Potential Due to a a Charged Length of Wire
\[(x,y)\]
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
\[\begin{equation} \begin{aligned} \phi &= \int^{\theta_2}_{- \theta_1} \frac{- \sigma x sec^2 \theta d \theta }{ \sqrt{x^2 + x^2 tan^2 \theta }} \\ &= \int^{\theta_2}_{- \theta_1} \frac{- \sigma x sec^2 \theta d \theta }{ x sec \theta } \\ &= - \sigma \int^{\theta_2}_{\theta_1} sec \theta d \theta \\ &= - \sigma [ln (sec \theta + tan \theta )]^{\theta_2}_{\theta_1} \\ &= \sigma ln (\frac{sec \theta_1 + tan \theta_1 }{sec \theta_2 + tan \theta_2 }) \\ &= \sigma ln (\frac{sec \theta_1 + tan \theta_1 }{sec \theta_2 + tan \theta_2 }) \\ &= \sigma ln (\frac{\sqrt{1+ tan^2 \theta_1} + tan \theta_1 }{\sqrt{1+ tan^2 \theta_2} + tan \theta_2 }) \\ &= \sigma ln (\frac{\sqrt{1+ ((y+a)/x)^2} + (y+a)/x }{\sqrt{1+ ((y-a)/x)^2} + (y+a)/x}) \\ &= \sigma ln (\frac{\sqrt{x^2+ (y+a)^2} + (y+a) }{\sqrt{x^2+ (y-a)^2} + (y-a)}) \end{aligned} \end{equation}\]
