Can the Electric Field Outside a Conductor Have a Tangential Component?

[Stephen Bulking]

Here I am going to include the proof provided by my book. It is quite a splendid explanation, though there are a few key points I have yet to fully understand. If the electric force by the electric field on the charge at the surface of the conductor is conservative (which it is), then why is there net work in moving it along a closed path, a loop? Hypothetically, if the electric field is not perpendicular immediately outside the conductor, how exactly is the charge moving in a loop?
Please read the following material to understand my question.

 

[etotheipi]

If the electric force by the electric field on the charge at the surface of the conductor is conservative (which it is), then why is there net work in moving it along a closed path, a loop? Hypothetically, if the electric field is not perpendicular immediately outside the conductor, how exactly is the charge moving in a loop?

That's the 'contradiction' in their argument that the field is orthogonal to the equipotential, i.e. they're saying suppose that the electric field outside of the conductor does have a tangential component, then there will be a non-zero line integral of the electric field around a small rectangular closed curve that is partly inside and partly outside the conductor (because $\int_C \vec{E} \cdot d\vec{x} = \int_{\text{outside}} \vec{E} \cdot d\vec{x} + \int_{\text{conductor}} \vec{E} \cdot d\vec{x} = \int_{\text{outside}} \vec{E} \cdot d\vec{x}$, since $\vec{E}$ is zero inside the conductor). That's nonsense in the static case, where the electric field is conservative, so you deduce there can be no tangential component.

N.B. However it's better to say something like, consider an equipotential curve $\vec{x} = \vec{x}(\lambda)$ with a tangent vector to the curve $\vec{T}(\lambda) = \vec{x}'(\lambda)$. The projection of the electric field onto the tangent vector is$$\vec{E}(\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = -\nabla \phi (\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = -\frac{d[\phi(\vec{x}(\lambda))]}{d\lambda} = 0$$because the curve is an equipotential, so $\phi$ is constant with $\lambda$. Hence $\vec{E}(\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = 0$ and the electric field is always orthogonal to the equipotential.

 

[kuruman]

The hand waving answer is this.
1. In a conductor electrons are free to move in response to electric fields.
2. If the entire conductor (not just the surface) were not an equipotential, then there ought to be two points A and B in the conductor with potential difference ΔV_AB between them.
3. If there is a potential difference between A and B there must be an electric field, say from A to B which you can imagine on the surface for a tangential component if you wish, but you don't have to.
4. If there is an electric field, the free electrons inside the conductor will move and keep on moving until the electric field is zero inside the conductor.
5. If the electric field is zero inside the conductor, there is no potential difference between any two points inside the conductor.
6. If there is no potential difference between any two points inside the conductor, the conductor is an equipotential.

Note that "all the charges are at rest" is a bit misleading. What the author means is a static case in which you leave the conductor alone and you don't subject it to, say, a changing magnetic field. In the static case the charge carriers inside a conductor have some thermal energy and bounce around. However within any time interval dt, on average, as many electrons enter some volume element dV as they exit.

 

[rude man]

Also: the word "conductor" is a bit misleading.

In actuality, all material behaves like a "conductor" unless it is purely an insulator (aka "dielectric").

The difference between say copper and wood is one of time, the time it takes for the charges to rearrange (if necessary) until the internal E field is net zero. In the case of copper this is practically instantaneous. In the case of wood it might take as much as hours or longer.

 

[Stephen Bulking]

That's the 'contradiction' in their argument that the field is orthogonal to the equipotential, i.e. they're saying suppose that the electric field outside of the conductor does have a tangential component, then there will be a non-zero line integral of the electric field around a small rectangular closed curve that is partly inside and partly outside the conductor (because $\int_C \vec{E} \cdot d\vec{x} = \int_{\text{outside}} \vec{E} \cdot d\vec{x} + \int_{\text{conductor}} \vec{E} \cdot d\vec{x} = \int_{\text{outside}} \vec{E} \cdot d\vec{x}$, since $\vec{E}$ is zero inside the conductor). That's nonsense in the static case, where the electric field is conservative, so you deduce there can be no tangential component.

N.B. However it's better to say something like, consider an equipotential curve $\vec{x} = \vec{x}(\lambda)$ with a tangent vector to the curve $\vec{T}(\lambda) = \vec{x}'(\lambda)$. The projection of the electric field onto the tangent vector is$$\vec{E}(\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = -\nabla \phi (\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = -\frac{d[\phi(\vec{x}(\lambda))]}{d\lambda} = 0$$because the curve is an equipotential, so $\phi$ is constant with $\lambda$. Hence $\vec{E}(\vec{x}(\lambda)) \cdot \vec{x}'(\lambda) = 0$ and the electric field is always orthogonal to the equipotential.

Thank you so much! I just keep forgetting that electric field inside the conductor is zero. Thanks for working out the math for me, I really appreciate it!