Why is this 0? Help Gauss's law

In summary, we discussed the problem of finding the electric field for a long straight metal rod with a given radius and charge per unit length. After solving for the electric field at a distance of 3.00cm, it was determined that the field inside the rod is zero due to the fact that the charge resides only on the surface of the metal. We also compared this to the case of an insulating sphere, where the charges would be distributed throughout the volume. Finally, we discussed the possibility of finding the electric field for an insulating cylinder using Gauss's theorem.
  • #1
flyingpig
2,579
1

Homework Statement



A long straight metal rod has a radius of 5.00cm and a charge per unit length of 30.0nC/m. Find the electric field

a) 3.00cm

The Attempt at a Solution



It says a) = 0

I did some math and I got

[tex]2k\frac{\lambda}{r}[/tex]

The rod is solid, so should a E-field be radially outward from its symmetrical axis?
 
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  • #2
I am assuming that, what you wrote as, "the electric field a) 3.00cm" means "the electric field at 3.00cm from the center of the wire"


Answer to this question first: If you have a conducting sphere, carrying a charge say Q, than what is the electric field at a point inside it??
 
  • #3
first check theory on electric field inside conductors.
 
  • #4
Mandeep Deka said:
I am assuming that, what you wrote as, "the electric field a) 3.00cm" means "the electric field at 3.00cm from the center of the wire"


Answer to this question first: If you have a conducting sphere, carrying a charge say Q, than what is the electric field at a point inside it??

[tex]k\frac{q}{a^3}r[/tex] for a < r
 
  • #5
it is a metal rod. so all the charge would reside on the surface only, none can stay in the bulk.

so the field inside the rod is zero.
 
  • #6
flyingpig said:
[tex]k\frac{q}{a^3}r[/tex] for a < r

its for symmetrical charge distribution throughout volume. metals can't have that, all of their charge resides on surface! Can you tell why?

and i believe it must be "r<a"
 
  • #7
graphene said:
it is a metal rod. so all the charge would reside on the surface only, none can stay in the bulk.

so the field inside the rod is zero.

So, but a metal sphere can too? Isn't that what my equation is?

Oh wait, I think I get it. The key here is that my equation is for an insulating sphere, not a conducting one. Hence.
 
  • #8
flyingpig said:
So, but a metal sphere can too? Isn't that what my equation is?

Oh wait, I think I get it. The key here is that my equation is for an insulating sphere, not a conducting one. Hence.

Yes now you're right
 
  • #9
cupid.callin said:
Yes now you're right

If this was not a "metal" rod, or "insulating rod", then my numbers in the above would be right, right?
 
  • #10
i can't get you
 
  • #11
cupid.callin said:
i can't get you

If it is insulating, all the charges would be stuck and won't be all on the surface
 
  • #12
flyingpig said:
[tex]k\frac{q}{a^3}r[/tex] for a < r

If you are saying that this formula is for insulating sphere then you are right!

And also for the explanation, the charges will not move to the surface!

can you find the same for insulating cylinder?
 
  • #13
flyingpig said:
[tex]k\frac{q}{a^3}r[/tex] for r<a

If you are saying that this formula is for insulating sphere then you are right!

And also for the explanation, the charges will not move to the surface!

can you find the same for insulating cylinder?
just draw a gaussian surface with r < radius of cylinder and coaxial with cylinder and use gauss theorm
 

FAQ: Why is this 0? Help Gauss's law

Why is Gauss's Law important in science?

Gauss's Law is an important concept in science because it is a fundamental principle in the study of electromagnetism and helps us understand the behavior of electric fields. It also has many practical applications in fields such as engineering and physics.

What does it mean for something to have a charge of 0 in Gauss's Law?

In Gauss's Law, a charge of 0 means that there is no net electric charge present in a given region. This can occur when there are equal amounts of positive and negative charges canceling each other out, or when there are no charges present at all.

How does Gauss's Law relate to electric flux?

Gauss's Law states that the electric flux through a closed surface is proportional to the net electric charge enclosed by that surface. This means that the total electric flux through a surface is directly related to the amount of charge present within that surface.

Can Gauss's Law be used to calculate the electric field at a point?

Yes, Gauss's Law can be used to calculate the electric field at a point if there is a known distribution of charges present. This can be done by applying the law to a small Gaussian surface enclosing the point of interest and solving for the electric field using the equation E = Q/ε0.

How is Gauss's Law expressed mathematically?

Gauss's Law is expressed mathematically as ∮SE•dA = Q/ε0, where ∮SE•dA represents the electric flux through a closed surface, Q is the net charge enclosed by the surface, and ε0 is the permittivity of free space.

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