TITLE: conventional notation for magnitude and vector
QUESTION [1 upvotes]: Suppose as an example I have the magnitude of an electric 
$\left \| \vec{E} \right \|=\frac{\lambda}{2\epsilon s\pi}$
This is the equivalent to $\vec{E}=\frac{\lambda}{2\epsilon s\pi}\hat{r}$.
Is this correct? But why? I haven't come across this signalling before.

REPLY [1 votes]: Notational Preliminaries:
The cylindrical coordinate variables are $(\rho,\phi,z)$, where $x=\rho \cos \phi$ and $y=\rho \sin \phi$ describe the transformation from cylindrical to Cartesian coordinates.  
Furthermore, $(\hat \rho,
\hat \phi, \hat z)$ represent the unit vector triad in cylindrical coordinates.  
Finally, the electric field in general can be written in cylindrical coordiates as $$\vec E=\hat \rho E_{\rho}(\rho,\phi,z)+\hat \phi E_{\phi}(\rho,\phi,z)+\hat  z E_{z}(\rho,\phi,z)$$ 

Exploit the symmetry of the problem, which you already have done when applying Gauss's Law (not Gaussian's Law).  
In your development, you assumed that $\vec E$ has only a radial component $\hat \rho$ and depends only on the radial variable $\rho$. 

This reasoning is based on the physics and geometry of this problem, namely an infinite line charge of uniform density $\lambda$ placed along the $z$ axis.  Since the field source has no dependence on $\phi$ or $z$, then the field is independent of those coordinate variables.  And understanding Coulomb's Law, we see that the field can have no azimuthal or axial components.

Thus, we can write the electric field as $$\vec E=\hat \rho E_{\rho}(\rho)\tag 1$$
Then, you constructed a cylindrical surface with height $L$, centered on the $z$ axis, encompassing part of the line charge.  
The total charge $Q$ enclosed is $Q=\lambda \, L$.  
Finally, Gauss's Law states $$\oint_S \vec E\cdot \hat n \,dS=Q/\epsilon_0\tag 2$$
Using $(1)$ in $(2)$ reveals
$$\int_0^L\int_0^{2\pi}\hat \rho E_{\rho}(\rho)\cdot \rho \rho d\phi\,dz=\lambda\,L/\epsilon_0\implies E_{\rho}(\rho)=\frac{\lambda }{2\pi\epsilon_0\,\rho} \tag 3$$

Note that there is no contribution to the flux on the ends of the cylinder since $\vec E$ has no axial component.

Finally, substituting $(3)$ into $(1)$ yields
$$\vec E=\hat \rho \frac{\lambda }{2\pi\epsilon_0\,\rho}$$
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