Given that,
Charge per unit length = λ
Point (x, y)=(0. d) parallel to the z axis
We know that,
The electric field due to the infinitely long wire is
[tex]E=\dfrac{\lambda}{2\pi\epsilon_{0}y}\hat{y}[/tex]
The electric potential is
[tex]V=-\int_{d}^{r}{\dfrac{\lambda}{2\pi\epsilon_{0}y}dy}[/tex]....(I)
Here, [tex]r=\sqrt{x^2+y^2}[/tex]
We need to calculate the potential due to this line charge
Using equation (I)
[tex]V=-\int_{d}^{r}{\dfrac{\lambda}{2\pi\epsilon_{0}y}dy}[/tex]
On integratinting
[tex]V=-\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{r}{d})[/tex]
[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{d}{r})[/tex]
Put the value of r
[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{d}{\sqrt{x^2+y^2}})[/tex]
[tex]V=\dfrac{\lambda}{4\pi\epsilon_{0}}ln(\dfrac{d^2}{x^2+y^2})[/tex]
Hence, The potential due to this line charge is [tex]\dfrac{\lambda}{4\pi\epsilon_{0}}ln(\dfrac{d^2}{x^2+y^2})[/tex]