Electromagnetism problem using Gauss's Law

AI Thread Summary
The discussion revolves around using Gauss's Law to calculate the electric field of an infinitely long insulating cylinder with a specific charge density. Participants express confusion about applying Gauss's Law and the relevant equations, particularly regarding the surface integral and the meaning of terms like "dA." A detailed solution is provided, showing that the electric field outside the cylinder is E = 3rR^2/2ε0 for r > R, and E = 3r^3/2ε0 for r < R. Additionally, a new participant inquires about applying similar reasoning to estimate the electromagnetic field near high voltage power lines, which is confirmed to be possible using the same principles. The conversation highlights the importance of understanding Gauss's Law and its applications in electromagnetism.
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Homework Statement


Using Gauss's law calculate the electric field everywhere for the infinitely long insulating cylinder of radius R and charge density of rho = 3r^2(nc/m^3). SHOW ALL YOUR WORK INCLUDING DIAGRAMS.


Homework Equations



I am lost.

The Attempt at a Solution



I don't know where to start
 
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Well...what is Gauss's Law? Write out. You should have had that in your "relevant eqtn"!
 
Electric Flux = qin/Eo
 
In integral form...?
 
I'm not sure what you mean.
 
Have you seen this before?

\oint \vec{E} \cdot d\vec{A} = \frac{\sum Q_{en}}{\epsilon_0}

I pretty much drilled this equation into the back of my head...
 
No. I'll take it the surface integral of E dot dA is equal to flux. I can remember that. I am still confused about how to use that equation to solve this problem.
 
Well, the cylinder is infinitely long. What is the E-field emitting? what happens if the cylinder is a wire?
 
Is it emitting charge? I don't know, what happens?
 
  • #10
You have a long wire (forget being cylinder for now), what is the E-field? What is the direction?
 
  • #11
If it helps, draw a Gaussian Surface over the wire/cylinder.
 
  • #12
I have absolutely no idea.
 
  • #13
Do you know what Gauss's Law means...? What it states? Don't just direct meback to the formula. What is "dA"?
 
  • #14
This was obviously posted a while ago, answering may help others though so:

Begin by constructing a cylindrical Gaussian surface of radius r and length l around the cylinder. Then the electric field outside of the cylinder is, by Gauss's law:

\ointE.da = Q / ε0 where Q is the total ENCLOSED charge.

The da on the left hand side refers to the area that the electric field passes through the Gaussian surface, not of the cylinder itself:

E x 2∏rl = Q/ε0

Now, the Q enclosed is the volume charge density multiplied by the volume in which Q is contained (i.e. the volume of the cylinder):

Q = ρ x ∏R^2 x l (R - the radius of the CYLINDER not the Gaussian surface is used)

Substitute back in:

E x 2∏rl = ρ x ∏ R^2 x l / ε0
E = ρ x R^2 / 2rε0

Then substitute ρ = 3r^2:

E = 3rR^2/2ε0 (outside; r>R)

For inside the enclosed charge has "the same r":

E x 2∏rl = ρl∏r^2 / ε0
E = ρr / 2ε0

Substitute ρ:

E = 3r^3 / 2ε0 (inside; for r < R)
 
Last edited:
  • #15
lmcelroy said:
This was obviously posted a while ago, answering may help others though so:

Begin by constructing a cylindrical Gaussian surface of radius r and length l around the cylinder. Then the electric field outside of the cylinder is, by Gauss's law:

\ointE.da = Q / ε0 where Q is the total ENCLOSED charge.

The da on the left hand side refers to the area that the electric field passes through the Gaussian surface, not of the cylinder itself:

E x 2∏rl = Q/ε0

Now, the Q enclosed is the volume charge density multiplied by the volume in which Q is contained (i.e. the volume of the cylinder):

Q = ρ x ∏R^2 x l (R - the radius of the CYLINDER not the Gaussian surface is used)

Substitute back in:

E x 2∏rl = ρ x ∏ R^2 x l / ε0
E = ρ x R^2 / 2rε0

Then substitute ρ = 3r^2:

E = 3rR^2/2ε0 (outside; r>R)

For inside the enclosed charge has "the same r":

E x 2∏rl = ρl∏r^2 / ε0
E = ρr / 2ε0

Substitute ρ:

E = 3r^3 / 2ε0 (inside; for r < R)
May I join the discussion?

Will this line of reasoning apply to calculating (estimating) the electromagnetic field (in Gauss) that one would expect to measure at a given distance from a high voltage power line? Since I would not have the charge available, but may have the voltage and line current, could the field be calculated in Gauss (or Teslas and then multiply by 10^4)? Will you show me the formula?
 
  • #16
SteveI46 said:
May I join the discussion?

Will this line of reasoning apply to calculating (estimating) the electromagnetic field (in Gauss) that one would expect to measure at a given distance from a high voltage power line? Since I would not have the charge available, but may have the voltage and line current, could the field be calculated in Gauss (or Teslas and then multiply by 10^4)? Will you show me the formula?

Hello Steve, welcome to PF!
It would be better if you start a new discussion, do you see the date of the last post? :smile:
 
  • #17
SteveI46 said:
May I join the discussion?

Will this line of reasoning apply to calculating (estimating) the electromagnetic field (in Gauss) that one would expect to measure at a given distance from a high voltage power line? Since I would not have the charge available, but may have the voltage and line current, could the field be calculated in Gauss (or Teslas and then multiply by 10^4)? Will you show me the formula?

Yes, the exact same method can be applied. A typical "text-book example" gives an infinitely long wire with a linear charge density λ. Just as with a cylinder a cylindrical Gaussian surface can be used for a wire (a wire in reality is really just a thin cylinder so this makes sense).
For your question however if you have a scalar potential (voltage, V) there is a simple formula connecting this to the E field:

E= -∇V - δA/δt , where A is the vector potential

If ∇xE = 0 (i.e. B field not changing with time) then this reduces to:

E= -∇V

Which in the one dimensional case of a wire can be treated as:

E= -dV/dx
 
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