# Gauss's Law question

1. Oct 19, 2008

### TFM

1. The problem statement, all variables and given/known data

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:

$$E_x = \frac{A_0x}{r^2} E_y = \frac{A_0y}{r^2} A_0 = \frac{\lambda}{2\pi\epsilon_0}$$

2. Relevant equations

Gauss Law:

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

3. 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:

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

a factor of a half out.

Any ideas where I could have gone wrong?

TFM

2. Oct 19, 2008

### TFM

Latex appears to be not working

1. The problem statement, all variables and given/known data

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}

2. Relevant equations

Gauss Law:

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

3. 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!!!

3. Oct 19, 2008

### Staff: Mentor

Yeah, it's being worked on.

Show how you got that.

4. Oct 20, 2008

### TFM

Well Gauss's law:

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

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

5. Oct 20, 2008

### 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!)

6. Oct 20, 2008

### 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

7. Oct 20, 2008

### 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.

8. Oct 20, 2008

### 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]

9. Oct 20, 2008

### 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.

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

10. Oct 20, 2008

### 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

11. Oct 20, 2008

### borgwal

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

12. Oct 20, 2008

### 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.

13. Oct 20, 2008

### borgwal

how?

14. Oct 20, 2008

### TFM

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

TFM

15. Oct 20, 2008

### Vuldoraq

Hey TFM,

That answer looks good to me.

16. Oct 20, 2008

### TFM

So

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

?

TFM

17. Oct 20, 2008

### 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.

Last edited: Oct 20, 2008
18. Oct 20, 2008

### TFM

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

TFM

19. Oct 20, 2008

### 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

Last edited: Oct 21, 2008
20. Oct 20, 2008

### 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

???