Find the potential a distance `z` above the center of a uniform line of charge with charge density `lambda ` and length `2L` .

For a continuous potential to equation is

`V=int (k*dq)/r`

Where `k=1/(4pi epsilon_0)` , `dq` is an infinitesimal piece of charge, and `r` is the distance away `dq` is from the point we are measuring.

We know the charge per unit length` lambda=(dq)/(dl)` where `dl` is an infinitesimal chuck of the wire. If we let the wire lay along the x-axis then `dl=dx` . Therefore,

`dq=lambda dx`

Now, looking at the picture we have a...

For a continuous potential to equation is

`V=int (k*dq)/r`

Where `k=1/(4pi epsilon_0)` , `dq` is an infinitesimal piece of charge, and `r` is the distance away `dq` is from the point we are measuring.

We know the charge per unit length` lambda=(dq)/(dl)` where `dl` is an infinitesimal chuck of the wire. If we let the wire lay along the x-axis then `dl=dx` . Therefore,

`dq=lambda dx`

Now, looking at the picture we have a right triangle. So we know ` r=sqrt(x^2+z^2)` . Then the integral becomes,

`V=int_(-L)^L (k*lambda dx)/sqrt(x^2+z^2)=k*lambda int_(-L)^L dx/sqrt(x^2+z^2)`

From a table of integrals we can find that

`V=k lambda ln(x+sqrt(x^2+z^2))|_(-L)^L`

`V=lambda/(4pi epsilon_0) ln((L+sqrt(L^2+z^2))/(-L+sqrt(L^2+z^2)))`

To simplify multiply the numerator and denominator by `L+sqrt(L^2+z^2)`

`V=lambda/(4pi epsilon_0)ln[(L+sqrt(L^2+z^2))^2/(-L^2+L^2+z^2)]`

`V=2*lambda/(4pi epsilon_0)ln[(L+sqrt(L^2+z^2))/z]`

`V=lambda/(2pi epsilon_0)ln[(L+sqrt(L^2+z^2))/z]`

This is the answer.