I was studying electrical potential and was caught in a loop. In the book shows that the electric field is path-independent by an example with a point charge. I needed more, I wanted some universal proof that the electric field was in fact a conservative field. For a vector field to be conservative, it should be a gradient of a scalar function. But to show that the electric field is a gradient of some scalar function, I need an expression for either the electric field or the scalar function. And for a general proof, a general expression is needed but this is not possible.
So I took an alternate path, a conservative vector field is path independent. Path independence is achived when the line integral of the vector field for path A to B to A is equal to zero. But again, I cannot find a general approach.
Is this something that I need to take in as a universal law like Coulomb's law which we have experimentally verified that it works to great distance and breaks in quantum scale?
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$\begingroup$ "Path independence is achieved when the line integral of the vector field for path A to B to A is equal to zero" It's zero if A=B. That is, over a closed loop. You can prove this to yourself mathematically. Another feature of conservative fields is that they are irrotational i.e., $\nabla\times E=0$ and this is something you can also do. $\endgroup$– joseph hCommented Nov 27, 2022 at 5:25
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1$\begingroup$ Any arbitrary charge distribution can be considered a collection of point charges. The fields of each point charge simply superpose. Since the field of a static point charge is conservative, the field of a static collection of point charges is conservative. $\endgroup$– GhosterCommented Nov 27, 2022 at 7:06
1 Answer
You can't show that the electric field is conservative without first postulating some rules for how it behaves. The usual starting place is Maxwell's equations, the second of which says that $$ \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}. $$ When the magnetic field is constant in time, the curl of the electric field is zero. That means that all the properties of the electric field that you mentioned (path-independent line integral, line integral around a closed loop is zero, existence of a scalar potential) hold. But when the magnetic field is changing, the curl of the electric field is nonzero, so it is not conservative.
To summarize, there is no "universal proof" that the electric field is conservative (and indeed, it isn't always conservative), only a derivation from the postulates of electrodynamics, which are in turn motivated by experimental observations.
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1$\begingroup$ Is integral form of what you wrote is $$\oint \vec E \times \vec{dr}=-\frac{d \phi}{dt} $$ $\endgroup$ Commented Nov 27, 2022 at 9:17
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$\begingroup$ Almost-- it's actually $$\int_{\mathcal{C}}{\mathbf{E}\cdot d\mathbf{r}} = -\frac{d}{dt}\int_{\mathcal{S}}{\mathbf{B}\cdot d\mathbf{a}},$$ where $\mathcal{C}$ is a closed loop and $\mathcal{S}$ is a surface bounded by $\mathcal{C}$. $\endgroup$ Commented Nov 27, 2022 at 23:13