In multivariable calculus the line integrals was parameterized and denoted: $$ \int_C \mathbf{F} \bullet \, d\mathbf{r}=\int_D\mathbf{F}(\mathbf{r}(t)) \bullet \frac{d \mathbf{r}(t)}{dt} \, dt $$ However in electromagnetism the line integrals are confusing. The electric field: $$ \mathbf{E}(\mathbf{r}(t))=\frac{1}{4\pi \epsilon_0} \int_{path} \frac{\lambda(\mathbf{r}')}{\lvert \mathbf{r}-\mathbf{r'} \rvert^2} \frac{\mathbf{r}-\mathbf{r'}}{\lvert \mathbf{r}-\mathbf{r'} \rvert} \, dl' $$ $\mathbf{r}$ = field point and $\mathbf{r'}$ = source point of the charge.

What is the difference between this two integrals? Is the second a line integral along a vector field?

In math the procedure was; parametrize the curve, take the derivative of it, dot it with the parameterized field.

Should a parametrize the electric field or not?

I also have problem setting up the differential line element $dl'$. What is $d\mathbf{r}$ equivalent to in the second integral?


$dl'$ is equivalent to "$d|\mathbf{r}|$", it is essentially a "scalar length measure".

The electrodynamics integral you wrote here is a vector-valued integral, so no dotting happens. If you use a linear coordinate system, it may be evaluated as three scalar line integrals, one for each coordinate. Vector valued integrals cannot really be evaluated using a curved coordinate system.

Your integral should be evaluated as follows: $\mathbf{r}'$ is actually a curve $\mathbf{r}'(t)$ (I am not going to use ' as a notation for derivatives), and your integral is $$\mathbf{E}(\mathbf{r})=\kappa\int_\gamma\frac{\lambda(\mathbf{r}'(t))}{|\mathbf{r}-\mathbf{r}'(t)|^2}\frac{\mathbf{r}-\mathbf{r}'(t)}{|\mathbf{r}-\mathbf{r}'(t)|}\left|\frac{d\mathbf{r}'(t)}{dt}\right|dt. $$

Note that whilst I do not know the full context, from the form of the integral I can infer that only $\mathbf{r}'$ is parametrized by $t$. You are asking the value of $\mathbf{E}$ at a single point ($\mathbf{r}$), and this value is given by integrating the $\mathbf{r}'$ variable over a curve.

Edit: To answer your comment, you are almost correct.

Now that I know the full case, you also need to parametrize the $\mathbf{r}$ variable, but only because you are interested in values of $E$ solely on the $z$ axis!

Let us calculate the relevant vector quantities:

$$\mathbf{r}=\mathbf{r}(z)=z\mathbf{e}_z$$ (I am using the generic $z$ as a parameter instead of $b$, since I think this way it is clearer.) $$\mathbf{r}'(t)=a\cos(t)\mathbf{e}_x+a\sin(t)\mathbf{e}_y, $$ $$ \mathbf{r}(z)-\mathbf{r}'(t)=-a\cos(t)\mathbf{e}_x-a\sin(t)\mathbf{e}_y+z\mathbf{e}_z, $$ $$|\mathbf{r}(z)-\mathbf{r}'(t)|^2=a^2+z^2, $$ $$ \frac{d\mathbf{r}'(t)}{dt}=-a\sin(t)\mathbf{e}_x+a\cos(t)\mathbf{e}_y, $$ $$\left|\frac{d\mathbf{r}'(t)}{dt}\right|=a. $$

Now the integral is (since $\lambda$ is constant) $$ \mathbf{E}(\mathbf{r}(z))=\frac{\lambda}{4\pi\epsilon_0}\int_\gamma\frac{1}{a^2+z^2}\left(-\frac{a\cos(t)}{\sqrt{a^2+z^2}}\mathbf{e}_x-\frac{a\sin(t)}{\sqrt{a^2+z^2}}\mathbf{e}_y+\frac{z}{\sqrt{a^2+z^2}}\mathbf{e}_z\right)a\ dt=\\\frac{\lambda}{4\pi\epsilon_0}\frac{a}{(a^2+z^2)^{3/2}}\int_0^{2\pi}z\ dt\mathbf{e}_z=\frac{\lambda}{2\epsilon_0}\frac{az}{(a^2+z^2)^{3/2}}\mathbf{e}_z, $$ where the $\mathbf{e}_x$ and $\mathbf{e}_y$ integrals are annihilated because $\sin(t)$ and $\cos(t)$ are integrated on their full period.

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  • $\begingroup$ Great, thanks! Shouln't $\mathbf{r}$ be parameterized (a function of $t$), i.e. $\mathbf{r}=\mathbf{r}(t)$? $\endgroup$ – JDoeDoe Jul 16 '16 at 21:18
  • $\begingroup$ I had problem with this question: A uniform line charge density $\lambda$ forms a circle of radius $a$ in the $xy$-plane ($z = 0$), with the center at the origin. Find the electric field at point $(0,0,b)$. Solution: ($\mathbf{r'}(t))$ is the source point, not the derivative) I don't really know there to start. Is the curve the same as the source point $\mathbf{r'}(t)$? Should I parametrize the circle in the $xy$-plane? If $x^2+y^2=a^2$, let $x(t)=\cos t , y(t)=\sin t, t\to[0,2\pi]$. $\endgroup$ – JDoeDoe Jul 16 '16 at 21:19
  • $\begingroup$ The position vectors are (parameterized) $$ \mathbf{r}(t)=b{\mathbf{\hat{z}}} $$ $$ \mathbf{r'}(t)=a(\cos t \,{\mathbf{\hat{x}}} +\sin t \, {\mathbf{\hat{y}}}) $$ $$ \mathbf{r}(t)-\mathbf{r'}(t)= b{\mathbf{\hat{z}}} - a(\cos t \,{\mathbf{\hat{z}}}+\sin t \, {\mathbf{\hat{y}}}) $$ Is this right? I'm kind of lost here. $\endgroup$ – JDoeDoe Jul 16 '16 at 21:20
  • $\begingroup$ @JDoeDoe I edited my post. $\endgroup$ – Bence Racskó Jul 16 '16 at 22:00
  • $\begingroup$ Thanks, really appreciate it! :) For surface charge density, is this correct (in terms of parameters and excluding parameterization of $\mathbf{r}$) $$ \mathbf{E(r)}=\frac{1}{4\pi\epsilon_0} \iint_{S} \frac{\sigma(\mathbf{r}'(s,t))}{|\mathbf{r}-\mathbf{r}'(s,t)|^2} \frac{\mathbf{r}-\mathbf{r}'(s,t)}{|\mathbf{r}-\mathbf{r}'(s,t)|} \biggl \lvert\frac{\partial\mathbf{r}'(s,t)}{\partial s} \times \frac{\partial\mathbf{r}'(s,t)}{\partial t} \biggl \rvert \, ds \, dt $$ $\endgroup$ – JDoeDoe Jul 17 '16 at 6:41

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