Understand the concept of Electric Field Flux

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Gauss’s Law Electric Flux
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Presentation transcript:

Chp 27 Goals Go to: http://www.aw-bc.com/knight Understand the concept of Electric Field Flux View charges as “sinks” and “sources” of Electric Field Understand “Gauss’ Law” Go to: http://www.aw-bc.com/knight

Chp 27 -- Gauss’ Law We’ve already discussed how charge can be thought of as a “source” or a “sink” of E-field line. It turns out that Electric Fields need to be symmetric around all spatial reflections for Gauss’s Law to be super useful. +++++++++++++++++++ Let’s see if this distribution of charge + hypothetical E-Field is symmetric in all dimensions. +++++++++++++++++++

The Flux of the E-field through a defined surface is:

The total electric flux through this box is 6 Nm2/C. 4 Nm2/C. 2 Nm2/C. 1 Nm2/C. 0 Nm2/C. STT27.3 Answer: C

Sample Problem: Find the E-field flux through this square. E= 20N/C

Sample Problem: Find the E-field flux through a 2-meter by 2-meter square where the E-field is given by the following equation. The square is located with one corner at the origin in the x-y plane.

Remember that charges can be thought of as “sources” and “sinks” of E-field. Let’s find the amount of E-field Flux “coming out of” a positive point charge.

Thus, the amount of flux created by a charge can be given as: When we combine the two ways we can think about flux we get “Gauss’ Law”:

Let’s use Gauss’ Law to find the E-field of an (infinitely) long straight wire.

Chapter 28 Summary Flux is defined as the normal component of E-field “poking” out of a defined surface. Gauss’ Law: The amount of E-field fluxing through a closed surface is equal to qenclosed.

You have a long straight wire that is charged up with a linear charge density of +. You then have a conducting cylinder of inner radius Rin and outer radius Rout. What is the E-field at: r<Rin, Rout>r>Rin, r>Rout? Also: what is the linear charge density on the inside surface and outside surface of the cylinder.

You have a hollow conductive sphere that is charged up to +5Q You have a hollow conductive sphere that is charged up to +5Q. Inside at the middle of the shell you place a metal sphere of charge of -2Q. Find the E-field everywhere.

Gauss’ Law ALWAYS works because it is a fundamental rule of the universe. HOWEVER, when trying to solve analytically you must have impressive symmetry for it to be useful. The three shapes of Gaussian surfaces that are “easy”: Cubes Spheres Cylinders These three shapes can help us solve problems that include: wires, spheres, very long cylinders, very large planes, point charge. If you don’t have a probably with appropriate symmetry, then you can use computational means when using Gauss’ Law

Also in Chp 25, we introduced the concept of an E-field. Chapter 25: Coulomb’s Law – The types of problems we could solve involved FBDs, anet, and other concepts that we discovered in Ph 211 and Ph 212. Also in Chp 25, we introduced the concept of an E-field. Chapter 26: We deeply investigated the definition of the E-field. We solved several examples where we had to find the E-field for a finite number of charges (such as a dipole) and/or a continuous distribution of charge. Chapter 27: We learned how to use Gauss’ Law To use this we first had to understand the concept of flux.

Find the E-field of a very large plate (l=w=2meters) charged to 5C Find the E-field of a very large plate (l=w=2meters) charged to 5C. Find out to what charge you need to charge a frog in order to levitate above the plate.