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The potential at a point  
\[(x,y)\]
  due to a charge  
\[dq\]
  at point  
\[(0,l)\]
  in the  
\[xy\]
  plane is  
\[d \phi = \frac{dq}{ \sqrt{x^2 + (y-l)^2 }}\]

Suppose that a piece of wire with charge per unit length  
\[\sigma\]
 is put along the  
\[y\]
  - axis from  
\[(0, - a)\]
  to  
\[(0, a)\]
.
The charge on a small length of wire at  
\[(0,l)\]
   
\[dl\]
  is  
\[\sigma dl\]
  and the potential is  
\[d \phi = \frac{\sigma dl}{ \sqrt{x^2 + (y-l)^2 }}\]

The potential at  
\[(x,y)\]
  due to the wire is
\[ \phi = \int^{a}_{- a} \frac{\sigma dl}{ \sqrt{x^2 + (y-l)^2 }} \]

To evaluate this integral substitute  
\[y-l=x tan \theta \rightarrow - dl =x sec^2 \theta d \theta \]

When  
\[l=a, \:y-a =x tan \theta_2 \]
  and when  
\[l=-a, \:y+a =x tan \theta_1 \]

The integral becomes
\[\begin{equation} \begin{aligned} \phi &= \int^{\theta_2}_{- \theta_1} \frac{- \sigma x sec^2 \theta d \theta }{ \sqrt{x^2 + x^2 tan^2 \theta }} \\ &= \int^{\theta_2}_{- \theta_1} \frac{- \sigma x sec^2 \theta d \theta }{ x sec \theta } \\ &= - \sigma \int^{\theta_2}_{\theta_1} sec \theta d \theta \\ &= - \sigma [ln (sec \theta + tan \theta )]^{\theta_2}_{\theta_1} \\ &= \sigma ln (\frac{sec \theta_1 + tan \theta_1 }{sec \theta_2 + tan \theta_2 }) \\ &= \sigma ln (\frac{sec \theta_1 + tan \theta_1 }{sec \theta_2 + tan \theta_2 }) \\ &= \sigma ln (\frac{\sqrt{1+ tan^2 \theta_1} + tan \theta_1 }{\sqrt{1+ tan^2 \theta_2} + tan \theta_2 }) \\ &= \sigma ln (\frac{\sqrt{1+ ((y+a)/x)^2} + (y+a)/x }{\sqrt{1+ ((y-a)/x)^2} + (y+a)/x}) \\ &= \sigma ln (\frac{\sqrt{x^2+ (y+a)^2} + (y+a) }{\sqrt{x^2+ (y-a)^2} + (y-a)}) \end{aligned} \end{equation}\]