A spherical shell, uniformly charged, exerts no field inside it ; the point being either at centre or off-centre. According to Gauss law if we construct a spherical Gaussian surface all inside the shell, it contains no charge inside and thereby zero net flux on that surface. Now, here we usually conclude that the field is zero as flux is zero. My question is how can we say the field is zero,  whereas the flux can also be zero in other configurations, where ingoing and outgoing field lines are equal in number. For instance, spherically symmetrical $1/r^2$ field and the source point being outside Gaussian surface.
81697
If you have a conducting hollow sphere with a uniform charge on its surface, then will the electric field at every point inside the shell be 0. The reason the electric field is 0 at the center is clear from the symmetry of the sphere, but for a point at a certain distance from the center, shouldn't a net electric field exist?  I saw a similar question on the site but it was vaguely explained.
34990
Would the field in a cavity inside an arbitrary conductor with some charge, which is NOT SYMMETRIC be zero?   My book only says that field inside a cavity is always zero when there is no charge in cavity. But they prove this by taking a sphere with a symmetrical cavity and using Gauss law. But how can I prove this in cases where I can't take $E$ (field) out of:   $$\oint\pmb{E}\cdot d\pmb{A}=\frac{Q_{enclosed}}{\varepsilon_0}.$$
21958
This is Griffiths, Introduction to Electrodynamics, 2.43, if you have the book.      The problem states Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express your answer in terms of the radius $R$ and the total charge $Q$. Note: I will say its uniform charge is $\rho$.   My attempt at a solution:   My idea is to find the field generated by the southern hemisphere in the northern hemisphere, and use the field to calculate the force, since the field is force per unit charge.   To do this I start by introducing a Gaussian shell with radius $r &lt; R$ centered at the same spot as our sphere. Then in this sphere,   $$\int E\cdot\mathrm{d}a = \frac{1}{\epsilon_0}Q_{enc}$$  Now what is $Q_{enc}$? I feel like $Q_{enc} = \frac{2}{3}\pi r^3\rho$ , since we're just counting the charge from the lower half of the sphere (the part thats in the southern hemisphere of our original sphere). (Perhaps here is my error, should I count the charge from the entire sphere?, if so why?)   Using this we get $$\left|E\right|4\pi r^2 = \frac{2\pi r^3\rho}{3},$$ so $$E = \frac{r\rho}{6\epsilon_0}.$$ Using these I calculate the force per unit volume as $\rho E$ or $$\frac{\rho^2 r}{6\epsilon_0}$$  Then by symmetry, we know that any net force exerted on the top shell by the bottom must be in the $\hat{z}$ direction, so we get $$ F = \frac{\rho^2}{6\epsilon_0} \int^{2\pi}_0\int^{\frac{\pi}{2}}_0\int^R_0 r^3\sin(\theta)\cos(\theta) \mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$  integrating we get $F = \frac{1}{4}\frac{R^4\rho^2\pi}{6\epsilon_0}$.   Now Griffiths requests us to put this in terms of the total charge, and to do so we write $\rho^2 = \frac{9Q^2}{16\pi^2R^6}$  Plugging this back into $F$ we get $$F = (\frac{1}{8\pi\epsilon_0})(\frac{3Q^2}{16R^2})$$  Now the problem is that this is off by a factor of $2$ ...   I tried looking back through and the only place I see where I could somehow gain a factor of $2$ is the spot I mentioned in the solution, where I could include the entire charge, however, I can't see why I should include the entire charge, so if that's the reason I would be very grateful if someone could explain to me why I need to include the entire charge.   If that is not the reason, and perhaps this attempt at a solution is just complete hogwash, I would appreciate if you could tell me how I should go about solving this problem instead. (but you don't need to completely solve it out for me.)
35613
On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to the surface at a point?
78539
Why is there no electric field in the hollow space of the dees of a cyclotron though it is open at one face? Is electrostatic shielding applicable for space(hollow space) inside open conductors(like dees) also?
8080
When we're dealing with problems in electrostatics (especially when we use Gauss' law) we often refer to the density of electric , which is inversely proportional to the radius in the case of a single point charge (all field lines are directed radially).   My question may sound dumb, despite the fact that this concept is quite intuitive, but if you think about it there is actually an infinite amount of field lines which we can draw everywhere (one through every point), so speaking of an area where field lines are more "dense" or "sparse" doesn't make much sense to me.  With this in mind, why can we still use such a concept? Why does it really work?
36038
I know from Gauss law, it is $\vec{E}=\dfrac{\sigma}{2 \epsilon_0}(\hat{n})$ at all points. But it doesn't make sense because of the inverse square nature of electric field which suggests if you move further away from the plane, electric field must reduce.   Then why is electric field of an infinite plate constant at all points?
85178
Suppose I have a sphere of radius $r$ with all the charge residing on the surface, distributed  uniformly i.e. charge density $\sigma$ is constant. I want to find the electric field created by this sphere.  All problems of this sort I have solved by enclosing the sphere in a larger imaginary sphere with radius $R$, and then using the assumption that the electric field lines are in a radial direction, in combination with the divergence theorem.   If I use that method for this question, I find that the Electric field inside the sphere is zero. I can't understand this intuitively. I would have thought that if you are in the proximity of charge, there will be an electric field. Why would the electric field lines extend outwards radially from the sphere, and not inwards too?
48633
In electromagnetism books, such as Griffiths or the like, when they talk about the properties of conductors in case of electrostatics they say that the electric field inside a conductor is zero.  I have 2 questions:    We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges? When the textbooks try to show why the electric field inside a conductor is zero they say let us put our conductor in an electric field. What happens then is that there will be an induced surface charge density which consequently induces an electric field within the conductor such that the total electric field within the conductor will be zero. That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic!
22128
Does a charged rod or a shell have a magnetic potential energy when it is kept in a uniform magnetic field?  Also, can anyone quantitatively explain why a wire loop carrying current would tend to be circular?
97012
I am self-studying Griffiths's Introduction to Electrodynamics (4th edition) and all has been smooth sailing except for section 4.2.3 in which Griffiths argues for why we can compute (at least at the elementary level of classical electrodynamics that I am currently studying) the field due to a polarized dielectric chunk of material by simply adding up the potential (or field) due to each tiny volume element times the polarization per unit volume $\mathbf{P}$, treating each volume element as containing a perfect dipole. I have attached the page and a half of the chapter here, as well as the Eq. 4.8 alluded to there. I'll explain what I understand and where I lose track of his argument, and I'm hoping someone can help me sort things out. After a very gentle qualitative intro, Griffiths essentially says that we are going to define the macroscopic field $\mathbf{E(r)}$ at the point $\mathbf{r}$ in space as the average electric field over a sphere of radius a &quot;thousand times the size of a molecule&quot; about the point $\mathbf{r}$, which I'll call $V$. Fine, this is just a definition, and he says that analogous definitions using ellipsoids etc. yield the same answer, which I'm willing to accept. Now by superposition, the average field on $V$ can be found as the average field due to charges inside and outside of the sphere. With (4.17), he gives the average potential over $V$ due to all charges (dipoles) outside of $V$, and with (4.18) he gives the average field inside $V$ due to the charges inside $V$. I'm still following up to here, but now he loses me. He says that what is left out of the integral &quot;corresponds&quot; to the field at the centre of a uniformly polarized sphere, but what does this correspondence mean? I see how the integral in (4.17) doesn't include $V$, but why does that mean that $\mathbf{E}_{in}$ adds back what is missing (what does it even mean for an average field to source potential? We can speak of charge sourcing potential, or taking a line integral of a field giving a potential, but this is just some field value?). The best I can come up with is to say that if you take (4.19), split the integral out into $V$ and outside $V$, and then apply the (negative) gradient to the integral, one should be able to show that $\mathbf{E}_{in}$ is recovered for the former term, and so that would seem to imply that our coarse graining calculation does indeed give the correct macroscopic field as we have defined it (but I couldn't see how to DO this derivative, so I couldn't prove it). I've tried to look at other resources on this (eg. Zangwill), but none were digestible for me unfortunately. Please let me know if this is unclear, and I can try to rephrase. But essentially I'm hoping someone can, in detail, walk me through Griffiths's arguments. Eq 4.8 (note that Griffiths's script vector r equals $\mathbf{r} - \mathbf{r'}$, and hats have their usual unit vector meaning):  Griffiths text:
7054
Flux, as I understand it, is the amount of substance passing through a particular surface over some time. So, from a simple perspective, considering photons that go through some virtual surface $A$ (or $S$, doesn't matter). They have a fixed speed in vacuum, $v=299,792,458$ $\text m/\text s$. To simplify even further, they're all hitting the surface head-on. So, if we wanted to figure out how many photons go through the surface, we conclude that at a constant velocity they will only pass through the surface if they are in the volume bounded by sweeping the surface area along the velocity vector (perpendicular to the surface, the opposite of its normal) a distance $d$ in the alloted time $t$: $d = vt$  So the flux volume is well-defined as $V(t) = Avt$. We could just look for a period of unit time and "drop" the dependency on time. But even then, it's useless if we cannot sample photon volume density $\rho_p$ to determine how many photons occupy a unit volume in order to determine the actual flux. That makes sense:  $$\Phi = \rho_pV(t)$$  And now comes along the electric flux and thwarts my understanding of the whole notion completely. An electric field is generated when a charge is dropped somewhere in space. Any other charge, especially idealized point charges, placed in its vicinity would experience a force exerted on them by the source charge, its magnitude modulated by the amount of charge. So, the electric field maps points in space with force vectors (ie. a vector field) whose direction and magnitude is parametrized by the interaction between the source charge and the point charge.  And this electric flux is defined as $\Phi = E \cdot S$ (I'll use $\Phi = E \cdot A$) and I just cannot interpret the semantics of this dot product, the product's dimensionality is not what I've come to expect from the notion of flux ($\text{Vm}^{-1}$ or any other). How does this in any way show how much electric field flow goes through a surface? Furthermore, what is this vague thing called electric field flow? It seems like it is completely disconnected.  I've tried expressing it in different ways from the derivation of the expression of an electric field, which makes sense (non-vector form, dropped unit vector):  $$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$  I've intentionally separated the inverse square of the distance and the $4\pi$ which is, I presume, a part of the normalization factor (steradians of sphere) -- but I noticed that together they forge the area of a sphere $A_r = 4\pi r^2$. This way, I could see the expression as the uniform charge density distribution on the surface of a sphere, scaled by the vacuum permittivity.  $$E = \frac{Q}{A_s\varepsilon_0}$$  And then, due to presumed uniformity, by multiplying by an arbitrary area, I could get the flux, the amount of charge(?) flowing through a particular surface in unit time(?):  $$\Phi = E \cdot A = \frac{Q}{A_s\varepsilon_0} A $$  I could see that as flux, but I'm really not sure can I really reinterpret parts of the normalization factor and the inverse-square of the distance into the area of a sphere. From the perspective of voltage over distance it makes absolutely no sense to me.  Any help would be appreciated.
23124
Consider a system of the two identical positive point charges situated in free space (isolated from the influence of any other external fields) as shown in the attached diagram. Particle $1$ is at $(a,a,0)$ and particle $2$ is at $(0,a,0)$. Their velocities, at the considered instant of time, are as shown in the diagram ($\mathbf{v}_1$ along the $+x$ axis, $\mathbf{v}_2$ along the $+z$ axis).    Now, by applying the Biot-Savart law, we find that the magnetic field due to particle $2$ at the position of particle $1$ is along $+y$ axis, which means that the force acting on particle $1$ is along $+z$ axis according to Fleming’s left-hand rule. A similar analysis shows that there is no magnetic force on particle $2$ as the magnetic field of particle $1$ should vanish at positions (relative to particle $1$) located along its velocity vector.  Now, if we observe the net torque on the system of two particles about the $y$ axis then it is non-zero and is directed along the $-y$ axis. Here, there are no external forces or torques acting on the isolated two-particle system, and yet the net torque, as well as the net force on the particles, are nonzero. Why? Also, Newton's third law of motion seems to be broken in this scenario. Why?    Edit $1$  I have come to know from the responses that the electromagnetic field itself takes away some momentum and angular momentum about the considered axis. However, I think that if I consider only two charged particles as my system then the Abraham-Lorentz force can be assumed to be acting upon the system, and that is sufficient to make sure that we have considered the momentum being carried away by the electromagnetic field itself.   Even after considering the action of the Abraham-Lorentz force for the two-particle system, the scenario breaks both Newton's third law and the conservation of linear and angular momentum. This is because the Abraham Lorentz forces do not exactly counterbalance the force and the torques, on the two-particle system under consideration, due to the magnetic field.    Edit $2$  The previous edit was a result of confusion and misunderstanding on my part. The force associated with the momentum carried away by the electromagnetic field as a result of the electromagnetic interaction described in the question is the Lorentz force itself which simply doesn't obey the third law of motion. The Abraham-Lorentz force is a different story. It is associated with the momentum carried away by the radiation emitted by the accelerated charged particles. This is an additional force apart from the Lorentz force and corresponds to an additional carriage of momentum by the electromagnetic field. The momentum carried away by the electromagnetic field in correspondence with the Lorentz force doesn't correspond to radiation.
27685
Let there be a hollow sphere (Throughout the question we will ignore the thickness of the sphere) which is positively uniformly charged,q of radius,  R. Suppose there to be a test positive charge, q' at point P which is r distance away from the hollow sphere(r>R, P is outside of hollow sphere).  Now I have an doubt on evaluating Electric Potential,V at point P.  Both source charge,q and test charge,q' are +ve, so source charge will repel the test charge which will make the charge, q' to go to infinity.I have to apply external force, F' to displace the charge, q' from infinity to point P against the equal but repulsive force, F.In short, F' is  displacing the q' from infinity to P.  So External Force, F' will do +ve work and Repulsive Force, F will do -ve work. This way the Electric Field,E produced by hollow sphere is Opposite to Displacement of test charge(i.e.,θ=180).  Now we know that E=k(q/(r^2))  [Please Notice that infinity is the lower limit of integration]  Also V=-∫r∞(E dr cosθ)=-∫r∞(E dr cos 180)=∫r∞(E dr)  At last I ended up to V=- k(q/r). I know that V=k(q/r).  I just want to know why I am having this shitty negative sign in my equation.
80508
Suppose, for sake of argument, we have a spherical grounded conductor at the origin. Additionally, let our reference voltage be at infinity. Now, I view the potential of a point in space as being the work that I personally would have to do to move a unit charge from the reference to the point in question. Thus, I interpret this grounded conductor as points in space at which I could bring in a unit charge from infinity and end up doing no work. However, that interpretation certainly cannot be true in this case. Suppose the charge that I am bringing in is unit positive. The entire time I bring that charge towards the conductor there will be a larger and larger negative charge build-up, so I will have to exert a larger and larger force in the opposite direction the entire time I bring it in. The net work that I do in moving this charge towards the conductor will clearly be negative, not zero as the potential value would seem to suggest.   Can someone resolve these conflicting interpretations? Much appreciated.
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?
116486
In my intro E&amp;M class, we were given an example problem to determine the voltage within a cylindrical capacitor as a function of radius. A diagram of such a cylindrical capacitor is below.  The solution from my instructor is as follows.  Let the outer, negatively-charged surface be denoted by “b” and the inner, positively-charged surface be denoted by &quot;a”. So, we have $$\begin{align}\Delta V &amp; = V_a - V_b\\&amp;= -\int_{r_b}^{r_a} \vec{E}\cdot\text{d}\vec{s}\\ &amp;= -\int_{r_b}^{r_a}E\cos\theta\text{ d}s\end{align}$$ Now we note that $\cos\theta = -1$ in this case and then make the substitution $\text{d}s = - \text{d}r$. $$\begin{align}\Delta V &amp;= -\int_{r_b}^{r_a} E\cos\theta\text{ d}s\\&amp;= \int_{r_b}^{r_a} E\text{ d}s\\&amp;= -\int_{r_b}^{r_a} E\text{ d}r\end{align}$$   When we actually substitute in the expression for the electric field as a function of $r$, we get the correct answer. However, my concern with this solution is in the nature of the order of substitution. When I approached this problem, instead of first making a substitution for $\cos\theta$, I first made the substitution $\text{d}\vec{s} = -\text{d}\vec{r}$. To me, this vector substitution made more sense than the scalar substitution of differentials since this would be more indicative of the opposite direction between the &quot;direction of integration&quot; and the radial vector. So, using this approach, we get $$\begin{align}\Delta V &amp; = V_a - V_b\\&amp;= -\int_{r_b}^{r_a} \vec{E}\cdot\text{d}\vec{s}\\ &amp;= \int_{r_b}^{r_a}\vec{E}\cdot\text{d}\vec{r}\\&amp;= \int_{r_b}^{r_a} E\cos\theta\text{ d}r\end{align}$$ Now, since the differential radial vector and the electric field are parallel, $\cos\theta = 1$, so we get $$\begin{align}\Delta V &amp;= \int_{r_b}^{r_a} E\cos\theta\text{ d}r\\&amp;= \int_{r_b}^{r_a} E\text{ d}r\end{align}$$ This is the negative of my instructor's solution, which obviously leads to the wrong solution. Where do I go wrong? Why can I not simply reverse the order in which my teacher makes substitutions?