Physical interpretation of the statement $\oint E\cdot dl=0$

Question

Can anyone provide me with a physical interpretation of $\oint E\cdot d\ell=0$ in electrostatics?

Answer 1

The vanishing of closed line integrals means that the field is conservative.

Since $\oint \vec E \cdot \mathrm{d}\vec l$ is equivalent to $\vec \nabla \times \vec E = 0$, the "physical interpretation" is the the electric field is irrotational, i.e. it has no "vortices". The, more valuable, mathematical implication is that there is a scalar potential whose gradient is the electric field.

Answer 2

Can anyone provide me with a physical interpretation

For an electric field satisfying the equation, the work associated with (slowly) moving a test charge around a closed path is zero.

To see this, recall that the electric force on a charge is

$$\vec F = q\vec E $$

The work associated with moving a particle along a closed path is

$$W = \oint \vec F \cdot d\vec l$$

Thus, the work associated with moving a test charge with charge $q$ along a closed path is

$$W = \oint q\vec E \cdot d\vec l = q \oint \vec E \cdot d\vec l = 0$$

Since this is the case, we call such a field conservative and it follows that the work done in moving a test charge between any two distinct points in the electric field is path independent; the work is the same regardless of the path taken between the points.

For this reason, we can define a potential difference between any two points which is just the work per unit charge associated with the two points.

Answer 3

Given a vector field $\mathbf F$ in $\mathbb R^3$ and a closed path $\gamma:\mathbb R\to\mathbb R^3$, the integral $$\int_\gamma \mathbf F\cdot\text d\mathbf r$$ represents the work done by $\mathbf F$ along $\gamma$ (this is true more generally, when $\gamma$ is not necessarily a loop). If this quantity is zero, then it can be proven that the integral $$\int_a^b\mathbf F\cdot\text d\mathbf r$$ depends only on the end points, but not on the path that connects them. Hence one can introduce a potential, that is a scalar field $U:\mathbb R^3\to\mathbb R$ of class $C^2$ that is such that $$\int_a^b\mathbf F\cdot\text d\mathbf r = U(b)-U(a).$$ The choice of $U$ is not unique, but any other choice $U'$ is related to $U$ by a mere additive constant, that is $U = U' + C$. Vector fields with this property are called conservative, since the work done by such fields along a certain path depends on the variation of the potential function, viz. $$W = \Delta U.$$ The relation between $\mathbf F$ and $U$ is simply $\mathbf F = \nabla U$, i.e. the vector field is the gradient of the scalar field.

In physics it makes more sense to define the potential $U$ in such a way that the relation $\mathbf F = -\nabla U$ holds for conservative vector fields. The relation between work done and the potential then becomes $$W + \Delta U = 0,$$ and the interpretation of this is that a variation of potential energy $\Delta U$ is converted into work $W$ (or that some work $W$ has caused a certain variation of potential energy $\Delta U$), but no energy gets lost in the process.

Answer 4

The electromotive force in a wire is the line integral $$\mathcal{E}=\int_\text{wire}\vec E\cdot d\vec\ell$$ (There may or may not be a negative.) So for a closed circuit the induced EMF is the loop line integral $$\mathcal{E}_\text{loop}=\oint_\text{loop}E\cdot d\vec\ell$$ The Maxwell-Faraday equation $$\nabla\times\vec E=\dot{\vec B}$$ leads to, upon volume integration of both sides and an application of Stokes' theorem $$\mathcal{E}_\text{loop}=\dot\Phi$$ where $\Phi$ is the magnetic flux. In statics, we have $\dot \Phi=0$. Thus the electromagnetic contribution to the EMF of the loop vanishes. You can amend the above definition of the EMF to include thermal and chemical terms as well, so the overall EMF must not necessarily vanish, just the E&M contribution.

Answer 5

It says (physically) that if you measure the field vector around a closed curve in space that you will get a zero sum. (You can think of this as moving a detector along the curve and measuring the total amount of work (the vector dot-product of force times delta-x) done along that path.) It implies that a field that does not change over time will not produce net work or change in energy to a charge after the charge has moved along such a path.

A closed loop conductor would be another way of measuring an electrical field, since the electrons are mobile and the protons are not. So this statement is saying that a constant electrical field (in time) will not induce a current in a conducting loop.

This also implies that if you do work in an electrical field to increase the potential energy of a charged particle that you can get all that energy back if you let the particle fall back, drift back or follow any path back to its starting point. (ACuriousMind has just stated my two paragraphs more mathematically.)