What Is the Potential a Distance z from an Infinite Line Charge?

[Old Guy]

Homework Statement

Find the potential a distance z from an infinite straight wire with uniform linear charge distribution.

Homework Equations

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

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.

 

[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)

 

[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?

 

[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.

 

[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

 

[cipher42]

Since you're using the equation
<br /> \text{V}(\textbf{r})=-\int_{\mathcal{O}}^{\textbf{r}}\textbf{E}\cdot\tex t{d}\textbf{l}<br />
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
<br /> \mathbf{E}\cdot d\mathbf{l}=E ds<br />
and on the second part \mathbf{E} and d\mathbf{l} will be perpendicular so that
<br /> \mathbf{E}\cdot d\mathbf{l}=0<br />

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)

 

[vaibhav1803]

@cipher42..pleasez simplify..the answer is
the bounds sre from -infinity to infinity
and the answer for electric field(gauss's law) is
\vec{E}=\frac{{2k}{\lambda}}{d}
where..
d= distance of point fom centre of the infinite wire
{\lambda}=linear charge density
k= dielectric constant of the medium
as you might have guessed..due to the extreme symmetry of an infinite wire..ie.all points on it are its midpoints...
the electric field is radially outward..having equal magnitude along a cylindrical surfae parallel to the wire