## Potential Due to a a Charged Length of Wire

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{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}

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