Gauss's Law long thin wire question

[TFM]

Homework Statement

A long thin wire carries a charge \lambda per unit length

a) Use Gauss's Law on a cylinder to find magnitude of the E-Field at a distance from the centre of the wire

b) Show that the electric field components (using cartesian coordinates) are of the form:

E_x = \frac{A_0x}{r^2}<br /> <br /> E_y = \frac{A_0y}{r^2}<br /> <br /> A_0 = \frac{\lambda}{2\pi\epsilon_0}<br /> <br />

Homework Equations

Gauss Law:

\int_sE.dA=\frac{q}{\epsilon_0}

The Attempt at a Solution

I have the asnwer for a) to be

E = \frac{q}{\pi r^2 \epsilon_0}

But for B:

E_x = \frac{\lambda x}{2\pi epsilon_0}

But I get:

q = \lambda x

giving:

E = \frac{q<br /> \lambda x}{\pi r^2 \epsilon_0}

a factor of a half out.

Any ideas where I could have gone wrong?

TFM

 

[TFM]

Latex appears to be not working

Homework Statement

A long thin wire carries a charge lambda per unit length


a) Use Gauss's Law on a cylinder to find magnitude of the E-Field at a distance from the centre of the wire

b) Show that the electric field components (using cartesian coordinates) are of the form:

E_x = \frac{A_0x}{r^2}

E_y = \frac{A_0y}{r^2}

A_0 = \frac{\lambda}{2\pi\epsilon_0}

Homework Equations

Gauss Law:

\int_sE.dA=\frac{q}{\epsilon_0}

The Attempt at a Solution

I have the asnwer for a) to be

E = q/(pi r^2 \epsilon_0)

But for B:

E_x = lambda x/2\pi epsilon_0

But I get:

q = \lambda x

giving:

E = \frac{\lambda x/(pi r^2 \epsilon_0)

a factor of a half out.

Any ideas where I could have gone wrong?

TFM

Edit: I don't think the non latex version helps!

 

[Doc Al]

Latex appears to be not working

Yeah, it's being worked on.

I have the asnwer for a) to be

E = q/(pi r^2 \epsilon_0)

Show how you got that.

 

[TFM]

Well Gauss's law:

<br /> \int_sE.dA=\frac{q}{\epsilon_0}<br />

intergral E.dA = q/epsilon-0

the area is a circle, pi r squared, so I get

E*pi r squared = q/epsilon-0

divide by pi r squared gives:

E = q/(epsilon-0 pi r squared)

E = \frac{q}{\epsilon_0 \pi r^2}

TFM

 

[borgwal]

You used the area of a circle, which is incorrect: you need the surface area of a cylinder. That has something to do with a circle indeed, but not with its area! Also, you have to calculate what the total charge inside the cylinder is, you can't just write q=\lambda x (wrong, too!)

 

[TFM]

So far, I now have:

E(2 pi r h + 2 pi r squared) = q/epsilon-0

E (2*\pi*rh + 2 \pi * r^2 = \frac{q}{\epsilon_0})

so E = q/(epsilon-0 [2 pi r h + 2 pi r squared])

How does this look now?

TFM

 

[borgwal]

You're halfway there:

The flux of E goes only through the 'side surface", not the top and bottom surfaces.
That is, the direction of E is perpendicular to the normal vectors pointing out of the top and bottom surfaces.

 

[TFM]

I see, so it will be

E(2 pi r h + 2 pi r squared) = q/epsilon-0

E (2*\pi*rh) = \frac{q}{\epsilon_0})

so E = q/(epsilon-0 2 pi r h)

as the flux isn't coming out the two circles at the top and bottom

TFM[/QUOTE]

 

[borgwal]

The left-hand side of your first equation still contains two contributions, flux through the sides, and flux through the top/bottom. Get rid of the second part.

Your final answer looks much better now, but what is q?

Also, insert unit vectors for the direction of the E-field.

 

[TFM]

I see, so it will be

E(2 pi r h) = q/epsilon-0

E (2*\pi*rh) = \frac{q}{\epsilon_0})

so E = q/(epsilon-0 2 pi r h)

as the flux isn't coming out the two circles at the top and bottom

Is lamda not the charge density? If it is, why isn't it lambda * d <- the length of the wire?

TFM

 

[borgwal]

You have an "h" in your answer: what's that quantity? Is q related to h?

 

[TFM]

h is the length of the wire, and the charge density is lambda - charge ;per unit length, so I'm guessing it is related.

 

[borgwal]

how?

 

[TFM]

I'd say lambda*d = q, for the wire

TFM

 

[Vuldoraq]

Hey TFM,

That answer looks good to me.

 

[TFM]

So

E = (lambda d)/(epsilon-0 2 pi r h)

?

TFM

 

[Vuldoraq]

Almost. If you make your Gaussian surface as long as the wire, what will happen to h?

Edit: Sorry I meant h not d.

 

[TFM]

h will be the same length as the wire, owuld it mnake the charge lambda?

TFM

 

[Vuldoraq]

Lambda would still be the charge per unit length, but h would drop out of the equation, leaving just lambda/(2*pi*r*epsilon0), which is the electric field at a distance r due to an infinitely long wire. In this case, since it is only a long wire, we would have to stipulate r<<h.

edit I meant r<<d

 

[TFM]

So:

E = lambda/(2*pi*r*epsilon-0)

but that doesn't seem to help with the next part, which is showing that the x-component is:

E_x = (lambda x)/2*pi*epsilon-0 r^2

and the y-component

E_y = (lambda y)/2*pi*epsilon-0 r^2

?

Daft-Vader

 

[Vuldoraq]

I'm stumped on that also, it is conceivable that I am mistaken in my solution to part a. All we can do is hope and pray that a nice Physicist will take pity on our mutual plight.

 

[borgwal]

No! You have to check how Gauss law is applied: q is not the total charge, it's the charge inside the volume you choose: you have chosen a cylinder of height h and radius r: what is q?

Vuldoraq also has to check his textbook.

 

[Vuldoraq]

q would be lambda*h? Isn't that right? And the integral becomes trivial because the electric field is always radial in this case. So it's just: magnitude-E*2*pi*r*h=(lambda*h)/epsilon-0, the 2*pi*r*h coming from the surface integral of the cylinder's curved surface?

So the electric field magnitude becomes: E=lambda/(2*pi*r*epsilon-0)? The direction is always orthogonal to the wire, provided r<<d, d being the length of the wire?

Did I go wrong in saying you let h tend to d? I see what you mean, you are only concerned with the segment of wire inside the Gaussian surface, so h and d would be equal.

 

[TFM]

I'm slightly confused now...

is the Electric Field:

E = lambda/(2*pi*r*epsilon-0)?

TFM

 

[Doc Al]

So:

E = lambda/(2*pi*r*epsilon-0)

Correct.

but that doesn't seem to help with the next part, which is showing that the x-component is:

E_x = (lambda x)/2*pi*epsilon-0 r^2

and the y-component

E_y = (lambda y)/2*pi*epsilon-0 r^2

?

I presume the cylinder is oriented along the z-axis. Draw yourself a picture of the electric field in the x-y plane. Hint: Use a little trig to express the x & y components of the field in terms of the x & y coordinates and r.

 

[borgwal]

Yep, correct

 

[TFM]

Surely on the axis themselves, x = r, y = r

TFM

 

[borgwal]

It seems you're having problems with elementary things that have nothing to do with physics. Check your trig, check cylindrical coordinates, check what vectors are and how you decompose them into components.

 

[TFM]

When you break vectors into components, they are usually vector sin or cos theta

I know it is a cylinder, but I thgought we were using cartesian x, y and z coordinates?

 

[Doc Al]

Surely on the axis themselves, x = r, y = r

That's certainly true. What about in general?

When you break vectors into components, they are usually vector sin or cos theta

Exactly!

I know it is a cylinder, but I thgought we were using cartesian x, y and z coordinates?

So express sin and cos in terms of the coordinates and r.