Potential on the axis of a uniformly charged ring

[Guillem_dlc]

Homework Statement

We have a uniformly charged ring of radius $R$ with a linear charge density $\lambda$. Determine the potential at a point on its axis at a distance $z$ from the plane of the ring.

Answer: $V=\dfrac{\lambda}{2\varepsilon_0}\dfrac{R}{\sqrt{z^2+R^2}}$

Relevant Equations

$V_z=\int E\cdot dl$

We know that
$$V_Z=\int_{\textrm{ring}} E\cdot dl$$
We therefore consider $E=\dfrac{\lambda}{2\pi \varepsilon_0}\cdot \dfrac1r$. Then,
$$V_Z=\int_{\textrm{ring}} \dfrac{\lambda}{2\pi \varepsilon_0}\cdot \dfrac1r\, dl = \dfrac{\lambda}{2\pi \varepsilon_0}\dfrac1r \int_{\textrm{ring}}dl=$$
$$=\dfrac{\lambda}{\cancel{2\pi} \varepsilon_0}\dfrac1r \cancel{2\pi} R=\dfrac{\lambda}{\varepsilon_0}\cdot \dfrac{R}{r}=\dfrac{\lambda}{\varepsilon_0}\dfrac{R}{\sqrt{z^2+R^2}},$$
where $r=\sqrt{z^2+R^2}$.
I have done this.

 

[nasu]

It's just an accident that your answer is close to the right one. To calculate the circulation of the electric field around the ring you will need the field at points on the ring not somewhere on the axis. This field is not directed tangential to the path, And anyway, this integral around the ring (closed path) should be zero. so it will be useless.
You need to calculate the potential by integrating the contributions of each element of the ring to the potential to the given point on the axis. Similar to what you did for the electric field of the ring.
Or, you could use the definitionn of potential based on the integral of the electric field but then you need to integrate from the given pont to infinity and use the right path (and right formula for the field).

 

[Guillem_dlc]

It's just an accident that your answer is close to the right one. To calculate the circulation of the electric field around the ring you will need the field at points on the ring not somewhere on the axis. This field is not directed tangential to the path, And anyway, this integral around the ring (closed path) should be zero. so it will be useless.
You need to calculate the potential by integrating the contributions of each element of the ring to the potential to the given point on the axis. Similar to what you did for the electric field of the ring.
Or, you could use the definitionn of potential based on the integral of the electric field but then you need to integrate from the given pont to infinity and use the right path (and right formula for the field).

Now I had done this and I would say it's fine. We have a charged ring:

$$V=\int_{\textrm{ring}} dV=\int_{\textrm{ring}} k\dfrac{dq}{r}=\int_{\textrm{ring}} k\dfrac{\lambda dl}{r}=k\dfrac{\lambda}{r}\int_{2\pi R}dl=$$
$$=k\dfrac{\lambda}{r}2\pi R=\dfrac{1}{4\pi \varepsilon_0}\dfrac{\lambda}{r}2\pi R=\dfrac{\lambda}{2\varepsilon_0}\dfrac{R}{\sqrt{z^2+R^2}}$$

 

[nasu]

It looks like you got the right answer. Doesn't it?