How Is Electric Potential Calculated for a Charged Rod?

[Screwdriver]

Homework Statement

Find an expression for the electric potential a distance z away from the center of a thin uniformly charged rod of length L on a line that bisects the rod.

Homework Equations

$$V=k\frac{Q}{r}$$

The Attempt at a Solution

Determine the electric potential due to one point charge i.

$$V_i=k\frac{\Delta Q}{r}$$

$$r=\sqrt{y^2+z^2}$$

Where y is some distance along the rod to i and z is the distance along the axis to the point away.

$$\Delta Q=\lambda \Delta y$$

Then,

$$V_i=k\lambda \frac{\Delta y}{\sqrt{y^2+z^2}}$$

Then,

$$V_{tot}=k\lambda \sum_{i=1}^{n}\frac{\Delta y}{\sqrt{y^2+z^2}}$$

Now take the limit as n goes to infinity and delta y goes to zero.

$$V_{tot}=k\lambda \int_{\frac{-L}{2}}^{\frac{L}{2}}\frac{d y}{\sqrt{y^2+z^2}}$$

$$V_{tot}=k\lambda \; ln({\frac{2\sqrt{{\frac{L^2}{4}}+z^2}+L}{2\sqrt{{\frac{L^2}{4}}+z^2}-L}})$$

Is this correct? I feel like the integration was too hard for this simple problem (I had to substitute y = ztan(x)).

 

[collinsmark]

It looks correct to me. [strike](Of course you could simplify it slightly by at least getting rid of the '2's)[/strike]

[Edit: Never-mind about the 2's'. They're fine the way they are. Sorry about that.]

 

[Screwdriver]

It looks correct to me.

Really? Well thank you!

Never-mind about the 2's'. They're fine the way they are. Sorry about that.

Okay