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AP Physics C: E & M

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Unit 1

1.4 Gauss' Law

2 min readโ€ขnovember 3, 2020

peter57616

Peter Apps


What the Flux?!?

Flux is a very useful concept to help describe a wide variety of physics concepts. We're going to apply it here for electric fields, and then use it to help describe magnetic fields. Basically flux describes how much of something goes through a given area.
We're going to imagine an area on the surface of a charged object. It doesn't matter what the object is. The electric flux is then described by how many electric field lines pass through the area. Generally, we define the area to be parallel to the electric field, since this simplifies the math. However, if we can't do that, we take the dot product between the area vector and the electric field to determine the flux.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-eiozGkYRfc0M.PNG?alt=media&token=8ac6123f-3c9c-44c3-ab84-2d74222318ee
Let's look at the total flux in this image below.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-vGU1r2KiTc54.png?alt=media&token=a219713f-0fb6-44ae-988f-cc41fa75b4f1

Image from opentextbookbc.ca

When field lines enter a closed surface, the flux is negative. When they exit the surface, the flux is positive.
In this case, we see that there are 11 field lines entering the area ABCD and 11 field lines exiting FGHK. The flux from these two areas cancel out, and, therefore, the total flux is 0. A simple way to think about this is that if there isn't a source of charge or a sink (something that absorbs charge) **inside the area, the net flux is 0.

Gauss' Law

Gauss's Law relates the flux on a closed surface to the amount of charge enclosed by the surface.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-fzFo5njWwXTe.PNG?alt=media&token=3c50f7a9-22a6-4b2c-829d-3bd072dc10d0
Let's break this formula down a bit and see where it comes from.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-731iSb05lEK1.PNG?alt=media&token=64408e90-357e-46f9-94cd-4a61a2e0b5e6
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-Ur0AAOWHhGUR.PNG?alt=media&token=f6f8ded5-7da9-45f7-9e8f-0d3d43d00980
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-smI1CfIgEx9Q.PNG?alt=media&token=6e00abf8-e161-479b-9bb2-b8c61f6409b4
Let's try to find the flux. Remember that E is constant across the entirety of the surface. Adding up all the partial areas of the sphere gives us the surface area. (A= 4*pi*r^2).
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-IJ4jNtf2SwnH.PNG?alt=media&token=414b2d77-8df3-4e13-aba2-2c04ca317bd9
In short, Gauss's Law states that sum of the charge sources within a closed surface is equal to the total electric flux through the surface.

Practice Problems

1.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-1BnO6wWbrcaM.png?alt=media&token=25305761-7992-4da5-a4a1-9182fce414e1

Image from AP Classroom

Answer:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-SmEtbcx74NpU.PNG?alt=media&token=2b94d63a-a002-43a3-9a40-c27135982240
2.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-8WQRfFQfwcKH.png?alt=media&token=0659f9da-950d-4f2f-a181-80e80653296a

Image from AP Classroom

Answer:

We know there is +Q enclosed in the inner cylinder. We can see 4 lines of flux leaving that cylinder. With the larger cylinder, there are 8 lines entering the cylinder. This means that the net charge enclosed by the larger cylinder is -2Q. So the outer cylinder must have a charge of -3Q.
3.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-YWv5lCO5hej5.png?alt=media&token=613a0482-6313-4754-9c18-bc9bb87a34ad

Image from AP Classroom

Answer:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-6petMcsXYF74.PNG?alt=media&token=bb720857-398e-4060-878f-5af4c29a5f37
4.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-FMmF5rL1gpDQ.png?alt=media&token=4e2e5ccb-3502-4f6f-9c92-5b51944bba0b

Image from AP Classroom

Answer:

The sphere is conducting, so any fields outside it can't penetrate inside and don't influence the internal field. The only charge enclosed is +Q, so choice A is the correct answer.

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