So lately I was speculating why nature choose the number 2 in Coulomb's law like why not 2.$10^{100}$ trailing zeros and then 1 or anything else. I find 2 a bit arbitrary  the given explanation being the cancellation of 2 while applying Gauss law in 3d space. So my question is, is there any fundamental explanation for this fact like from QED or something like that?
19392
Coulomb's Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges. Gauss's law implies that the total flux through a surface completely enclosing a charge is proportional to the total amount of charge. If we imagine a two-dimensional world of people who knew Gauss's law, they would imagine a surface completely enclosing a charge as a flat circle around the charge.  Integrating the flux, they would find that the electrostatic force should be inversely proportional to the distance of the charges, if Gauss's law were true in a two-dimensional world. However, if they observed a $\frac{1}{r^2}$ fall-off, this implies a two-dimensional world is not all there is. Is this argument correct? Does the $\frac{1}{r^2}$ fall-off imply that there are only three spatial dimensions we live in? I want to make sure this is right before I tell this to my friends and they laugh at me.
119651
Recently I learned that electrostatic potential energy of a system of charges can be calculated like so:  $$E = \int \frac{1}{2} \epsilon_0 \mathbf E^2 dV.$$  However, for a system of two point charges Q1 and Q2, we have  $$E = \int \frac{1}{2} \epsilon_0 (\mathbf E_1 + \mathbf E_2)^2 dV = \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_1^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_2^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 * 2 \mathbf E_1 \cdot \mathbf E_2 dV$$  from which we subtract the first two terms because they are already properties of the charges themselves.  Question:  Why do we not need to do something like this for charge distributions? Aren't charge distributions made up of many many little charges of magnitude $e$? Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?
156
My textbook says the field inside a conductor must be zero in order for the system to be equilibrium and therefore there must be no excess charge inside.  Their proof:  1) Place a gaussian surface inside the conductor. Since the system is at equilibrium, all points on the surface must have an electric field of zero.  2) Therefore the net flux is zero, implying the charge inside is zero.  3) If there is no charge inside, all excess charge must lie on the surface.  My problem:  "Since the system is at equilibrium, all points on the surface must have an electric field of zero."  Why?  Consider an conductor with no excess charge. To simplify things, say the conductor consists of a 100 free electrons and a 100 ions of charge 1$e^+$. The conductor is neutral and at equilibrium.  Now consider the field on one specific electron. There a total of 99 other electrons and 100  ions. So there would be a net field acting on that specific electron. So even though the system is at equilibrium, the net field on the individual particles is non-zero.  So a system can be at equilibrium, even if the net field inside is non-zero. What am I misunderstanding?  Another example: Consider a system consisting of one electron and one proton. If nothing is holding them back they will attract each other until they touch. After they touch, the force acting on them is still non-zero, yet they wouldn't accelerate.
82362
Actually my question comes from Gauss law which says charges outside a closed surface, doesn't effect on the whole electric flux of the surface.  The reason of that is that the solid angle of a closed surface while measuring it from outside of the surface, is zero. ( on the other hand, the solid angle of a closed surface from inside the closed surface is $4\pi$)  My question is why solid angle of a closed surface from exterior  is zero? Please don't use Stokes' theorem!
98951
There are so many available bases. Why is the strange number $e$ preferred over all else?  Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
109325
When we have an uncharged conductor with a cavity and apply an external electric field, the charges rearrange to keep the electric field inside the metal zero. The driving force is the need to keep electric field inside the metallic layer, zero. Then why is the field inside the cavity also zero? I think this is because the charges would arrange in the same way as in the case of a cavity free conductor but why can't they arrange in a different way?  Why is this arrangement not possible which clearly has a field inside the cavity but not inside the metal?
85691
I'd like to understand  and its contents as to what it implies in a bit simpler terms. I am familiar with mathematics unto Calculus 1,2,3 and some linear algebra and group theory. I am familiar with that for each symmetry there is a conservation law.    Now is there a one-one correspondence between these two always? What about laws like conservation of charge?  And conservation of energy?  And conservation of mass?  Considering our view of nature was changed as Special Relativity was introduced and thus, what earlier the theorem would have asserted as mass and energy to be originating from different symmetries then how did we reconcile these two as just one symmetry?
41277
Here are some depictions of electromagnetic wave, similar to the depictions in other places:      Isn't there an error? It is logical to presume that the electric field should have maximum when magnetic field is at zero and vise versa, so that there is no moment when the both vectors are zero at the same time. Otherwise one comes to a conclusion that the total energy of the system becomes zero, then grows to maximum, then becomes zero again which contradicts the conservation law.
43769
In MIT's 8.02 course, it is shown in lecture 3 that we can derive Gauss's Law from Coulomb's to get  $ \phi = \oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_{0}} $  However, in the lecture, it was assumed that there were no charges outside the surface. Later, Gauss's Law was used and the charges outside the sphere were ignored.  I've been thinking about it and came across .  The first answer by user levitopher helped a bit. I think that the lecture's argument still holds when there are charges outside because $ \vec{E} $ is added vectorially.  Suppose we have $ Q_{enc} $ in a sphere and $ Q_{out} $ outside the sphere. Then,  By definition, $ \phi = \oint (\vec{E_{Q_{enc}}} + \vec{E_{Q_{out}}}) \cdot \vec{dA} $ $ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + \oint \vec{E_{Q_{out}}} \cdot \vec{dA}$ $ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + 0 $ $ = \frac{Q_{enc}}{\epsilon_{0}} $  This assumes that $ \oint \vec{E_{Q_{out}}} \cdot \vec{dA} = 0 $ because any field lines that go in must come out (hand wavy but I'm just going to accept it for now).  I'd just like to know if this argument is sound.
79793
As far as I know, Coulomb's law of electrostatic force is applicable on two different charges situated in same medium. But if two individual charges are in different media (say one charge on a iron surface another on a plastic surface), then how does one calculate the force between them?  And if I have wrong conception , please correct me.
8113
gives a table of certain constants in different systems of units. I noticed that in Gaussian and electrostatic CGS the value of $\epsilon_0$ equals the dimensionless $1$.  I wondered whether anyone could provide justification for this? I was under the impression that $\epsilon_0$ is not even necessary in these units since it's essentially just a dimensional conversion factor in the SI, however I suppose it would still be possible to convert it into CGS units. How is this done?
5956
If the charge $q_1$ has to repel the charge $q_2$, the electric field has to go inside the conductor which contradicts the fact that electric field inside conductors is zero. Then why do the charges distribute themselves in a particular manner? Why can't they be distributed over a small place on the conductor?
105537
My question may be very basic, but I can't think of a reasonable explanation for this.  Consider a solid charged sphere. Now, we have an electric field inside the solid sphere, but at any particular point there are charges infinitesimally close to the point. As I see it, these charges should contribute to an infinite value of the electric field.  One might possibly argue that at any given point, infinitely-close charges are distributed uniformly everywhere and hence their effects cancel out. But at the surface of the sphere the uniformity of infinitely-close charges is not present and hence here we must have infinite field, but that is not the case as well! Where are we going wrong?  Also, if we think of electric field at that point in continuous charge distribution, the contribution of electric field might exist finitely from nearby charges with both charge and distance between them tending to zero, but for the very small charge present exactly there, the electric field contribution from itself must be infinite with charge being finitely small and distance being exactly 0. What's wrong here?  I have strong suspicions, is it sensible and correct to define electric fields at such a close vicinity of other continuous charges?
36150
If we write out Maxwell's equations with magnetic charges, we get $$ \begin{align} \nabla \cdot \mathbf{E} &amp;= 4 \pi \rho_e \tag{1}\\ \nabla \cdot \mathbf{B} &amp;= 4 \pi \rho_m \tag{2}\\ -\nabla \times \mathbf{E} &amp;= \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3}\label{Eq:Faraday}\\ \nabla \times \mathbf{B} &amp;= \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4}\label{Eq:Ampere} \end{align} $$ In particular, Faraday's law \eqref{Eq:Faraday} contains a minus sign that Ampere's law \eqref{Eq:Ampere} does not. This always struck me as odd because it's often said the fields are dual to each other (i.e. you can replace E with B and &quot;get the same result&quot;), but that requires a bit of mental recalibration to accommodate that minus. So I'm curious what the origin of that negative is and what it means. Are there any intuitive explanations for how to think about it?
47102
My textbook says that when a conductor is placed in an electric field, the electrons in it realign so that the net electric field inside the conductor is zero. There isn't a proof for this. It merely states that the electrons drift opposite to the applied electric field and create an opposite electric field inside the conductor canceling the external one.   I imagine Gauss's Law would provide a simple proof. If we try enclose a point inside the conductor in a gaussian surface, and note that the enclosed charge is zero, so the flux through the surface is zero and so is the electric field at the surface.   When I looked back at my reasoning, I realized that I had assumed that the charges induced on the conductor always stay at the surface, which I had no way of proving. How do I prove this formally?
74471
Why do we use the ?  Why did it come to be the electron volt and not, say, just a prefix of the joule, like the nanojoule?  Does the electron volt represent anything particular as far as the mathematics goes? I am guessing that it does, and if so, what is it that the electron volt exactly represents in terms of the mass of a particle, as I have seen it used for both the energy of a photon and the mass of subatomic particles?
112897
I've been wondering, in Coulomb's Law, $k_e = \dfrac{1}{4\pi\epsilon_0}$. Therefore, why do we use $G$ in Newton's Law of Gravitation? What if the constant is more like Coulomb's Law, e.g. $G = \dfrac{1}{4\pi G_0}$ where $G_0$ is some constant.  This would make Newton's Law of Gravitation look like the following: $$\bf{F}_{12} = -\dfrac{m_1 m_2}{4\pi G_0 |\bf{r}_{12}|^2} \bf{\hat{r}}_{12}$$  $GM = \mu$ is used for calculations but so could $\dfrac{M}{4\pi G_0} = \mu$.  If this is not the case, what is the significance of this definition?