Electric potential linear charge

[deedsy]

Homework Statement

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

Homework Equations

$V = \frac{\lambda}{4 \pi \epsilon_0 r}$

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)

 

[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.

 

[deedsy]

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.

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?

 

[Make7UpYours]

Yes, that is correct about √r²

 

[HalfLostAndSearching]

- so I'm not sure how to plug in the correct values for those variables.

Dear student,

Thank you for your response. It seems like you have made a good attempt at solving the problem. However, as you have correctly pointed out, your solution is only valid for points on the x-axis, and the problem requires a solution for any point around the line of charge.

To generalize your solution, you can use the distance formula to calculate the distance (r) between any point (x,y,z) and the center of the charge (0,0,0). This will give you a new expression for 1/r in the integral. Additionally, you will need to change the limits of the integral to take into account the fact that the charge is distributed over a length of a, rather than just at the center point.

As for the book's solution, it seems like they have set up their axis in a different way, possibly with the center of the charge at a different point than the origin. The variables r and z likely correspond to the distance from the center of the charge and the distance along the line of charge, respectively. Without a picture or more information, it is difficult to say for sure.

I hope this helps guide you in finding the correct solution. Keep up the good work!