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
![E=(\lambda)/(2\pi\epsilon_(0)y)\hat{y}](https://img.qammunity.org/2021/formulas/physics/college/e67g6wdcch43yj6jwjqufhy1jbe3ljar7s.png)
The electric potential is
....(I)
Here,
![r=√(x^2+y^2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/dw5cb02frlnrmlqwuxltzq4wgaxsg7p5yu.png)
We need to calculate the potential due to this line charge
Using equation (I)
![V=-\int_(d)^(r){(\lambda)/(2\pi\epsilon_(0)y)dy}](https://img.qammunity.org/2021/formulas/physics/college/9etzri8ldxv0wgucpebmxhcy0ynjqtekbw.png)
On integratinting
![V=-(\lambda)/(2\pi\epsilon_(0))ln((r)/(d))](https://img.qammunity.org/2021/formulas/physics/college/lbfw03qp58l3n6y51pd57emgq24h8qmo06.png)
![V=(\lambda)/(2\pi\epsilon_(0))ln((d)/(r))](https://img.qammunity.org/2021/formulas/physics/college/ge4wq7s5h3sybxkasfvsixnhn178w63ob6.png)
Put the value of r
![V=(\lambda)/(2\pi\epsilon_(0))ln((d)/(√(x^2+y^2)))](https://img.qammunity.org/2021/formulas/physics/college/4zplgs0k7nuvf3p0l74sfesmwc0m5idzo3.png)
![V=(\lambda)/(4\pi\epsilon_(0))ln((d^2)/(x^2+y^2))](https://img.qammunity.org/2021/formulas/physics/college/jdenbfbftlylcssl10ayk0nel0hn01t29z.png)
Hence, The potential due to this line charge is
![(\lambda)/(4\pi\epsilon_(0))ln((d^2)/(x^2+y^2))](https://img.qammunity.org/2021/formulas/physics/college/d2s9t869ew0kkaxjv6qrufar2myo71nxe0.png)