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I was told a proof that the electric field was conservative (without using $\nabla$) which used a point charge and showed the following:

$$w.d.=\int_c{\vec F \cdot \mathrm{d} \vec l}=\int_c{\vec F\cos(\theta) \mathrm{d} l}$$ where $c$ is a path from a to b and $\theta$ is the angle between $\vec F$ and $\mathrm{d}\vec l$. In the case of a point charge it can be shown that in the limit (i.e. the limit of small elements of path getting infinitesimal) we have $dr=cos(\theta)dl$ where $dr$ is the radial distance from the point charge. It is now this next step that I have problems with, every where I have seen this proof they do the following:

$$\int^{r_b}_{r_a}{F\cos(\theta)\mathrm{d}r}$$ This is no longer a line integral, why? (they then go onto say that this is path independent and there the force is conservative). So why have we suddenly jumped from a line integral to not a line integral. (a specific and general reason would be appreciated)

Here is a link to the proof.

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So why have we suddenly jumped from a line integral to not a line integral

For a point charge, the electric field has a radial component only

$$\mathbf E = \frac{kQ}{r^2}\hat{\mathbf r}$$

Thus, the dot product of the electric field and the infinitesimal displacement vector is

$$\mathbf E \cdot d\mathbf l = \frac{kQ}{r^2}\hat{\mathbf r}\cdot \left( dr\;\hat{\mathbf r} + r\;d\theta \; \hat{\boldsymbol {\theta}} + r\;\sin \theta \;d\phi \;\hat{\boldsymbol{ \phi}}\right) = \frac{kQ}{r^2}\;dr$$

so, only displacement in the radial direction contributes to the line integral.

If the start and end radial coordinates are the same, the line integral is zero.

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I guess that by line integral you mean integral along the path whose differential element is d$\vec \ell$. Well, in fact you don't need to mention the path, it's $\theta$ that "knows" the path. If the path between the points $a$ and $b$ is along $\vec F$, the $\theta = 0$. If the path is some undulated line, $\theta$ will vary along the path.

Now, if $a$ and $b$ are very close as you say, $\theta$ will be constant.

Though I don't have the feeling that this is a sufficient proof for the conservative nature of the field. I would prefer to see a proof that shows that for two arbitrary points in the field, $a$ and $b$, we get : $\int_a^b F \ cos(\theta) \ d \ell = const.$ for whatever path $\ell$ chosen between $a$ and $b$.

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  • $\begingroup$ I have added a link in the question like I said it is the step going from the line Intergral to the normal Intergral that I don't get, it seems to me that even though we have reduced it only to r it could still be path dependent, I.e. Going back and forth in the direction could still do work $\endgroup$ – Quantum spaghettification Jan 20 '15 at 19:00
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An electric field is the force that fills the space around every electric charge or group of charges. Electric fields are caused by electrical forces. Electrical forces are similar to gravitational forces in that they act between things that are not in contact with each other. Electric fields are also analogous to magnetic fields resulting from forces acting upon magnetic substances or magnet poles. Electromagnetic waves have both an electric field and a magnetic field that are coupled to each other. Mathematically, the magnitude (or the strength) of an electric field at any point is defined by the force experienced by the charge at that point divided by the charge. This concept is written mathematically as E = F / q. Electric field strength is measured in units of newtons/coulomb. Electric fields are either static or dynamic.

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    $\begingroup$ So, how does this answer the question on why electrostatic fields are conservative? $\endgroup$ – user36790 Apr 11 '16 at 16:00

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