Consider an infinitely long wire with charge per unit length ????centered at (x, y) = (0, d) parallel to the z-axis.A) Find the potential due to this line charge referenced to the origin so that ϕ=0.

Respuesta :

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]

Ver imagen CarliReifsteck