User:Jinhohah

Claimed by Jinho Hah

Patterns of Field in Space

The two major components in Physics II are interactions between electric fields and magnetic fields. This broad subject focuses from a simple topic, electric fields to a much complex idea, electromagnetic radiation. Though this course is very broad in terms of materials that it covers, each topic is very important in understanding the phenomena of electric and magnetic interactions between particles (protons, electrons, dipoles, point charge, capacitor, and, etc), as omitting one concept out of hundreds of concept could lead one approaching the problem differently. This Wiki Page will discuss Chapter 21 of the Matters and Interactions Text Book, 4th edition (Patterns of Field in Space), specifically Gauss's Law and Electric Flux. In order to understand these concepts, one first need to understand the definition of electric field and know each component of Gauss's Law.

Electric Flux

"Electric Flux" is a quantitative measure of the amount and direction of electric field over an entire surface of a specified object. There are two components in electric flux: direction of the electric field and magnitude of the electric field. These two sums up and give us the value, electric flux, which has a unit of Vm. In order to determine the direction of the electric field of an object, one need to figure out the x,y,z coordinates of the faces of an object and then calculate the normal vector that comes out of the surface. Secondly, to determine the direction of the electric field of an object, one first need to know the number of dimensions of an object (i.e: 6 faces in a rectangular prism) and areas for each face of the object. Finally, one should be able to calculate the electric flux of an object by multiplying the electric field at a location on each surface of the box by corresponding normal vector and multiplying this value by the area of the surface that was just calculated. One must repeat this process the remaining surfaces (faces) and by adding up all electric flux, that will be the electric flux of the object one wanted to calculate. This value is essential because it will be useful for calculating total charged enclosed inside the object later on. The above written method of calculating electric flux may be confusing at first, but knowing the Gauss's Law, being able to apply this Law to the real problem, and by going through the example below should make sure understanding of this concept.

Gauss's Law

The Gauss's Law simplifies definition of "Electric Flux" into a one simple equation.

Gauss's Law

Gauss's Law Examples

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

Electric flux on the disk, by using Gauss's Law, is the multiplication between Electric Field normal to the disk's surface and surface area of the disk.

Electric Field Normal to the Surface: E x sin(40) = 327 V/m x sin(40)

Surface Area of the disk: 3.14 x 0.02 x 0.02 m^2

Electric Flux on the Disk: 327 V/m x sin(40) x 3.14 x 0.02 x 0.02 m^2 = 0.264 Vm

Middling

The electric field has been measured to be vertically upward everywhere on the surface of a box 20 cm long, 4 cm high, and 3 cm deep, shown in the figure. All over the bottom of the box E1 = 1100 V/m, all over the sides E2 = 950 V/m, and all over the top E3 = 750 V/m.

Since E1, E2, and E3, are all measured to be vertically upward everywhere on the surface of a box, only the bottom surface and the top surface will be focused (multiplying the normal vector of other surfaces than the bottom and top surfaces will result in zero electric flux). The normal vector of the bottom surface is known to be <0,-1,0> and that of the top surface is known to be <0,1,0> assuming vertically upward is in +y direction.

E1 = <0, 1100, 0> V/m, E2 = <0, 950, 0> V/m, E3 = <0, 750, 0> V/m,

E1 corresponds to the bottom surface according to the diagram. Multiplying vector E1 with its normal vector of the bottom surface equals -1100 (dot product), and multiplying this vector by the area equals -1100 V/m x 0.20 m x 0.03 m = -6.6 Vm.

E3 corresponds to the top surface according to the diagram. Multiplying vector E3 with its normal vector of the top surface equals 750 (dot product) and multiplying this vector by the area equals 750 V/m x 0.20 m x 0.03 m = 4.5 Vm.

Therefore, the sum of the electric flux in this box equals -6.6 Vm + 4.5 Vm = -2.1 Vm

To determine the amount of charge enclosed by the box, we use Gauss's Law. Since we know the sum of the electric flux, in order to find q (inside the box) we just have to multiply summed electric flux and epsilon naught (8.85 e-12 unit)

Total Charge: -2.1 x 8.85 x 10 ^ (-12) = -1.8585 x 10 ^ (-11) C

Difficult

To calculate the electric flux you will have to know the normal vector or the electric field in normal direction of each side of the cube.

There are only two sides affecting the net electric flux on this cubical surface, bottom face and right-handed side face (rest will be zero).

Normal Electric Field (Bottom Face): E1 x sin(27) = 400 V/m x sin(27) x 0.35m x 0.35m = 22.2455 Vm

Normal Electric Field (Right-Handed Side Face): E1 x sin(27) = 400 V/m x sin(27) x 0.35m x 0.35m = 22.2455 Vm

Net Electric Field = 44.49 Vm

Net Charge inside the cubical surface = 44.19 Vm x 8.85 x 10 ^ (-12) = 3.94 x 10 ^ (-10) C

Connectedness

1. How is this topic connected to something that you are interested in?

1. How is it connected to your major?

I am majoring in Materials Science and Engineering and going further on from Electric Flux, there is a way to calculate Electric Flux Density (D). To obtain D, you have to multiply epsilon (permittivity) of the material to the Electric Field, E. Of course, different materials will have different Electric Flux. For example there are three major types of materials: insulator, semi-conductor, and conductor. Depending on their uses, you would want to make sure that the material of the desired product does not contain electric flux due to excess charge on the surface. This is the reason that electric flux and total charge of the closed surfaced objects are useful when as an engineer you want to think what type of properties you would desire for your product.

1. Is there an interesting industrial application?

History

Thermodynamics was brought up as a science in the 18th and 19th centuries. However, it was first brought up by Galilei, who introduced the concept of temperature and invented the first thermometer. G. Black first introduced the word 'thermodynamics'. Later, G. Wilke introduced another unit of measurement known as the calorie that measures heat. The idea of thermodynamics was brought up by Nicolas Leonard Sadi Carnot. He is often known as "the father of thermodynamics". It all began with the development of the steam engine during the Industrial Revolution. He devised an ideal cycle of operation. During his observations and experimentations, he had the incorrect notion that heat is conserved, however he was able to lay down theorems that led to the development of thermodynamics. In the 20th century, the science of thermodynamics became a conventional term and a basic division of physics. Thermodynamics dealt with the study of general properties of physical systems under equilibrium and the conditions necessary to obtain equilibrium.

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

Electric Field and Gauss's Law

External links

Gauss's Law

Gauss's Law and Examples

Electric Flux Examples

Internet resources on this topic

References

Chabay, Ruth W. Matter and Interactions Volume II: Electric and Magnetic Interactions. Hoboken, NJ: Wiley, 2015. Print.

"Electric Flux Density." SpringerReference (2011): n. pag. Web. 1 Dec. 2015. <http://www.colorado.edu/physics/phys1120/phys1120_sp08/notes/notes/Knight27_gauss_lect.pdf>.

"WebAssign." WebAssign. N.p., n.d. Web. 01 Dec. 2015. <http://webassign.net/>.