Potential of an infinite line of charge

[th5418]

Homework Statement

Find a potential a distance r from an infinitely long straight wire that carries a uniform line charge $$\lambda$$. Compute the gradient of your potential and check that it yields the correct field.

Homework Equations

$$V=$$$$\frac{KQ}{R}$$
$$\oint E \bullet dS=\frac{Q}{episolon}$$

The Attempt at a Solution

I tried doing it the V=kq/r way, but then I realized it doesn't work, since for that it assumes that potential is zero infinitely far away. But it isn't. The problem doesn't want me to find the e-field first, which would be a lot easier, any tips?

Do i need to go to the most basic definition of potential?

 

[gabbagabbahey]

V=kq/r gives the potential of a point charge.

For a continuous linear charge distribution,

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int \frac{\lambda(\textbf{r}')dl'}{|\textbf{r}-\textbf{r}'|}$$

 

[th5418]

Hm, yeah I got that.
This is what I have right now.

$$V=\frac{1}{4\pi\epsilon_o}\int\frac{\lambda dx}{\sqrt{r^{2}+x^{2}}}$$

But the integral blows up when I integrate from negative to positive infinity.

 

[gabbagabbahey]

But the integral blows up when I integrate from negative to positive infinity.

Yes, I suppose it does.

So, you can either compute the field first using Gauss' law and then find the potential using

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

Or, if you aren't allowed to do that, use separation of variables to solve Laplace's equation in cylindrical coordinates.

 

[th5418]

I can do that? Please advise on that second part.

This is using $$\nabla^{2} V = \frac{\rho}{\epsilon_o}$$??

 

[gabbagabbahey]

I can do that? Please advise on that second part.

This is using $$\nabla^{2} V = \frac{\rho}{\epsilon_o}$$??

Yes (sorry , I meant Poisson's equation, not Laplace's), based on symmetry, what variables would you expect $V$ to depend on? The distance from the axis? The axial coordinate? The azimuthal coordinate? You can use that to express $\nabla^2V$ in terms of ordinary derivatives, giving you a second order ODE to solve.

 

[th5418]

That's out of the league of the class. I could do it though...

I know the potential field is suppose to be...

$$V = -\frac{\sigma}{2\pi \epsilon_o} ln(r)$$

Is there any other way to solve this? I've exhausted all possibilities.

QUESTION:
For the
$$V=\frac{kq}{r}$$
does that assume that potential is 0 $$\infty$$ far away?

 

[gabbagabbahey]

That's out of the league of the class. I could do it though...

I know the potential field is suppose to be...

$$V = \frac{\sigma}{2\pi \epsilon_o} ln(r)$$

Is there any other way to solve this? I've exhausted all possibilities.

If you haven't been taught this method yet, then I would guess you are expected to first calculate the electric field and then integrate...the question doesn't explicitly tell you not to do that does it?

QUESTION:
For the
$$V=\frac{kq}{r}$$
does that assume that potential is 0 $$\infty$$ far away?

Yes,

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int \frac{\lambda(\textbf{r}')dl'}{|\textbf{r}-\textbf{r}'|}$$

is a solution (in integral form) of Poisson's equation, subject to the boundary condition $V\to 0$ at $|\textbf{r}|\to \infty$.

 

[th5418]

Well, I'm just going to go to the professor's office hour, thanks though!