Gauss's Flux Theorem - Revision history

Cmarston7: Fixed broken link to simulation.

2025-04-13T19:30:29Z

Fixed broken link to simulation.

← Older revision		Revision as of 15:30, 13 April 2025
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	</math>, which visually shows the Gauss Flux Theorem in action.		</math>, which visually shows the Gauss Flux Theorem in action.

	<b>Access the simulation here: </b> https://trinket.io/glowscript/~~31d0f9ad9e~~		<b>Access the simulation here: </b> https://trinket.io/glowscript/ed6ad4894b08

	You can experiment with changing the position of the point charge inside the cube (at line 19) to see how the local electric field vectors change but the total charge will always remain the same as		You can experiment with changing the position of the point charge inside the cube (at line 19) to see how the local electric field vectors change but the total charge will always remain the same as

Cmarston7: Added a Glowscript model of the Gauss Flux Theorem with a cube enclosing a point charge. The previous page had only descriptions of observations instead of an actual Glowscript model.

2025-04-13T19:24:03Z

Added a Glowscript model of the Gauss Flux Theorem with a cube enclosing a point charge. The previous page had only descriptions of observations instead of an actual Glowscript model.

← Older revision		Revision as of 15:24, 13 April 2025
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	[[File:sign of flux.jpg\|600px]]		[[File:sign of flux.jpg\|600px]]

			We've just discussed how the electric flux depends on the electric field <math> \vec{E} </math> and the unit normal vector <math> \hat{n} </math> of a surface, and that electric flux is defined by the integral
			<math>
			\Phi_E = \oint_C \vec{E} \cdot d\vec{A}
			</math>,
			where <math>d\vec{A} = \hat{n}\,dA</math> is a small area with a normal that faces outward from the enclosed surface. Because of this relationship, the only the part of the electric field perpendicular to the surface contributes to the net flux.
			This is computational model, we take a cube to be a Gaussian surface enclosing a point charge inside at its center. Each face of the cube is subdivided into a small area where the contribution of the total flux is calculated to be the sum of all the electric fields on the areas with
			<math>
			d\Phi_E = \vec{E} \cdot d\vec{A}
			</math>.
			This model demonstrates that the sum of the electric field vectors at the surface of the cube is equal to
			<math>
			\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}
			</math>, which visually shows the Gauss Flux Theorem in action.

			<b>Access the simulation here: </b> https://trinket.io/glowscript/31d0f9ad9e

			You can experiment with changing the position of the point charge inside the cube (at line 19) to see how the local electric field vectors change but the total charge will always remain the same as
			<math>
			\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}
			</math> in accordance with the Gauss Flux Theorem.

	==Examples==		==Examples==

Cmarston7 at 23:17, 12 April 2025

2025-04-12T23:17:38Z

← Older revision		Revision as of 19:17, 12 April 2025
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	Edited by Anna Seppings (Spring 2019)		Edited by Anna Seppings (Spring 2019)
	Edited by Robbie van Zyl (Spring 2020)		Edited by Robbie van Zyl (Spring 2020)
	Edited by Cameron Marston (Spring 2025		Edited by Cameron Marston (Spring 2025)

	==The Main Idea==		==The Main Idea==

Cmarston7 at 23:17, 12 April 2025

2025-04-12T23:17:27Z

← Older revision		Revision as of 19:17, 12 April 2025
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	Edited by Anna Seppings (Spring 2019)		Edited by Anna Seppings (Spring 2019)
	Edited by Robbie van Zyl (Spring 2020)		Edited by Robbie van Zyl (Spring 2020)
			Edited by Cameron Marston (Spring 2025

	==The Main Idea==		==The Main Idea==

Robbievz: /* See also */

2020-04-21T21:31:08Z

Robbievz: /* Electric Field inside Cylinder of Charge */

2020-04-21T21:28:05Z

Electric Field inside Cylinder of Charge

← Older revision		Revision as of 17:28, 21 April 2020
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	This answer can also be in a different form if the linear charge density is used: <math>{\lambda} = \frac{Q}{L} </math>		This answer can also be in a different form if the linear charge density is used: <math>{\lambda} = \frac{Q}{L} </math>

	~~So, <math>Q = {\lambd[[File:Example.jpg]]a}L</math>~~

	Finally, the <math>\frac{Q}{L}</math> can be replaced by <math>{\lambda}</math> in the electric field equation:		Finally, the <math>\frac{Q}{L}</math> can be replaced by <math>{\lambda}</math> in the electric field equation:

Robbievz: /* Multiple Charges/Objects */

2020-04-21T21:17:45Z

Multiple Charges/Objects

← Older revision		Revision as of 17:17, 21 April 2020
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	One of the most challenging kinds of problems for gauss's law is when one is asked to find either the electric flux, electric field, or charge inside a given gaussian surface or enclosed volume. In the following example, I am going to calculate the electric field at a specified radius r using gauss's law and my knowledge regarding the law:		One of the most challenging kinds of problems for gauss's law is when one is asked to find either the electric flux, electric field, or charge inside a given gaussian surface or enclosed volume. In the following example, I am going to calculate the electric field at a specified radius r using gauss's law and my knowledge regarding the law:

	[[File:~~Example~~.jpg\|600px]]		[[File:GaussExampleMulti1.jpg\|600px]]

	Above are the question description and a supporting diagram. Notice that the first and third cases will be relatively easy because the radius of our gaussian surface(r) does not fall partially within a charged object. The second case is far more complex because the radius of the gaussian surface includes some portion of the charge from the spherical shell.		Above are the question description and a supporting diagram. Notice that the first and third cases will be relatively easy because the radius of our gaussian surface(r) does not fall partially within a charged object. The second case is far more complex because the radius of the gaussian surface includes some portion of the charge from the spherical shell.
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	Let's jump into some algebra:		Let's jump into some algebra:

	[[File:~~Example~~.jpg\|600px]]		[[File:GaussExampleMulti2.jpg\|600px]]

	The key to solving the second case was to realize that the amount of 4Q included in the gaussian surface can be found using a ratio describing how much volume the gaussian surface includes out of the total volume of the spherical shell. The ratio ended up looking something like this: (volume up till r - volume of space up to R1) / (volume up to R2 - volume of space up to R1 ) and you multiply this ratio by 4Q and then subtract Q(center charge) to get the total charge inside the gaussian surface.		The key to solving the second case was to realize that the amount of 4Q included in the gaussian surface can be found using a ratio describing how much volume the gaussian surface includes out of the total volume of the spherical shell. The ratio ended up looking something like this: (volume up till r - volume of space up to R1) / (volume up to R2 - volume of space up to R1 ) and you multiply this ratio by 4Q and then subtract Q(center charge) to get the total charge inside the gaussian surface.
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	And case three to round it up:		And case three to round it up:

	[[File:~~Example~~.jpg\|600px]]		[[File:GaussExampleMulti3.jpg\|600px]]

	==Differential Form==		==Differential Form==

Robbievz: /* Examples */

2020-04-21T18:21:27Z

Examples

@@ Line 187: / Line 187: @@
 This answer can also be in a different form if the linear charge density is used: <math>{\lambda} = \frac{Q}{L} </math>
-So, <math>Q = {\lambda}L</math>
+So, <math>Q = {\lambd[[File:Example.jpg]]a}L</math>
 Finally, the <math>\frac{Q}{L}</math> can be replaced by <math>{\lambda}</math> in the electric field equation:
@@ Line 198: / Line 198: @@
 [[File:NetExample.jpg|600px]]
 ==Differential Form==

Robbievz: /* Multiple Surfaces and Uniform Electric field */

2020-04-21T18:09:26Z

Multiple Surfaces and Uniform Electric field

← Older revision		Revision as of 14:09, 21 April 2020
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	An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.		An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

	[[File:Ex-2-2.JPG]]		[[File:Ex-2-2.JPG\|600px]]

	Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>		Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

Robbievz: /* Middling */

2020-04-21T18:09:10Z

Middling

← Older revision		Revision as of 14:09, 21 April 2020
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	===Middling===		===Middling===
	[[File:flux.jpg]]		[[File:flux.jpg\|600px]]

	===Multiple Surfaces and Uniform Electric field===		===Multiple Surfaces and Uniform Electric field===

← Older revision		Revision as of 17:31, 21 April 2020
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	http://phys.libretexts.org/TextMaps/Map%3A_Conceptual_Physics_(Crowell)/10%3A_Fields/10.7_Gauss'_Law_In_Differential_Form		http://phys.libretexts.org/TextMaps/Map%3A_Conceptual_Physics_(Crowell)/10%3A_Fields/10.7_Gauss'_Law_In_Differential_Form


			===Video Tutorials===
			https://www.youtube.com/watch?v=q1eor6oIuUo <------ SUPER HELPFUL

	https://www.youtube.com/watch?v=zn5ObHtGWeg		https://www.youtube.com/watch?v=zn5ObHtGWeg

			https://www.youtube.com/watch?v=QNIJC1emss8

			https://www.youtube.com/watch?v=EojpjIED_qg


	===External links===		===External links===