# Potential for infinite line charge

1. Jul 8, 2009

### Old Guy

1. The problem statement, all variables and given/known data
Find the potential a distance z from an infinite straight wire with uniform linear charge distribution.

2. Relevant equations

$$\int\textbf{E}\cdot\text{d}\textbf{l}$$

3. The attempt at a solution
The direction of the field is clearly radial from the wire, and dl is in the direction of the wire. Since these are perpendicular, the dot product is zero, which leaves nothing to integrate and implies the potential should be zero, which is clearly wrong. Now, I understand that this is saying that the potential will be same (relative to some reference point) at any point a distance z from the wire, but the math doesn't seem to go there. Can someone please explain? Thanks.

2. Jul 8, 2009

### gabbagabbahey

No, $\text{d}\textbf{l}$ in the above equation does not point in the direction of the wire. In this context, $\text{d}\textbf{l}$ is not an infinitesimal piece of the wire (source), but rather an infinitesimal piece of the path from whichever reference point you choose for your potential (remember, the potential is only defined up to a constant of integration, which is fixed by choosing an appropriate reference point $\mathcal{O}$ ; usually an infinite distance from the source) to your field point $\textbf{r}$ (the point at which you measure the potential relative to your chosen reference point)

Also, the integral you've posted needs to have appropriate limits and a negative sign:

$$\text{V}(\textbf{r})=-\int_{\mathcal{O}}^{\textbf{r}}\textbf{E}\cdot\text{d}\textbf{l}$$

Since the path integral above is actually path independent for electrostatic fields, it doesn't matter which path you choose. For simplicity, I recommend you choose a reference point at $s=a$ (where I'm using $s$ to represent the radial coordinate in cylindrical coords, and $a$ is any non-zero constant) and integrate directly along a radial path to your field point (due to the symmetry of the source, it doesn't matter where you are along the axial direction)

Last edited: Jul 8, 2009
3. Jul 8, 2009

### Old Guy

Thanks for your reply, but I remain a bit confused. You say that dl is in the direction of the path, but if potential is independent of the path, which seems to imply that dl could be in ANY direction (or none?). Also, and equipotential line would be perpendicular to the field, so in that case the dot product concept would work, but only then. What am I missing here?

4. Jul 8, 2009

### gabbagabbahey

No, $\text{d}\textbf{l}$ is just a tiny piece of the path, whichever path you choose, you may have some pieces of the path pointing whichever way you like, but many pieces will have to have at least some component pointing along the line between the two endpoints of the path (otherwise, your path would never get from the reference point to the field point. The integral is path independent, but whichever path you choose has to go from the reference point to your field point; whether it goes in a straight line between the points, or some strange curved shape is what doesn't matter.

5. Jul 16, 2009

### vaibhav1803

find the electric field..by gauss' law..create a gaussian surface..
OR choose "dl" element at "l" from the centre/midpt of the wire..and find net field (dE)due to the element and integrate it with proper bounds/limits
it will have to be normal to the wire as each point on the infinite line is its midpoint...(u can justify this argument by symmetry..)
if you need more help or this is a bit too ambiguous..do tell me..its fun solving probs 4 me..xD

6. Jul 16, 2009

### cipher42

Since you're using the equation
$$\text{V}(\textbf{r})=-\int_{\mathcal{O}}^{\textbf{r}}\textbf{E}\cdot\tex t{d}\textbf{l}$$
to calculate the potential, this implies that you have already computed the electric field, $\textbf{E}$ for this configuration, correct? Now what you need to do is pick a reference point, that is to say a point where you are doing to define the potential to be zero (since only differences in potential have any physical significance, we are free to do this). Now you need to pick a path between the origin, $\mathcal{O}$ an the general point $\mathbf{r}$ where you want to know the potential. Because of the conservative nature of electric force, it doesn't matter which path you pick, you'll get the same answer, [bold]but you still have to pick some specific path to be able to do the calculation[/bold]. Now $d\mathbold{l}$ is an infinitesimal displacement along this path.

Because of the cylindrical symmetry of the electric-field (which we're assuming you have already calculated, if not, you're dong to need to get on that first) a good choice for a path would be one that moves radially away from the wire until it has the same s-coordinate as the point your interested in, and then perpendicular to the wire. We choose this path because on the first part $\mathbf{E}$ and $d\mathbf{l}$ will be parallel so that
$$\mathbf{E}\cdot d\mathbf{l}=E ds$$
and on the second part $\mathbf{E}$ and $d\mathbf{l}$ will be perpendicular so that
$$\mathbf{E}\cdot d\mathbf{l}=0$$

Do you think you can compute the integral from here? (Remember what gabbagabbahey said and let 'a' be the radial component of your reference point)

7. Jul 17, 2009

### vaibhav1803

$$\vec{E}$$=$$\frac{{2k}{\lambda}}{d}$$
$${\lambda}$$=linear charge density