How can we prove that surface integral of the electric fieldfor a point charge that is outside a gaussian surface, $$\int\mathbf E\cdot\,\mathrm d\mathbf S,$$ without actually using the concept of flux?
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?
7607
Let us define the magnetic field $$\vec{B} = g\frac{\vec{r}}{r^3}$$ for some constant $g$. How can we show that the divergence of this field correspond to the charge distribution of a single magnetic pole (monopole)?  EDIT:  If I calculate the divergence I get $$ \begin{align} \nabla\cdot\vec{B} &amp;= \nabla\cdot\left(g\frac{\vec{r}}{r^3}\right) \\ &amp;= g\left(\nabla\frac{1}{r^3}\right)\cdot\vec{r}+\frac{g}{r^3}\left(\nabla\cdot\vec{r}\right)\\ &amp;=g\left(-3\frac{\vec{r}}{r^5}\right)\cdot\vec{r}+\frac{g}{r^3}(1+1+1)\\ &amp;=-3g\frac{1}{r^3}+3g\frac{1}{r^3}\\ &amp;= 0 \end{align} $$ which contradicts that there is a sink or source of magnetic flux.
122155
I understand that with an infinite plane, as you get closer, the infinitesimal contributions to the electric field become greater in module. The direction of the vectors become less perpendicular to the plane as you get closer, hence reducing the overall electric field in the perpendicular direction. Both effects offset each other such that the electric field is the same at whatever distance.  However, wouldn't this same argument apply to an infinite line? Gauss's law shows that the electric field of an infinite line depends on the distance, but supposedly the same would happen as with the infinite plane, and yet it doesn't.  Does anyone have an explanation for this?
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.
11522
One of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this:    the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are assumed to be large enough that edge effects can be neglected i.e. they are effectively infinite.  In practice these problems are solved by the (frequently tedious) application of some simple rules, one of which is:     The charge densities on the two outermost surfaces (labelled $Q_{ext}$ in my diagram) are identical.   My question is how we can prove the statement that the two charge densities, $Q_{ext}$, are equal?  The problem is that our system of plates can have an arbitrary number of plates with arbitrary charges and spacings. Presumably one could with enough effort grind through a massive calculation where the plate charges and positions are all variables, but this seems an awfully complicated way of approaching an exceedingly simple observation. Surely there has to be a better way?  I am guessing that there must be some general argument one can make that means we get the same external charges regardless of the details of plates charges and spacings. Can anyone describe such an approach?  For bonus marks another of the starting points students are taught is that if any plate in the setup is earthed then the exterior charges are zero. If this doesn't emerge naturally from whatever approach is used to answer my first question I would also be curious to see how this could be proved.
101527
I was teaching kids about how to find electric field using the superposition principle for continuous charge distributions. I thought maybe I should derive the formula for electric field due to a finite rectangular sheet of charge of charge on the surface $S$, where $$ S = \left\{(x,y,z)\in \mathbb{R}^3 \mid -a/2&lt; x &lt; +a/2; -b/2&lt; y &lt; +b/2 ; z = 0 \right\} .$$ However, I got stuck at the following integration. $$ E(0,0,r) = \frac{\sigma r}{4\pi\epsilon_o}  \int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}}, $$ where $\sigma$ is the surface charge density. Note: This integration can be done if $a$ or $b$ or both are very large i.e. $\infty$ in which case we get usual result of $E=\frac{\sigma}{2\epsilon_o}$ So my question is, Can this integral be calculated? If not then what method would I use to find the electric field in this case. Also It would be greate if anyone can comment on how to find the electric field by directly solving the poisson equation.  Consequently if we take case of finite disk the following is the resulting integration. $$ E = \frac{\sigma r}{2\epsilon_o}  \int_{\xi=0}^{\xi=R} \frac{\xi d\xi}{(\xi^2+r^2)^{3/2}} $$ which can be solved as $$ E = \frac{\sigma}{2\epsilon_o} \left(1- \frac{r}{\sqrt{r^2+R^2}}\right) $$ Now by taking the limit $R \rightarrow \infty$ we can show that $E \rightarrow \frac{\sigma}{2\epsilon_o}$.
84517
The electric energy stored in a system of two point charges $Q_1$ and $Q_2$ is simply $$W = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{a}$$ where $a$ is the distance between them.  However, the total energy can also be calculated through the volume integral of magnitude squared of the electric over all space: $$W = \frac{\epsilon_0}{2}\int_{\mathbb{R}^3} E^2\,dV$$  Suppose that $Q_1$ sits on the origin and $Q_2$ a distance $a$ away on the $z$-axis.  Then the electric field is $$\vec{E}(x,y,z) = \frac{1}{4\pi\epsilon_0}\left[\left(\frac{Q_1}{r^3} + \frac{Q_2}{d^3}\right)x\hat{x} +\left(\frac{Q_1}{r^3} + \frac{Q_2}{d^3}\right)y\hat{y} + \left(\frac{Q_1z}{r^3} + \frac{Q_2\left(z-a\right)}{d^3}\right)\hat{z}\right]$$ where \begin{align} r &amp;= \sqrt{x^2+y^2+z^2} \\ d&amp;= \sqrt{x^2+y^2+\left(z-a\right)^2}  \end{align}  Calculating $W$ through $\int_{\mathbb{R}^3} E^2\,dV$ seems extremely difficult; Mathematica, for example, appears stumped.  Yet its result should simply be $\frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{a}$, correct?  The integral formula still applies to point charges, correct?
78131
I know we can find out the electric field using the electric field $$E=\frac{KQ}{R^2}$$ taking small element $dq$ and finding the electric field by integrating the value of $dE$ over the circumference which will be $$E=\frac{kxQ}{\sqrt {(a^2+x^2)^3}}$$ where $a$ is radius of ring and $x$ is distance of point $p$ on the axis of the ring. Can we find the same using the Gauss law?
2645
How to prove  $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
8745
Say you have a hollow sphere with a uniformly distributed charge on the surface. Why is the electric field everywhere inside the sphere zero?  For the centre, its easy to add the vectors from the surface charges and show they sum to zero because of symmetry.   But can you show me how the field cancels out to zero for points other than the centre using vector addition?
109752
I have a problem understanding Stokes' Theorem and Gauss' Divergence Theorem. Suppose the following:  Let $F$ be a vector field in $\Bbb R^3$. Let $S$ be an oriented closed smooth Surface enclosing a volume $V$ and let $C$ be a positively-oriented closed curve surrounding $S$  Stokes' Theorem says: $$  \int_C F·dr=\iint_S (\nabla \times F) · dS $$  Then, by the Divergence Theorem: $$ \iint_S (\nabla \times F)·dS = \iiint_V \nabla·(\nabla \times F)  dV $$  But $ \nabla·(\nabla \times F)=0$. So everything is $0$  What is it that I am not seeing?
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?
78037
I have, for instance, a problem with a spherically symmetric charge distribution. I deduce here, in order to solve the problem easily, that the corresponding electric field must be symmetric. How is this type of deduction rigorously justified?
110881
Taking the curl of the electric field must be possible, because Faraday's law involves it: $$\nabla \times \mathbf{E} = - \partial \mathbf{B} / \partial t$$ But I've just , where it says  The curl of the gradient of any twice-differentiable scalar field $\phi$ is always the zero vector: $$\nabla \times (\nabla \phi)=\mathbf{0}$$  Seeing as $\mathbf{E} = - \nabla V$, where $V$ is the electric potential, this would suggest $\nabla \times \mathbf{E} = \mathbf{0}$. What presumably monumentally obvious thing am I missing?
103133
In the usual bookwork treatment, it is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. I have always thought that neither version is more fundamental than the other and each has its place in solving problems. (See also  )  But: I have a conceptual problem with applying the integral forms of these equations in cases where there is time-dependence and the "size" of the loop or area means there is a significant light travel time across the regions considered compared with the timescale on which fields vary.  e.g. Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire. My first instinct is that this ought to be solved using the inhomogeneous wave equations to give A- and V-fields that are dependent on the retarded time - hence leading to the E- and B-fields.  But what about using Ampere's law in integral form? What is the limit of its validity? If we write $$ \oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \int \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$$ then presumably the $t$ that is defined on each side of the equation cannot be the same, since a change in $I$ at time $t$, presumably leads to a change in $B(r)$ at a time $t + r/c$? I suppose one does not care about this so long as the timescale for a current change is $\gg r/c$.  So my question is: Are the integral forms of Maxwell's equations inherently limited by this approximation, or is there a way of formulating them so that they take account of the finite size of a region in cases where the fields are time-variable?
116788
Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$  Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side.  How we can solve such integrals?I need your help.  Without using the Gauss Divergence theorem.
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?