• Support PF! Buy your school textbooks, materials and every day products Here!

Gauss's law and coulombs law

  • Thread starter Fanta
  • Start date
  • #1
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, couldnt 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?
 

Answers and Replies

  • #2
Doc Al
Mentor
44,877
1,129
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
38
0
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
Doc Al
Mentor
44,877
1,129
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 lets you say that the field is uniform on that surface, then Gauss's law can be used.
 
  • #5
38
0
got it, thanks!
 
  • #6
38
0
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
Doc Al
Mentor
44,877
1,129
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
38
0
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
Doc Al
Mentor
44,877
1,129
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
38
0
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 paralel to the sheets, as you mentioned) the area is not dependent from the displacement, so the field wouldnt be either?
 
  • #11
Doc Al
Mentor
44,877
1,129
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
38
0
opposite.
 
  • #13
Doc Al
Mentor
44,877
1,129
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 paralel to the sheets, as you mentioned) the area is not dependent from the displacement, so the field wouldnt be either?
That's true.
 
  • #14
38
0
so, outside of the sheets the electric field equals zero. Okay, got it.
Thanks!
 

Related Threads for: Gauss's law and coulombs law

Replies
2
Views
6K
Replies
4
Views
4K
Replies
4
Views
797
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
22K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
8
Views
2K
Top