Gauss's law and coulombs law

In summary, we discussed the use of Gauss's law and Coulomb's law to calculate electric fields in various situations. For an infinitely long charged rod, Coulomb's law can be used to find the field at an arbitrary point, while Gauss's law is only applicable in cases of high symmetry. For example, between two infinite planes with uniform charge distributions, Gauss's law can be used to find the electric field between them. However, outside of the planes, Coulomb's law must be used. We also discussed the importance of choosing an appropriate gaussian surface when using Gauss's law.
  • #1
Fanta
38
0
When should I use one and when should I use the other?
For example, suppose I have a rod of length 2L, with an edge on the point -L on the X axis and another on L.
The rod is uniformly charged, with total charge Q>0.
having that said, if i wanted to calculate the electric field in an arbitrary (0,y,0) point, couldn't I use Gauss's law, using a cilinder as my gauss surface?

And if the field is not specifically radial, in order to calculate it for a (x,0,0) point, i would have to integrate using coulomb's law, considering it a continuous distribution, correct?

So, the problem lies within the first problem:
It gives me two different results wether i use the method described(Gauss's law), or if I use coulomb's law for continuous distributions of charge, and integrate from -L to L.

which of the methods is wrong, and why?
 
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  • #2
For a finite rod, Gauss's law won't help since the field will not be uniform on your cylindrical gaussian surface. You can always use Coulomb's law to find the field if you have the charge distribution. Gauss's law only helps in cases of high symmetry--like an infinitely long charged rod or a charged sphere.
 
  • #3
so, for example, for two infinite planes with symmetrical charges (if the charge is uniform throughout them) i can use a gaussian surface to determine the electric field?
 
  • #4
Fanta said:
so, for example, for two infinite planes with symmetrical charges (if the charge is uniform throughout them) i can use a gaussian surface to determine the electric field?
Sure. As long as symmetry let's you say that the field is uniform on that surface, then Gauss's law can be used.
 
  • #5
got it, thanks!
 
  • #6
for example, the electric field between two symmetrically charged infinite planes would be given by :

[tex]\oint E(r) \cdot n \cdot dS = \frac{Q_{interior}}{\epsilon_{0}}[/tex]

so, between both planes, E(r) = 0, and for r > a or r < a:

[tex]E(r) = \frac{\sigma}{\epsilon_{0}}[/tex]

with sigma being the surface density of charge.

but this way the field is not a fuction of the displacement (r), so i think i might be doing something wrong.

The problem is perhaps with my choice of gaussian surface as a cube. Since it's side area is not a fuction of it's distance to the center.
 
  • #7
Fanta said:
for example, the electric field between two symmetrically charged infinite planes would be given by :

[tex]\oint E(r) \cdot n \cdot dS = \frac{Q_{interior}}{\epsilon_{0}}[/tex]

so, between both planes, E(r) = 0, and for r > a or r < a:

[tex]E(r) = \frac{\sigma}{\epsilon_{0}}[/tex]

with sigma being the surface density of charge.
That's the field between two oppositely charged infinite sheets.

but this way the field is not a fuction of the displacement (r), so i think i might be doing something wrong.
The field from an infinite sheet of charge does not depend on distance from the sheet.

The problem is perhaps with my choice of gaussian surface as a cube. Since it's side area is not a fuction of it's distance to the center.
But there's no flux through the sides that are perpendicular to the sheets.
 
  • #8
between them?
So how would you calculate the field outside of them?

And why is there a field between the sheets, if the charge in the interior of the surface equals zero?
 
  • #9
Fanta said:
between them?
So how would you calculate the field outside of them?
Use Gauss's law, of course. What's the net charge enclosed in your gaussian surface in that case?

And why is there a field between the sheets, if the charge in the interior of the surface equals zero?
I don't know what you mean. The field from each sheet adds up in between them.

Just to make sure we are talking about the same problem, try and state it completely.
 
  • #10
Consider two infinite and plane uniform distributions of electric charge. One of the distributions is in the plane x = -a and the charge density sigma > 0. The other is in the plane x = a and the density is symmetric.

I think we are. However i remember to have learned that when applying the gauss's law we consider the charge inside of the gaussian surface, not the external charges.
If the gaussian surface is between the sheets it has no charge inside it, right?
So I thought the electric field would be 0, but it obviously doesn't make much sense, since a uniform field must be created since they are infinite and symmetrically charged.

The net charged enclosed would be the double, since the surface catches both sheets. However how is the field dependent to the displacement (outside of the sheets - r > a or r < -a)?

If I use gauss's law with a square cubic surface (I will only consider the two surfaces parallel to the sheets, as you mentioned) the area is not dependent from the displacement, so the field wouldn't be either?
 
  • #11
Fanta said:
Consider two infinite and plane uniform distributions of electric charge. One of the distributions is in the plane x = -a and the charge density sigma > 0. The other is in the plane x = a and the density is symmetric.
Do both charge distributions have the same or opposite signs?
 
  • #12
opposite.
 
  • #13
Fanta said:
Consider two infinite and plane uniform distributions of electric charge. One of the distributions is in the plane x = -a and the charge density sigma > 0. The other is in the plane x = a and the density is symmetric.
OK, the sheets are oppositely charged.

I think we are. However i remember to have learned that when applying the gauss's law we consider the charge inside of the gaussian surface, not the external charges.
If the gaussian surface is between the sheets it has no charge inside it, right?
That's true for any gaussian surface that doesn't include the sheets.
So I thought the electric field would be 0, but it obviously doesn't make much sense, since a uniform field must be created since they are infinite and symmetrically charged.
All it means is that the net flux is zero.

The net charged enclosed would be the double, since the surface catches both sheets. However how is the field dependent to the displacement (outside of the sheets - r > a or r < -a)?
If your gaussian surface includes both sheets, the net charge enclosed will be zero.

If I use gauss's law with a square cubic surface (I will only consider the two surfaces parallel to the sheets, as you mentioned) the area is not dependent from the displacement, so the field wouldn't be either?
That's true.
 
  • #14
so, outside of the sheets the electric field equals zero. Okay, got it.
Thanks!
 

What is Gauss's Law?

Gauss's Law is a fundamental law of electromagnetism that relates the distribution of electric charges to the resulting electric field. It states that the electric flux through a closed surface is proportional to the enclosed charge.

What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two point charges. It states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

How are Gauss's Law and Coulomb's Law related?

Gauss's Law is essentially a mathematical consequence of Coulomb's Law. It can be derived from Coulomb's Law by considering the electric field due to a point charge and integrating it over a closed surface.

What is the significance of Gauss's Law and Coulomb's Law?

Gauss's Law and Coulomb's Law are important laws in electromagnetism that help us understand and predict the behavior of electric charges and electric fields. They have numerous applications in various fields such as engineering, physics, and chemistry.

Are Gauss's Law and Coulomb's Law applicable in all situations?

No, Gauss's Law and Coulomb's Law are valid only in electrostatic situations where charges are at rest. They do not apply to situations involving moving charges or changing magnetic fields.

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