From Gauss's law we find that E in any point near an infinite charged metal plate is:  $$E = \frac{\sigma\epsilon_0}{2}$$  which indicates that $E$ is same for any distance, but it seems weird and untrue from the concept of Coulomb's law. If distance from the plate is decreased a charged particle should feel more force on it. Can you please explain?
24177
For an infinite line charge, $E$ falls off with $1/r$; for an infinite sheet of charge it's independent of r! The infinitesimal contributions to $E$ fall off with $1/r^2$, so why doesn't the total $E$ fall off the same way for the infinite line and sheet charges?
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!
6733
The Electric field lines coming out of a charge are infinite, So the lines passing through a closed surface should be infinite , right?
101441
I am really confused in Gauss law.  Why do E3 and E2 pointing up? and also E1 pointing down?  The lecture note said infer from symmetry and you will get the following but I dont really understand. Could anyone explain to me? I am new to gauss law.
90824
When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass. However, had they been electrically charged bodies and we needed to compute electrostatic attraction, there is no "centre of charge" notion available to us. ( We would evaluate a 2-dimensional integral, or equivalently, apply Gauss's law. )  Is there any simple reason for the absence of such a centre of charge? Frequently, I've got the answer: "charge just doesn't work that way".
119170
Lets say I have a positively charged conducting sphere of Q. The electric field outside the sphere is  $$E = \frac Q{4\epsilon\pi r^2}.$$  Now suppose this sphere is enclosed inside another hollow conducting sphere of -Q charge. By Gauss law, the electric field is still the same?  Won't the negatively charged outer sphere reinforce the electric field, causing it it be larger?  In another example,  The field from a infinite sheet of charge is  $$ E = \frac \sigma{2\epsilon}. $$  But the field between 2 oppositely charged infinite sheets is  $$ E = \frac \sigma{\epsilon}. $$  Why is this not the case above?
84564
I am going over Coulomb's law and there is something that is a bit confusing for me:  According to Coulomb's law, if I have a charge $q_{1}$ at a point $\vec{r_{1}}$ and a charge $q_{2}$ at a point $\vec{r_{2}}$ then the force that the first charge applies to the second charge is given by  $$ F_{1,2}=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{2}}(\vec{r_{2}}-\vec{r_{1}})=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{3}}\hat{(\vec{r_{2}}-\vec{r_{1}})} $$  I see that in the first expression is does look like the force is proportional to $\frac{1}{r^{2}}$, but I don't understand why it isn't the second expression that matters, giving us that the force is proportional to $\frac{1}{r^{3}}$:  Say both charges are on the $x-$axis, then $\hat{(\vec{r_{2}}-\vec{r_{1}})}=\hat{x}$. Increasing the distance between the charges to two times what is by moving the second charge on the $x$- axis was, would decrease the force $2^{3}=8$ times, since $\hat{(\vec{r_{2}}-\vec{r_{1}})}$ is still $\hat{x}$.  What is the mistake in my reasoning ?
40103
As written in Principles of Physics by Halliday, Resnick and Walker:     The electric field is a vector field; it consists of a distribution of vectors,one for each point in the region around a charged object. In principle, we can define the electric field at some point near the charged object. We first place a positive charge $q_0$, called a test charge,at the point. We then measure electrostatic force $\vec{F}$ that acts on the test charge. Finally, we define the electric field $\vec{E}$ at that point due to the charged object as $$\vec{E} = \frac{\vec{F}}{q_0}$$ .   So, this is the way they derive the electric field. But one thing that perplexes me is that as test charge has its own field, so when the test charge is introduced in the field of concerned charged object, say another point charge $+q$ , won't it get shifted due to the repulsion from the test charge $q_0$ (which is, by convention, positive)? Then consequently the field of $+q$ will also be shifted. In a word, it will get disturbed by that test charge. But the book derived the formula as if the positive point charge $+q$ remained fixed i.e. the test charge had no effect on the field of the point charge. But how can it be? If the test charge $q_0$ is a charge, then obviously it has also field which will exert force on $+q$. So how can the formula be true? Why did the book neglect this?  This phenomenon can be compared to earth's gravitational field. When a body moves upward, it exerts force on the earth downward. But as the earth has huge mass, its shifting can be neglected and hence it doesn't affect its gravitational field.   But here it is a point positive charge; it has no huge mass. So why did the book neglect its shifting and how can the formula be true? Please help.
13750
Coulomb gave the law for the force between two static charges while considering them to be points in space. But the differential form of Gauss' Law talks about charge densities, a thing possible only if charges are smeared out in space.   Even Feynman addresses to the problem in his lectures when he says that on solving for the electrostatic energy in the field of a point charge we get infinity as the limit.  So do we know now that whether charges are point-like or smeared out?
18648
Imagine a negatively charged conductor in a vacuum. The excess electrons will be spread out over the surface such that the net electric field inside the conductor is zero.  What keeps these extra electrons bound to the conductor?   For the case of a neutral conductor, I could intuitively understand some sort of Coulomb force being responsible. But for a negatively charged conductor, I don't understand what (net?) positive entity is exerting the attractive force.   Is there an explanation based only on classical electrodynamics (say, at the level of Jackson). If not, is the explanation purely quantum mechanical?   Basically, how does one model this phenomenon?
17265
Let's say we have a cube of side length $a$ as a Gaussian surface in an electric field (in vacuum) and then we raise the strength of the electric field. As I understand, this rise in the field strength would travel through the field at the speed of light. This means that for a duration of $\frac{a}{c}$, the field strength, and thus the flux, would differ at the 2 opposing faces of the cube. This implies that the flux through the surface is non-zero and thus the charge inside it must also be non-zero by Gauss' Law. How is this possible?
81840
A metal rod of length $l$ is placed (as shown in figure) in a cylindrical time varying magnetic field. Find the potential difference across it.   My answer:  As electric fields as forming circles around the centre with magnitude $$\frac12r\frac{dB}{dt}$$ whose magnitude is same along any chord at a fix distance from centre: $$\Delta V=E.l=\frac12r\frac32at^{\frac12}.l$$ $$\Delta V=\frac34arlt^{\frac12}$$  My question: Is $\Delta V =V_M-V_N$ or $\Delta V=V_N-V_M$ $\bf OR$ equivalently which side is at high potential?  Possible    $(i)$ As Field lines are circular from $M$ to $N$, $M$ must be high potential. $(ii)$ As this field line apply a force on electrons- in the rod- towards left, then $M$ must have excess $e^-$ and $N$ must have deficiency; so $N$ would be high potential as $M$ is $(+)$ and $N$ is $(-)$.
63141
E = electric field V = electrostatic potential r = distance ' = derivative   E = - (V)'   therefore if E increases V decreases Also if r increases E decreases                                                      because  E=kq/r^2 From 1 &amp; 2, if r increases then V increases But V=kq/r, so if r increases V decreases   Why does the 3rd point contradict with the 4th ?
85489
I understand how, and why, an accelerating charge emits radiation, and loses energy in the process, as well as the  formula for the power, and its derivation.  However, in classical mechanics, when we want to find the motion of a particle, the process is (theoretically) simple - find all forces on the particle as a function of position and time, plug in Newton's second law and solve the differential equation. It seems to me as if simply because a particle is charged, this process just ceases to work.  When talking about a charged particle, it is simply not enough to find all forces and solve Newton's law - now we need to somehow take into account the energy lost by the particle to radiation. However no one seems to mention how to take that into account, as a term in an equation of motion. It seems like all we know is the total power radiated by the particle - there is no concrete, complete description of how that loss affects its motion.  Of course, in some simple cases we can easily guess how the particle will be affected: for example, for an electron circling in a uniform magnetic field, it is obvious that energy loss due to radiation will cause the electron to spiral inward. You can, at least theoretically, write the Lorentz force acting on the particle, due to the uniform field, plug it into Newton's second law, and also write the Larmor formula, take that loss into account and find a complete description of the particle's motion.   However still, when we write simply Newton's second law here, with the only force being the Lorentz force, we find a circular, not spiral, solution; we have to synthetically add in the Larmor formula to truly recover the particle's motion. Contrary to classical mechanics where Newton's law is a complete description of the motion.   Why is Newton's law no longer enough, or put another way, what term am I missing that will fix the electron's motion, add an additional "force" the will cause it to spiral inward? Is there a general expression for that force? And how does one in general write the equation of motion for a charged particle (or a system of)?   I suspect the answer has something to do with the EM field changing not instantaneously but at the speed of light, since Newton's law of gravitation clearly yields stable, closed orbits and propagates instantaneously. Perhaps I should phrase my question this way:   How does the fact that the EM field propagates at a finite speed affect the way we have to write the equations of motion  for a charged particle (as opposed to the equations of motion for a massive particle, acted upon by an instantaneous, newtonian gravitational field)?   One last note: perhaps this has something to do with Maxwell's equations being Lorentz-invariant and not Galilean? However I find this unlikely to be the case since the electron doesn't have to move in a relativistic speed in order to experience this effect. In any case, this is also why I allowed myself to use Newton's second law and not its (special) relativistic expression - would using that fix the problem?
2983
In classical mechanics, we had Newton's law of gravity $F \propto \frac{Mm}{r^2}$.  Because of this, all laws of classical electrostatics applied to classical gravity if we assumed that all charges attracted each other due to Coulomb's law being analogous.  We can "tweak" classical electrostatics to fit gravity.  In modern physics, does the reverse work?  Can we "tweak" General Relativity to accurately describe electrostatics or even electromagnetism?
95455
"Now we turn our attention to what happens in $S'$, in which the particle is at rest and the wire is running past (toward the left in the figure) with the speed $v$. The positive charges moving with the wire will make some magnetic field $\ B'$ at the particle. But the particle is now at $rest$, so there is no magnetic force on it! If there is no magnetic force on the particle, it must come from an electric field. It must be that the moving wire has produced an electric field. But it can do that only if it appears $charged$-it must be that a neutral wire with a current appears to be charged when set in motion."  So i got this from Feynman's lectures in physics volume 2 pg 13-7 and 13-8. I find it hard to believe that the particle experiences no magnetic force due to the moving wire although it's stationary. Since this is classical and not quantum electrodynamics if we assume the charged particle to have inertia, then by Newton's second law, the particle will remain at rest until acted upon by an external force, which i assume is the $magnetic$ $wind$ created by the moving wire past the stationary charged particle. Could someone please clarify??
15077
One of my physics books has a nice example on how to use Gauss's Law to find the electric field of a long (infinite) charged wire.  However, at the very end of the example, the author ends by saying Gauss's Law cannot be used to find the electric field of finite-length charged wire.  I could not understand why not.    Please this is not a homework problem or nothing like that. This is just a question I have and I have not being able to solve it in my mind yet.