# Electric potential linear charge

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1. Oct 21, 2014

### deedsy

1. The problem statement, all variables and given/known data
A charge Q is uniformly distributed along a straight line of length a. Calculate V by integration, choosing the origin of the coordinates as the center of the charge.
Next, expand this value of V up to terms in 1/r^2
2. Relevant equations
$V = \frac{\lambda}{4 \pi \epsilon_0 r}$

3. The attempt at a solution
I let the line of charge sit along the z-axis, with its center at z=0, in the z-x plane
I know the charge density $\lambda=\frac{Q}{a}$

$V = \frac{Q}{4\pi a \epsilon_0} \int_{-a/2}^{a/2} \frac{1}{\sqrt{z^2 + x^2}} \, dz$

Solving the integral, this gives me the solution $V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{x^2 + \frac{a^2}{4}}+ \frac{a}{2}}{\sqrt{x^2 + \frac{a^2}{4}}- \frac{a}{2}}$

However, this is not the right answer. My book says the answer is $V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{r^2 - az + \frac{a^2}{4}}+ \frac{a}{2} -z}{\sqrt{r^2 + az + \frac{a^2}{4}}- \frac{a}{2} -z}$

So, now I realize I solved for the potential only straight above the center point of the line, on the x-axis... So, I'm guessing my book wants a solution for the potential anywhere, and that is why their answer looks more complicated, with more variables...

So, I'm having trouble generalizing my V formula to work for any point around the line of charge. I know I need to change $\frac{1}{\sqrt{z^2 + x^2}}$, but I'm not sure how exactly to change it..

Also, I'm not sure what r and z correspond to in the book's solution - I'm not even sure how they set up their axis (no picture)

2. Oct 22, 2014

### Make7UpYours

You are correct in your assumptions, but you do need to find V for every point on the 3D plane.

Try using the generic form for electric potential where you would replace 1/√z2 + x2
with 1/√(x - x')2 + (y - y')2 + (z - z')2

and think about which values would be zeroed due to the location of the rod.

3. Oct 22, 2014

### deedsy

Yes I'm glad to see that suggestion, as it was something i was considering this morning. Because the line of charge is on the z-axis, x'=y'=0. So I just need to replace $\frac{1}{\sqrt{x^2 + z^2}}$ with $\frac{1}{\sqrt{x^2 +y^2 + (z-z')^2}}$ and solve this integral with dz'

Then I suppose the book just replaced $\sqrt{x^2 + y^2 + z^2}$ with $\sqrt{r^2}$ in the answer?

4. Oct 22, 2014

### Make7UpYours

Yes, that is correct about √r2