Electromagnetism problem using Gauss's Law

[hagisucks]

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

 

[flyingpig]

Well...what is Gauss's Law? Write out. You should have had that in your "relevant eqtn"!

 

[hagisucks]

Electric Flux = qin/Eo

 

[flyingpig]

In integral form...?

 

[hagisucks]

I'm not sure what you mean.

 

[flyingpig]

Have you seen this before?

[tex]\oint \vec{E} \cdot d\vec{A} = \frac{\sum Q_{en}}{\epsilon_0}[/tex]

I pretty much drilled this equation into the back of my head...

 

[hagisucks]

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.

 

[flyingpig]

Well, the cylinder is infinitely long. What is the E-field emitting? what happens if the cylinder is a wire?

 

[hagisucks]

Is it emitting charge? I don't know, what happens?

 

[flyingpig]

You have a long wire (forget being cylinder for now), what is the E-field? What is the direction?

 

[flyingpig]

If it helps, draw a Gaussian Surface over the wire/cylinder.

 

[hagisucks]

I have absolutely no idea.

 

[flyingpig]

Do you know what Gauss's Law means...? What it states? Don't just direct meback to the formula. What is "dA"?

 

[lmcelroy]

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:

[itex]\oint[/itex]E.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)

 

[SteveI46]

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:

[itex]\oint[/itex]E.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?

 

[Saitama]

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?

 

[lmcelroy]

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