Can a Gaussian Surface Confirm Zero Electric Field Between Two Charges?

In summary, the goal is to find the electric field at point A, which is halfway between two positive point charges (+q). By using a gaussian surface, we can prove that the electric field at point A is zero. This is because external charges do not affect the flux through the entire surface, even though they may affect individual portions of the surface. Using symmetry and the formula for electric field created by a point charge, we can easily calculate the electric field at point A.
  • #1
Gamezwn
2
0
Warning: Homework Template Must Be Used
1. We have two positive point charges (+q) at a distance from each other
2. Goal is to Find electric field at point A halfway between the point charges
3.By logic and summation of fields we should get ZERO
4. But how would I use a gaussian surface (sphere i assume ) to prove at point A E=0 n/c
5. Personally, I'm attempting a spherical surface around one of the points with radius of :(half distance between charges)
https://drive.google.com/file/d/0B0TaDjU5g_GAVm9hUjliTHhSM1BSbXoyRVp3dXd6MDR4TzE4/view?usp=sharing
 

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  • #2
I think you are making it harder than it really is. There is a formula for the electric field created by a point charge. Use that together with symmetry arguments since an electric field is a vector field.
 
  • #3
Avatrin said:
I think you are making it harder than it really is. There is a formula for the electric field created by a point charge. Use that together with symmetry arguments since an electric field is a vector field.
Yeah I know It is E= kQ/r^2 ...i just want to understand the effect of external charges on a gaussian surface...like in my book they said that external charges don't affect the gaussian surface because all the external field lines come in and come out...no idea what that means tbh
 
  • #4
Okay, I understand. The idea is that, for a gaussian surface, the same amount of field lines will be pointing into the surface as will be pointing out.

Lets take a very simple example; Think of a perfect sphere in an electric field that is constant everywhere. You are looking directly at the sphere, and let's say the field vectors are pointing towards your right. Using symmetry, it should be easy to see that the net electric flux through the sphere is zero (as much is "coming in" as is "leaving"). Gauss's law says that is true in general; If the electric field is generated by something outside the gaussian surface, the electric flux due to that thing through the surface will be zero.

So, the external charges do affect individual portions of the surface. However, they do not affect the flux through the entire surface. That will remain zero.
 

1. What is a Gaussian surface?

A Gaussian surface is an imaginary surface or boundary used in Gauss's law to simplify the calculation of electric fields. It is a closed surface that surrounds a group of charges and allows for the integration of the electric field over the entire surface.

2. How does a Gaussian surface help in calculating electric fields?

By using a Gaussian surface, the calculation of the electric field can be simplified to a single parameter, the total charge enclosed by the surface. This allows for the application of Gauss's law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

3. What are some common examples of Gaussian surfaces?

Some common examples of Gaussian surfaces include a spherical surface, a cylindrical surface, or a cubical surface. The choice of the Gaussian surface depends on the symmetry of the charge distribution being considered.

4. Can a Gaussian surface be used for any charge distribution?

No, a Gaussian surface can only be used for charge distributions with a high degree of symmetry, such as a spherical, cylindrical, or planar distribution. For more complex charge distributions, other methods, such as numerical integration, may be necessary.

5. How is the electric field found using a Gaussian surface?

The electric field can be found by first choosing a Gaussian surface that encloses the charge distribution and has a high degree of symmetry. Then, using Gauss's law, the electric field can be calculated by dividing the total charge enclosed by the surface by the permittivity of free space. This can be written as E = Q/ε0, where E is the electric field, Q is the total charge enclosed, and ε0 is the permittivity of free space.

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