# Homework Help: Electric potential calculation

1. Sep 25, 2007

### indigojoker

A thin wire has charge +Q uniformly distrubuted along it's length L. It's lying on the x-axis with its midpoint at the origin. Calculate the potential at a point P on the x axis, where P>L/2.

$$dV= \frac{k dq}{r}$$
$$V = k \lambda \int \frac{1}{r}dr$$
$$V = k \lambda \int^{P+L/2} _{P-L/2} \frac{1}{r}dr$$

So when i use that integral, i get:

$$V = k \lambda \ln {\left( \frac{P-\frac{L}{2}}{P+\frac{L}{2}} \right)}$$

If i wanted to find the E field, I know that $$E = - \nabla V$$

but do I take the derivative w.r.t. P?

2. Sep 26, 2007

### Proggle

Try changing the limits of your integral so that it works for any x before making x=P.

3. Sep 26, 2007

### ehild

You made a small mistake, replace P+L/2 first, so the potential becomes

$$V = k \lambda \ln {\left( \frac{P+\frac{L}{2}}{P-\frac{L}{2}} \right)}$$

And yes, you take the derivative with respect to P (the coordinate of point P).

ehild

4. Sep 26, 2007

### Mindscrape

There are two ways to go about this problem. You chose the best way, the way that assumes a zero potential infinitely far away, but I'll show you the second way too because hopefully it will be instructive. The first way is to use

$$V = k \int \frac{\lambda}{\pi} dl$$

But be careful because pi (I just renamed r to be something that resembled script r, which pi sort of does, since script r is used everywhere), is supposed to be the magnitude of your vector. You did not account for this in your equation, maybe you thought that r was just some vector. I think if I explain the second way that you might get an idea, without me actually telling you the answer.

So, the second way to do it is find the potential given the electric field. I present the electric field without derivation, you should do it, as

$$E = k \frac{2 \lambda L}{x \sqrt{x^2 + L^2}} \hat{x}$$

where s is the distance away from the line of charge.

The potential is then defined as
$$V = - \int_{special}^r \mathbf{E} \cdot dl$$

Where the special part is the reference point where you define the potential to be zero, typically defined at infinity, but there are cases where it doesn't work (like if your line charge were infinite). Since the field is only in the x direction the potential would become

$$V = -\int k \frac{2 \lambda L}{x \sqrt{x^2 + L^2}} dx$$

Now you chose your bounds strangely for your first go at it. Remember that you don't want V for a specific point P, you want it for all points x past L/2. Also, the first way will not produce the exact same integral (though they will be similar) because it has already assumed the reference point to be infinity, so you can try out this integral, or redo the first one, or do both and see if they match up. Even though they are different, they have the same answer.

Both ways have their advantages and disadvantages. Sometimes E is easy to find, and you can use the line integral to find V, and sometimes E is really hard to find so you have to use the charge density integral to find V.

Hah, probably more information than you wanted, but it's a learning experience and I'm not just going to correct your mistakes.

Last edited: Sep 26, 2007
5. Sep 26, 2007

### indigojoker

I'm not sure why
$$V = k \lambda \ln {\left( \frac{P+\frac{L}{2}}{P-\frac{L}{2}} \right)}$$

I dont see why the signs changed

6. Sep 26, 2007

### Mindscrape

That's still not right. You made the mistake of not making r the magnitude of r.

7. Sep 26, 2007

### indigojoker

well, that's why ehild did and I was wondering why the signed changed and why ehild suggested that i should "replace P+L/2 first"

8. Sep 26, 2007

### Mindscrape

Nevermind about my magnitude stuff actually, I thought you were finding it above the line, not along its axis. Your bounds are still off though, you want to be integrating over the line of charge to find the function of x.

Last edited: Sep 26, 2007
9. Sep 26, 2007

### indigojoker

My limits are:

$$V = k \lambda \int^{P+L/2} _{P-L/2} \frac{1}{r}dr$$

From the start of the line on the far left (-L/2), to the end of the line (L/2)

I just took into account that it's w.r.t. P so i have from P+L/2 to get the start of the line, to P-L/2 to get the end of the line.

10. Sep 26, 2007

### Proggle

Try x instead of P :P

11. Sep 26, 2007

### indigojoker

x=P-L/2

12. Oct 1, 2007

### Mindscrape

In case you were wondering how the solution went. You want to use the eqn

$$V = \frac{1}{4 \pi \epsilon_0} \int \frac{\lambda(\mathbf{r'})}{\mathbb{R}} dx'$$

So if your origin is at the center of the rod, you want to integrate over the whole rod, which is from -L/2 to L/2.

$$V_x = \frac{\lambda}{4 \pi \epsilon_0} \int_{-L/2}^{L/2} \frac{1}{((x-x')^2)^{(1/2)}}dx' \hat{x}$$

The distance away is simply x-x', as you could have easily jumped straight to, so do the integration

$$V_x = \frac{\lambda}{4 \pi \epsilon_0} \int_{-L/2}^{L/2} \frac{1}{(x-x')}dx' \hat{x}$$

$$V_x = - \frac{\lambda}{4 \pi \epsilon_0} [ln(x+L/2) - ln(x-L/2)] \hat{x}$$

Let's check to see if this makes sense, so look at x>>L, and we should see the electric field, which is the gradient of V, go as 1/x^2

$$E = -\nabla V = \nabla \frac{\lambda}{4 \pi \epsilon_0} ln(x^2) \hat{x}$$

Do the gradient, which in this case is simply d/dx

$$E_x = \frac{\lambda}{4 \pi \epsilon_0} \frac{1}{x^2}$$

Which is what we expect if we go really far away, and the line looks more like a point charge.

Last edited: Oct 2, 2007
13. Oct 2, 2007

### ehild

There are two variables here, one the position of the charge element along the wire, let's denote it by l, and the other is the coordinate of point P, denoted by x. -L/2 <= l <= L/2 and x>L/2, so x-l cannot be negative.

The charge element at l contributes to the potential at P by

$$dV = k \lambda \frac{1}{|x-l|}dl$$

You have to integrate this contibution with respect to l along the wire. You denoted x-l by r, the distance of point P from the charge element, this is all right, and changed the integration variable to r. The distance r changes from x-L/2 at the right end of the wire to x+L/2 at the left end. You are wrong in what you said but your integral is correct .

(To be more rigorous, when you switch to r=x-l, you have to multiply the integrand with dr/dl=-1, and replace the upper limit to x-L/2 and the lower limit to x+L/2. If you integrate in the other way, from r=x-L/2 to x+L/2, the integral changes sign and we arrived at your formula.)

Remember how to get a definite integral: you replace the upper limit first and then subtract the value of the integral at the lower limit. It results in
ln(x+L/2)-ln(x-L/2) = ln ((x+L/2)/(x-L/2)).
This expression is easy to differentiate with respect to x to get the electric field intensity.