~~ the ~~same~~ amount of charge ~~will give~~ the ~~same answer~~		Electric flux through an area is the electric field multiplied by the area of a plane that is perpendicular to the field. Gauss's Law relates the electric flux through an area to the amount of charge enclosed in that area. Gauss's law can be applied to any closed surface and calculates the amount of charge enclosed based on the electric field at that closed surface.