Electric potential at point above a uniformly charged finite rod

[sb_4000]

Hi,
Im having a little problem doing a question in mastering physics, I was wondering if anyone can help.

here is the problem:
A finite rod of length L has total charge q, distributed uniformly along its length. The rod lies on the x -axis and is centered at the origin. Thus one endpoint is located at (-L/2,0), and the other is located at (L/2,0). Define the electric potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression 1/4*pi*epsilon_0.

What is Va, the electric potential at point A (see the figure), located a distance d above the midpoint of the rod on the y axis?
Express your answer in terms of k, q, d, and L.

Here is what I have so far:

r =sqrt(d^2+x^2)
dV= k*(q/(2*(L/2)))*(dx/(sqrt(d^2+x^2)))
V=K*(q/2*(L/2)))*(ln*((sqrt(d^2+x^2)+a)/(sqrt(d^2+x^2)-a)) *I got this from book*

my problem is putting it all together, using k,q , d and L. especially the (ln*((sqrt(d^2+x^2)+a)/(sqrt(d^2+x^2)-a)) part, which does not want x in the answer. I am really lost.

Thanks.

 

[StatusX]

Are you sure you copied that formula for V correctly? It's not V, it's the antiderivative of dV/dx, and to get V you must evaluate it at the endpoints. In other words, after doing the indefinite integral of dV/dx with respect to x, you get what the book gives as V. But V is actually the definite integral over the portion of x where there is charge (-L/2, L/2). Also, where did 'a' come from?

Edit: actually, that isn't the antiderivative of dv/dx. I don't know what that is.

 

[Galileo]

The expression for dV looks correct.

$$V=\int_{-L/2}^{L/2}dV=\frac{kq}{L}\int_{-L/2}^{L/2}\frac{dx}{\sqrt{x^2+d^2}}$$

Just evaluate it and put in the right limits (not 'a' or 'x').