Electrodynamics: electrostatic potential

[Angela G]

Homework Statement

Consider a circular ring with radius R and uniform longitudinal charge density λ as in the figure at the left, below.

Determine the electrostatic potential in the center of the circle.

Relevant Equations

V = - ∫ E · dl
∇·E
F = qE

Hi!

I tried to solve it by using the equation of the electric potential above and as we see it requires the electric field, but the electric field at the center of the ring is zero. Then I tried by using the equation [text] V = \frac{1}{4\pi\epsilon_0r} \int \lamda dl [\text] and I got [text] V = \frac{\lamda}{2\epsilon_0} [\text]. It feels like i did something wrong, Is my solution right? if not where did I wrong?

 

[Delta2]

I think your second approach and result is correct. Why do you feel you did something wrong?

If I tell you to calculate the potential at distance $R$ from the center things get a lot more complex, because the distance of each element $\lambda dl$ of the ring from the point at which we want to calculate the potential, changes as the element changes and the integral cannot be simplified that easily.

BTW if you still can edit your post, put the latex code inside "tex" brackets instead of "text".

 

[PeroK]

I tried to solve it by using the equation of the electric potential above and as we see it requires the electric field, but the electric field at the center of the ring is zero.

You can't calculate the potential from the field at a single point. To calculate the potential at the centre you would first need to calculate the field inside the circle and use $\vec E = -\vec \nabla V$.

In this case, it must be simpler to calculate the potential directly.

 

[Delta2]

In this case, it must be simpler to calculate the potential directly.

Yes I think that's what she did, though she doesn't write full details of the calculation. I think her result $$V(0)=\frac{\lambda}{2\epsilon_0}$$ is correct, what do you think?

 

[PeroK]

Yes I think that's what she did, though she doesn't write full details of the calculation. I think her result $$V(0)=\frac{\lambda}{2\epsilon_0}$$ is correct, what do you think?

Potential is a scalar, and you have a total charge of $2\pi R \lambda$ all a distance $R$ from the centre. So, yes, it's correct.

 

[Angela G]

I think your second approach and result is correct. Why do you feel you did something wrong?

If I tell you to calculate the potential at distance $R$ from the center things get a lot more complex, because the distance of each element $\lambda dl$ of the ring from the point at which we want to calculate the potential, changes as the element changes and the integral cannot be simplified that easily.

BTW if you still can edit your post, put the latex code inside "tex" brackets instead of "text".

Thank you, I'm not sure on my solution, I think I missed something aboout the field.

 

[Delta2]

Thank you, I'm not sure on my solution, I think I missed something aboout the field.

Finding the field first and then finding the potential is the hard way of solving this. There are some threads in this forum that deal with this subject of finding the E-field in the interior of a charged ring, it is not so easy task. And you need to know the E-field at every point, not only at the center, to be able to calculate the potential as the line integral $V=\int \mathbf{E}\cdot d\mathbf{l}$.

 

[PeroK]

In any case, a general technique is to find the potential first and then calculate the field from that. Again, however, you need the potential in a region so that you can calculate its gradient.

 

[Angela G]

ok, so using the equ. that I used was right? I mean I used $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl$ and got $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl = \frac{ \lambda }{4\pi\epsilon_0r} \int dl = \frac{ \lambda }{4\pi\epsilon_0r} 2 \pi r = \frac{\lambda}{2\epsilon_0}$ Is it right?

 

[Delta2]

ok, so using the equ. that I used was right? I mean I used $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl$ and got $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl = \frac{ \lambda }{4\pi\epsilon_0r} \int dl = \frac{ \lambda }{4\pi\epsilon_0r} 2 \pi r = \frac{\lambda}{2\epsilon_0}$ Is it right?

Yes this is correct, just to emphasize one thing, that if we were being asked for the potential at a point other than the center of the ring, then you wouldn't be able to pull the $\frac{1}{r}$ out of the integral, you would have to calculate the integral $$\int\frac{\lambda}{r}dl$$ where $r$ now is the distance of the element $\lambda dl$ from the point at which we want to calculate the potential, and this $r$ varies inside the integral for the different elements $\lambda dl$ around the circumference, so you just can't pull it out of the integral.

 

[PeroK]

ok, so using the equ. that I used was right? I mean I used $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl$ and got $V = \frac{1}{4\pi\epsilon_0r} \int \lambda dl = \frac{ \lambda }{4\pi\epsilon_0r} \int dl = \frac{ \lambda }{4\pi\epsilon_0r} 2 \pi r = \frac{\lambda}{2\epsilon_0}$ Is it right?

If you look again at post #5, you'll see that you didn't need to integrate in this case.

 

[Delta2]

The general formula for the electrostatic potential (with the only boundary condition being that it vanishes at infinity) is $$V(\vec{r})=\iiint \rho(\vec{r'})\frac{1}{|\vec{r}-\vec{r'}|}d^3\vec{r'}$$ so strictly speaking one has always to integrate in order to find the potential. It is just that depending on the charge density $\rho$ and the point $\vec{r}$ in which we wish to evaluate the potential, that integral might simplified a lot, so it would be like doing no integration.

 

[Angela G]

ok, I think I understand thank you both for your help