Physics Book - User contributions [en]

Electric Flux

2015-12-02T00:39:33Z

Ericabush:

== The Main Idea ==

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.

'''A Mathematical Model:'''

<math display="block">\text{Electric Flux:}</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.

Combining these two equations gives:

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

== 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 all sides. What is the total electric flux of the box?

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

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= E*cosθ*A </math>

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= E*0*A</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*-1*3*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= 1400*1*3*4 = 16,800</math>

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

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

'''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?

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

Pulling out the constant E and theta, and integrating dA give:

<math display="block">Φelectric= E*A*cos(θ) </math>

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.

<math display="block">Φtop= 900*cos(60)*3*4= 5400 </math>

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= 900*cos(60)*3*4= 5400 </math>

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>

<math display="block">Φtotal= 5400 +5400 = 10800 </math>

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> 10800 = {Q\over ε_0} </math>

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

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

'''Difficult:'''

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.

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

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> E*cos(θ)*4\pi r^2 = 700* cos(90)*4\pi r^2 = 0 </math>

<math display="block"> 0 = {Q\over ε_0} </math>

<math display="block">Q= 0 </math>

== 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.

== References ==

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

Electric Flux

2015-12-02T00:30:34Z

Ericabush:

== The Main Idea ==

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.

'''A Mathematical Model:'''

<math display="block">\text{Electric Flux:}</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.

Combining these two equations gives:

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

== 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 all sides. What is the total electric flux of the box?

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

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= E*cosθ*A </math>

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= E*0*A</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*-1*3*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= 1400*1*3*4 = 16,800</math>

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

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

'''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?

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

Pulling out the constant E and theta, and integrating dA give:

<math display="block">Φelectric= E*A*cos(θ) </math>

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.

<math display="block">Φtop= 900*cos(60)*3*4= 5400 </math>

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= 900*cos(60)*3*4= 5400 </math>

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>

<math display="block">Φtotal= 5400 +5400 = 10800 </math>

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> 10800 = {Q\over ε_0} </math>

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

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

'''Difficult:'''

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.

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

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> E*cos(θ)*4\pi r^2 = 700* cos(90)*4\pi r^2 = 0 </math>

<math display="block"> 0 = {Q\over ε_0} </math>

<math display="block">Q= 0 </math>

== Connectedness ==

Is there an interesting industrial application?

== 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.

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

== References ==

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

Electric Flux

2015-12-02T00:28:10Z

Ericabush:

== The Main Idea ==

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.

'''A Mathematical Model:'''

<math display="block">\text{Electric Flux:}</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.

Combining these two equations gives:

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

== 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 all sides. What is the total electric flux of the box?

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

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= E*cosθ*A </math>

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= E*0*A</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*-1*3*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= 1400*1*3*4 = 16,800</math>

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

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

'''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?

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

Pulling out the constant E and theta, and integrating dA give:

<math display="block">Φelectric= E*A*cos(θ) </math>

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.

<math display="block">Φtop= 900*cos(60)*3*4= 5400 </math>

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= 900*cos(60)*3*4= 5400 </math>

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>

<math display="block">Φtotal= 5400 +5400 = 10800 </math>

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> 10800 = {Q\over ε_0} </math>

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

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

'''Difficult:'''

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.

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

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> E*cos(θ)*4\pi r^2 = 700* cos(90)*4\pi r^2 = 0 </math>

<math display="block"> 0 = {Q\over ε_0} </math>

<math display="block">Q= 0 </math>

== Connectedness ==

Is there an interesting industrial application?

== 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.

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

== References ==

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

Electric Flux

2015-12-02T00:07:20Z

Ericabush:

== The Main Idea ==

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.

'''A Mathematical Model:'''

<math display="block">\text{Electric Flux:}</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.

Combining these two equations gives:

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

== 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 all sides. What is the total electric flux of the box?

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

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= E*cosθ*A </math>

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= E*0*A</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*-1*3*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= 1400*1*3*4 = 16,800</math>

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

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

'''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?

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

Pulling out the constant E and theta, and integrating dA give:

<math display="block">Φelectric= E*A*cos(θ) </math>

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.

<math display="block">Φtop= 900*cos(60)*3*4= 5400 </math>

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= 900*cos(60)*3*4= 5400 </math>

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>

<math display="block">Φtotal= 5400 +5400 = 10800 </math>

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> 10800 = {Q\over ε_0} </math>

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

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

'''Difficult:'''

== Connectedness ==

Is there an interesting industrial application?

== 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.

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

== References ==

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

Electric Flux

2015-12-02T00:02:45Z

Ericabush:

== The Main Idea ==

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.

'''A Mathematical Model:'''

<math display="block">\text{Electric Flux:}</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.

Combining these two equations gives:

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

== 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 all sides. What is the total electric flux of the box?

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

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= E*cosθ*A </math>

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= E*0*A</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*-1*3*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= 1400*1*3*4 = 16,800</math>

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

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>
<math display="block">Φtotal= -16800 +16800 +0 +0 +0 +0 = 0</math>

'''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?

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

Pulling out the constant E and theta, and integrating dA give: electric flux= E*A*cos(θ)
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.

<math display="block">Φtop= 900*cos(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:

<math display="block"> Φbottom= 900*cos(60)*3*4= 5400 </math>

<math display="block">Φtotal= Φtop + Φbottom + Φright + Φleft + Φfront + Φback</math>

<math display="block">Φtotal= 5400 +5400 = 10800 </math>

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

<math display="block"> 10800 = {Q\over ε_0} </math>
ε_0 = 8.84e-12

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

'''Difficult:'''

== Connectedness ==

Is there an interesting industrial application?

== 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.

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

== References ==

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

Electric Flux

2015-12-01T21:25:42Z

Ericabush:

Electric Flux

2015-12-01T02:01:59Z

Ericabush:

Electric Flux

2015-12-01T02:00:27Z

Ericabush:

Electric Flux

2015-12-01T01:16:17Z

Ericabush:

Electric Flux

2015-12-01T01:14:05Z

Ericabush:

Electric Flux

2015-12-01T01:10:29Z

Ericabush:

Electric Flux

2015-12-01T01:04:34Z

Ericabush:

Electric Flux

2015-12-01T01:02:24Z

Ericabush:

== The Main Idea ==

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 related 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

'''A Mathematical Model'''

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

<math display="block">\text {Electric Flux:} Φelectric= \int \ {\vec{E}cosθdA} </math>
Where theta is the angle between the electric field vector and the surface normal.

Combining these two equations into Gauss's law gives:
<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

'''A Computational Model '''

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

== Examples ==

Be sure to show all steps in your solution and include diagrams whenever possible

== Simple
Middle
Difficult ==

== Connectedness ==

Is there an interesting industrial application?

== History ==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also
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Electric Flux

2015-12-01T01:00:54Z

Ericabush:

== The Main Idea ==

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 related 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

'''A Mathematical Model'''

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

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

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

'''A Computational Model '''

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

== Examples ==

Be sure to show all steps in your solution and include diagrams whenever possible

== Simple
Middle
Difficult ==

== Connectedness ==

Is there an interesting industrial application?

== History ==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

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Electric Flux

2015-12-01T00:48:34Z

Ericabush: /* The Main Idea */

== The Main Idea ==

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 related 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

'''A Mathematical Model'''

<math display="block">\text{Gauss's Law for Electric Fields:} \oint{ \vec{E} \cdot d\vec{A} } ⃗= {Q\over ε_0} </math>

'''A Computational Model '''

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

== Examples ==

Be sure to show all steps in your solution and include diagrams whenever possible

== Simple
Middle
Difficult ==

== Connectedness ==

Is there an interesting industrial application?

== History ==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

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

References
This section contains the the references you used while writing this page.

Electric Flux

2015-12-01T00:33:41Z

Ericabush:

== The Main Idea ==

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 related 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

'''A Mathematical Model'''

∮E ⃗∙dA ⃗ =Q/ε_0 .

'''A Computational Model '''

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

== Examples ==

Be sure to show all steps in your solution and include diagrams whenever possible

== Simple
Middle
Difficult ==

== Connectedness ==

Is there an interesting industrial application?

== History ==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

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

References
This section contains the the references you used while writing this page.

Electric Flux

2015-12-01T00:24:39Z

Ericabush:

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 related 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.

Gauss's Law:

File:Capture.jpg

2015-12-01T00:23:39Z

Ericabush: Gauss's Law for Electric Fields

Gauss's Law for Electric Fields

Electric Flux

2015-12-01T00:23:10Z

Ericabush: Created page with "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 related the electric flux through an area..."