# Gauss's Law: Field of a line charge and cylinder

1. May 29, 2010

### tim37123

1. The problem statement, all variables and given/known data
A long, thin straight wire with linear charge density \lambda runs down the center of a thin, hollow metal cylinder of radius R. The cylinder has a net linear charge density 2(\lambda). Assume \lambda is positive.

Find expressions for the magnitude of the electric field strength outside the cylinder, r > R.

2. Relevant equations
q1=\lambda
q2=2(\lambda)

3. The attempt at a solution
Enet = E_line + E_cylinder

Eline = 1/4(\pi)(\epsilon_0) * ((2(\lambda))/r)
Ecylinder = ?

Having trouble determining the field of the cylinder, and how that effects the overall field based on a gaussian surface. Any tips come to mind?

2. May 29, 2010

### pgardn

All you need is a gaussian cylinder outside of the charged cylinder and wire I would think. There is nice circular symmetry. Unless I am missing something obvious that makes this more complicated. The only problem with any of this is that you dont have Q in so you have to play around with linear charge density and surface density, lamda and sigma.

3. May 29, 2010

### tim37123

would the formula for the cylinder be the same as a line charge since the gaussian surface causes the cylinder to have a similar field?

4. May 30, 2010

### pgardn

Yes, but when you calculate Q inside your gaussian surface, you are going to use different equations for the line and the charged cylinder. The line is just lambda = Q/l = dq/dl where l is the length of the wire and we must assume it is long enough and the gaussian surface is close enough to the wire to where length is not significant. If it were a short wire you would have a more difficult problem and not use Gaussian methods.

The charged cylinger would be sigma = Q/SA of cylinder = Q/ 2*pi*r = dq/dA. So because one is a line of charge and the other is a surface area of charge you will get diff Q inside for each. And it makes sense as the total charge would be more spread out on a surface than on a line.

Oh I just realized something after reading the problem again. They give you linear charge density for the cylinder. So you will have to multiply this by the circumference of the cylinder. Sort of pretending like the cylinder is a whole bunch of long wires all together arranged in a cricle.

Anyhow the bottom line is you have to get all the charge from the line and the cylinder inside your gaussian surface and make sure the field lines are perpendicular to the gaussain surface. So the correct gaussian surface to "capture" all the field lines would be a cylinder outside of both.

I have an answer in terms of lamda, R, r, epsilon and pi. If you can sort of show me what you did with yours I will show you mine.

Last edited: May 30, 2010
5. May 30, 2010

### RoyalCat

You don't even have to go into surface charge density, since the linear charge density of the cylinder is given in the question.

6. May 30, 2010

### pgardn

yes that why I added this in my answer...

Oh I just realized something after reading the problem again. They give you linear charge density for the cylinder. So you will have to multiply this by the circumference of the cylinder. Sort of pretending like the cylinder is a whole bunch of long wires all together arranged in a cricle.

But you would have to multiply by 2pi*R correct, the cylinder is 2pi*R worth of long wires?

7. May 30, 2010

### RoyalCat

No, you've misunderstood. :)
Linear charge density doesn't refer specifically to thin wires. Linear charge density simply means, how much charge there is per unit length of the system.

Whether the other dimensions of the system are negligible, or not!

Let's say you had a big block of wood, with cross-section area $$A$$ and length $$\ell$$, and that you charged it with a uniform volume charge density, $$\rho\equiv \frac{Q}{A\ell}$$

That means that you have $$\rho$$ charge per unit volume of the block of wood.

Now let's define a linear charge density, $$\lambda$$ as the amount of charge per unit length of the system. $$\lambda \equiv \frac{dq}{dx}$$

Let's say then, that we want to look at a part of the block of wood of length $$x$$, the charge on that block of wood would be $$\rho Ax$$
$$\lambda = \frac{Q}{\ell}$$

I hope that's cleared up a bit of the misconception you've been facing.

Linear charge density, just as surface charge density, do not necessarily deal with thin wires or thin sheets, but simply confine our discussion of the system to some of its dimensions.

8. May 30, 2010

### pgardn

Aha... Thanks.
I hope the other poster reads this.
I love this site...

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