What's wrong? Electric potential of a point on a ring

• yucheng
In summary, the problem with the electric potential of a point on a ring lies in the fact that it is not a constant value. This is due to the varying distance between the point and different points on the ring, resulting in different electric potential values. Additionally, the potential can be affected by the presence of other charges or electric fields in the surrounding area. This makes it more challenging to calculate and predict the electric potential at a specific point on the ring accurately.
yucheng
Homework Statement
N/A
Relevant Equations
N/A

So I have a ring(red) of uniform charge ##\lambda## per unit length, and I want to calculate the electric potential at the origin (actually on any point of the ring). It is clear that the ring is given by the equation $$r=2 R \sin \theta$$, in polar coordinates, where R is the radius of the ring. Since potential is given by $$V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}$$, and $$dq = d\ell \lambda$$, but $$d\ell = \sqrt{(\frac{dr}{d\theta})^2 + r^2} d\theta = 2R d\theta$$. Therefore, $$V = \frac{1}{4 \pi \epsilon_0} \int \frac{d\ell \lambda}{r} = \frac{1}{4 \pi \epsilon_0} \int^{\pi}_{0} \frac{2R\lambda}{2R \sin \theta} d\theta = \frac{1}{4 \pi \epsilon_0} \lambda \int^{\pi}_{0} \frac{1}{\sin \theta} d\theta$$, which does not converge... what's wrong?

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Delta2
Yes because the ring is infinitesimally thin and has finite charge density ##\lambda## the electric field (and potential) is infinite for points on the ring circumference. Electric field and Potential are well defined for points on the interior or the exterior of the ring though.
Hold on for a minute while I locate a previous thread where this was discussed in more detail.

Here it is

My explanation is that we know from Gauss's law in differential form that $$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$ and $$\nabla^2 V=-\frac{\rho}{\epsilon_0}$$.

Where ##\rho## is the volume (not the linear) charge density. Because the ring is infinitesimally thin the ##\rho## of the ring become infinite for points on the ring, hence from the above two equations, electric field and potential become infinite too.

yucheng
Delta2 said:
My explanation is that we know from Gauss's law in differential form that $$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$ and $$\nabla^2 V=-\frac{\rho}{\epsilon_0}$$.

Where ##\rho## is the volume (not the linear) charge density. Because the ring is infinitesimally thin the ##\rho## of the ring become infinite for points on the ring, hence from the above two equations, electric field and potential become infinite too.
Uh oh! I spent quite some time wondering where did I get it wrong... Luckily I asked. thanks!

Delta2
Delta2 said:
Because the ring is infinitesimally thin the ρ of the ring become infinite for points on the ring
Oh yes, just curious, for surface charge density, like a thin sheet of charge, we also assume that it's infinitesimally thin. Does this also mean that ##\rho## is infinite?

Delta2
Delta2 said:
My explanation is that we know from Gauss's law in differential form that $$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$ and $$\nabla^2 V=-\frac{\rho}{\epsilon_0}$$.

Where ##\rho## is the volume (not the linear) charge density. Because the ring is infinitesimally thin the ##\rho## of the ring become infinite for points on the ring, hence from the above two equations, electric field and potential become infinite too.
This argument is unfortunately invalid as it does not work for a surface charge where the volume charge density becomes formally infinite at the surface, but the potential is well defined.

Instead, we need to argue how the potential behaves as we get closer to the wire. As we approach a line charge and the distance ##\ell## to the charge is much smaller than the curvature radius of the line, the porential behaves as proportional to ##\ln(\ell)## (it must because of the divergence theorem). Therefore, the potential diverges and the field diverges as ##1/\ell##.

Delta2
Yes ok I see now, infinity of the divergence of a field at a point doesn't necessarily imply infinity of the field at that point.

Delta2 said:
Yes ok I see now, infinity of the divergence of a field at a point doesn't necessarily imply infinity of the field at that point.
Indeed, but almost always. The exception is when the source is spread over a hypersurface of one dimension lower than the space itself. The reason is that the one-dimensional Green function of the Laplace operator does not diverge, but for any higher dimension it does

Delta2
Orodruin said:
Indeed, but almost always
Thanks good to know that .

Orodruin said:
This argument is unfortunately invalid as it does not work for a surface charge where the volume charge density becomes formally infinite at the surface, but the potential is well defined.

Instead, we need to argue how the potential behaves as we get closer to the wire. As we approach a line charge and the distance ##\ell## to the charge is much smaller than the curvature radius of the line, the porential behaves as proportional to ##\ln(\ell)## (it must because of the divergence theorem). Therefore, the potential diverges and the field diverges as ##1/\ell##.

Orodruin said:
Indeed, but almost always. The exception is when the source is spread over a hypersurface of one dimension lower than the space itself. The reason is that the one-dimensional Green function of the Laplace operator does not diverge, but for any higher dimension it does

Are there any discussions (in books, articles) about these? The closest I found is Jackson pg. 32 the paragraph below equation (1.23) "For colume or surface distributions of charge, the potential is everywhere continuous... from the fact that E is bounded". Maybe one can carry out the limiting procedures themselves to verify, as suggested by Jackson...

@Orodruin? Do you know where to read up on the green's function thing?

1. What is electric potential?

Electric potential is a measure of the electrical potential energy per unit charge at a specific point in space. It is a scalar quantity and is measured in volts (V).

2. What is a point on a ring?

A point on a ring refers to a specific location on a circular loop or ring. In the context of electric potential, it is a point at a certain distance from the center of the ring.

3. How is electric potential calculated for a point on a ring?

The electric potential at a point on a ring can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the ring, and r is the distance from the center of the ring to the point.

4. What factors affect the electric potential of a point on a ring?

The electric potential of a point on a ring is affected by the charge of the ring, the distance from the center of the ring, and the surrounding electric field. It is also affected by the presence of other charges or conductors nearby.

5. How is electric potential different from electric field?

Electric potential is a measure of the potential energy per unit charge at a point, while electric field is a measure of the force per unit charge at a point. Electric potential is a scalar quantity, while electric field is a vector quantity. Additionally, electric potential is dependent on the location, while electric field is dependent on the charge distribution in the surrounding space.

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