Solve Gauss's Law and Flux: 55.3M Excess Electrons

In summary, the problem involves finding the net electric flux through a closed surface with 55.3 million excess electrons. The equations \Phi= E . A = Qin/\epsilon can be used, and the enclosed charge can be solved for by using Q/epsilon. The elementary charge of an electron and the number of electrons are known.
  • #1
dals2002
12
0
[SOLVED] Gauss's Law and Flux

ok so i have the biggest question of all i tried solving the following problem:

55.3 million excess electrons are inside a closed surface. What is the net electric flux through the surface?

now i tried the following equations:

[tex]\Phi[/tex]= E . A = Qin/[tex]\epsilon[/tex]

but i have the biggest question when there is excess electron does it mean that the electron field is negative, and if it does then that goes in Qin where then [tex]\Phi[/tex] would be equal to Qin/[tex]\epsilon[/tex], and epsilon being 8.85*10 (permittivity constant)

well any help in solving this problem is very grateful
 
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  • #2
You can solve for the enclosed charge, since the elementary charge of an electron (-1.60*10^-19 C) and the number of electrons is known. Just use Q/epsilon to solve for flux.
 
  • #3
oh wow that was easy, i always see things harder then they are, thanks terbum
 

Related to Solve Gauss's Law and Flux: 55.3M Excess Electrons

1. What is Gauss's Law and how does it relate to flux?

Gauss's Law is a fundamental law in physics that describes the relationship between electric charge and electric field. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface. This means that the amount of electric flux, or flow of electric field lines, through a surface is directly proportional to the amount of electric charge inside that surface.

2. What is the significance of having 55.3M excess electrons in this scenario?

In this scenario, having 55.3M excess electrons means that there is a large amount of negative charge present. This can have significant effects on the electric field and flux, as the presence of excess electrons can create a stronger electric field and increase the amount of electric flux.

3. How do you solve for Gauss's Law and flux in this situation?

To solve for Gauss's Law and flux in this scenario, you would need to know the total charge enclosed by the closed surface and the electric field at every point on the surface. Using the equation for Gauss's Law, you can calculate the electric flux through the surface by multiplying the electric field at each point by the surface area and summing them all together.

4. What factors can affect the calculation of Gauss's Law and flux?

The calculation of Gauss's Law and flux can be affected by several factors, including the shape and size of the closed surface, the distribution of the excess electrons, and the strength and direction of the electric field. Other external factors such as the presence of other charges or materials can also impact the calculation.

5. What practical applications does Gauss's Law and flux have in the real world?

Gauss's Law and flux have many practical applications in the real world, such as in the design and analysis of electrical circuits, the development of electronic devices, and in understanding the behavior of lightning and other atmospheric electricity phenomena. It is also used in electromagnetic theory to study the behavior of electric and magnetic fields and their interactions with each other.

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