Gaussian Surface: Electric Field & dA Visualization

In summary, when calculating electric field due to an infinitely long line charge, we assume cylinder to be a gaussian surface and electric field and small area element of the cylinder acts along the same direction at the curved surface and they act perpendicular direction at the plane faces. However, in the case of electric field due to an infinite plane sheet of charge, we again assume cylinder to be a Gaussian surface but in this case at the plane faces electric field and surface element dA acts along the same direction and they act perpendicular along the curved surface. Area element dA has a direction perpendicular to the surface element. If you know the direction of the electric field, a good choice of the dA vector orientation is in the same direction as the
  • #1
logearav
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Homework Statement


In the case of calculating electric field due to an infinitely long line charge, we assume cylinder to be a gaussian surface and electric field and small area element of the cylinder acts along the same direction at the curved surface and they act perpendicular direction at the plane faces.
in the case of electric field due to an infinite plane sheet of charge we again assume cylinder to be a Gaussian surface but in this case at the plane faces electric field and surface element dA acts along the same direction and they act perpendicular along the curved surface. i am confused. please let me know how to visualise area element dA. i know electric field vector acts perpendicular to the surface but how to know the direction of the area element dA? help in this regard is greatly appreciated

Homework Equations





The Attempt at a Solution


 
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  • #2
Remember, dA represented as a vector has a direction perpendicular to the surface element. If you know the direction of the electric field, a good choice of the dA vector orientation is in the same direction as the (local) electric field which makes the surface element perpendicular to the local electric field. Then the dot product E.dA becomes just the product of magnitudes E dA and you don't have to worry about the cosine of the angle between them because it is 1.

So if you have a positive point charge that radiates electric field lines radially outward, what would be a good choice of dA? And when you have many such dAs next to one another, what kind of Gaussian surface do you get?
 
  • #3
but sir, if i draw a cylinder the plance face is elliptical in shape. dA should be perpendicular to the surface. i can draw dA in top direction, right direction and left direction. all these directions are perpendicular to the surface. what is the correct choice of direction.
 
  • #4
The correct choice of direction is perpendicular and away from the cylinder. Please look up "Gaussian surface" in wikipedia. There is a decent drawing of a cylindrical Gaussian surface for a linear charge distribution with elements dA drawn in. A picture is worth a thousand words. If you still have questions after you study the wikipedia article, please ask again.
 
  • #5
I think people are glossing over the fact that there is a radial ds vector as well. Just because the flux is zero there doesn't mean it isn't a valid direction. All three surfaces need to be taken into account, the two caps and the cylinder itself. When you do the calculations you will see that only the two caps contribute to the field and I guess this is why everyone is ignoring the cylinder itself.
 
  • #6
Thanks a lot mr.kuruman and mr. cyosis. now i understood the concept. thanks for sparing your valuable time.
 

1. What is a Gaussian surface and how is it related to electric fields?

A Gaussian surface is an imaginary surface that is used to study the electric field in a given region. It is a closed surface that encloses a charge or a group of charges. The electric field at every point on the surface is perpendicular to the surface and the magnitude of the field is the same at every point on the surface.

2. How is the electric field calculated using a Gaussian surface?

The electric field is calculated by using Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of the medium. This means that the electric field can be determined by calculating the electric flux through the Gaussian surface and dividing it by the charge enclosed.

3. What is the significance of using a Gaussian surface to visualize the electric field?

Using a Gaussian surface allows us to simplify the calculation of the electric field by considering only the charge enclosed by the surface. This makes it easier to analyze and understand the behavior of electric fields in different situations. Additionally, using a Gaussian surface can help us determine the direction and strength of the electric field at various points in space.

4. How does the shape and orientation of the Gaussian surface affect the calculation of electric fields?

The shape and orientation of the Gaussian surface do not affect the calculation of electric fields as long as the surface encloses the charge or charges of interest. The size of the surface, however, can affect the accuracy of the calculation. A smaller surface will give a more accurate result, but it may also be more difficult to calculate the electric flux through it.

5. Can a Gaussian surface be used to calculate the electric field for any charge distribution?

Yes, a Gaussian surface can be used to calculate the electric field for any charge distribution as long as the surface encloses the charge or charges of interest. This includes point charges, line charges, surface charges, and volume charges. However, the shape and size of the Gaussian surface may need to be adjusted depending on the charge distribution to get an accurate result.

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