Can anyone please explain how to know whether a vector field has  or not by seeing its diagram?  I have read that a vector field must change for having divergence but why is divergence zero in middle of electric field between dipole?
73183
What is ? I was learning about Maxwells equations and don't understand the divergence part of it. Can someone give an intuition of what divergence is in relation to maxwells equation. To give you a sense of language to use I'm 17 and don't do maths or physics at school (I've just recently started loving the two.)
99105
I'm wondering, for no particular reason: are there differentiable vector-valued functions $\vec{f}(\vec{x})$ in three dimensions, other than the constant function $\vec{f}(\vec{x}) = \vec{C}$, that have zero divergence and zero curl? If not, how would I prove that one doesn't exist?  I had a vague memory of learning some reason that such a function doesn't exist, but there's a pretty good chance my mind is just making things up to trick me ;-) But I thought about it for a little while and couldn't think of another divergence-free curl-free function off the top of my head, so I'm curious whether I was thinking of real math or not.
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?
1073
I need to find a vector field as described in the title.  I was given a couple hints, and this is what I have so far. Let $\varphi:S^2\setminus\{N\}\to\mathbb{R}^2$ be stereographic projection ($N$ is the north pole). Let $Y$ be a smooth, non-zero vector field on $\mathbb{R}^2$ that dies off to $0$ "at infinity." Since $\varphi$ is a diffeomorphism, we can push $Y$ forward along $\varphi^{-1}$ to get a smooth vector field $(\varphi^{-1})_*Y_{\varphi(p)}$ on $S^2$. But this vector field is not zero at the north pole... it's undefined. So can I just define my vector field on $S^2$, call it Z, to be the pushforward of $Y$ (as above) away from $N$ but $0$ at $N$? And if so, do we still have smoothness?
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!
13117
As we know that if we put any free test charge in an electric field which is moving horizontally then it will experience an repulsive force and move in that direction so by that we can say that for Straight lines the above statement is true that path followed by a free test charge in electric field represent the electric lines of forces which will be straight line. So we can say that it is TRUE.  But after that I read one exception case which states that the above statement is not true for Curved path.   Why is that so ? Means why it is for curved path and what is the exception ?Why it is not true?
22203
Consider the vector field $$\vec{\mathrm{F}} = \frac{\hat{\mathrm{r}}}{r^{2}},$$ then the divergence of this field is: $$\vec{\nabla}\cdot\left(\frac{\hat{\mathrm{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathrm{\vec{r}})$$  What is the proof of this relation?
53343
Can anyone explain where the following expression for the electric field vector comes from? $$ \mathbf E(\mathbf r,t) = -\nabla \phi(\mathbf r,t) - \frac{\partial}{\partial t}\mathbf A(\mathbf r,t)  $$  where $\phi$ and $\mathbf A$ are the scalar and vector potentials, respectively.  Presumably it can it be derived from Maxwell's Equations?
84745
The electric force (Coulomb's law) on a point charge $Q_2$ due to $Q_2$: \begin{gather*} \mathbf{F}_{12}=\frac{Q_1Q_2}{4\pi\epsilon_0}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3} \end{gather*}  The seperation vector between the charges is $\mathbf{r}_{12}=\mathbf{r}_2-\mathbf{r}_1=(x_2-x_1)\hat{\textbf{i}}+(y_2-y_1)\hat{\textbf{j}}=(z_2-z_1)\hat{\textbf{k}} $.  Does this mean that the force is a vector field or just a vector?   Is the force a function of $\mathbf{r}_{12}$, so $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?
72031
As far as I know, unpolarized light is defined as light which has no clear axis of polarization, but electric field vectors in all directions. However, I also know that electric fields can be superpositioned. So this leads me to the conclusion that if unpolarized light has equal $\vec E$ vectors in all directions, they should cancel.   Obviously, though, this is wrong (we've got light!). What's the explanation?
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.
93717
What does it mean to take the gradient of a vector field? $\nabla \vec{v}(x,y,z)$? I only understand what it means to take the grad of a scalar field.
42534
Applying the divergence to Eq. $47$, we obtain $$ \mathbf{\nabla} \cdot \mathbf{B} = \frac{\mu_{0}}{4\pi} \int \nabla \cdot \left( \mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right) d\tau^{'}. \tag{48}$$ Invoking the product rule number $6$, $$\nabla \cdot \left(\mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right)=\frac{\hat{\mathbf{r}}}{r^2}\cdot(\nabla \times \mathbf{J})-\mathbf{J}\cdot\left(\mathbf{\nabla} \times \frac{\hat{\mathbf{r}}}{r^2}\right)\; .\tag{49} $$ But $\nabla \times \mathbf{J}=0$, because $\mathbf{J}$ doesn't depend on the unprimed variables, while $\mathbf{\nabla} \times \frac{\hat {\mathbf{r}}}{r^2}=0$, so $$\mathbf{\nabla} \cdot \mathbf{B}=0\;.\tag{50}$$  I have the Fourth Edition of Introduction to Electrodynamics by David J. Griffiths, published by Pearson. I have found this erratum in Chapter 5 Magnetostatics, page 234 . It's just a normal triple product manipulation, i.e. $$ \vec a \cdot \left(\vec b \times \vec c \right) = \vec b \cdot \left( \vec a \times \vec c \right) = \vec c \left(\vec b \times \vec a\right)$$ Therefore, I think that an &quot;equal to&quot; sign is missing between the expressions on the RHS of Eq. $(49)$. I searched for  but couldn't find it.
24767
In electromagnetism, we can re-write the electric field in terms of the electric scalar potential, and the magnetic vector potential.  That is:  $E = -\nabla\phi - \frac{\partial A}{\partial t}$, where $A$ is such that $B = \nabla \times A$.  I have an intuitive understanding of $\phi$ as the electric potential, as I am familiar with the formula $F = -\nabla V$, where $V$ is the potential energy.  Therefore since $E = F/q$, it is easy to see how $\phi$ can be interpreted as the electric potential, in the electrostatic case.  I also know that $F = \frac{dp}{dt}$, where $p$ is momentum, and thus this leads me to believe that $A$ should be somehow connected to momentum, maybe like a "potential momentum".  Is there such an intuitive way to understand what $A$ is physically?
79351
I learned vector analysis and multivariate calculus about two years ago and right now I need to brush it up once again. So while trying to wrap my head around different terms and concepts in vector analysis, I came to the concepts of vector differentiation, gradient, divergence, curl, Laplacian etc.   The reference that I'm using is very inadequate to give any geometric/physical interpretetions of these (almost) new concepts. I don't have much problem with their formulae and working rules, but I want to look at them in a little more geometric way. For example, the geometric significance of gradient that I got out of my book is the following: if $f:\mathbb{R}^3\to \mathbb{R}$ be a differentiable function then $\nabla f (x,y,z)$ is the vector perpendicular to the level surface $f(x,y,z)=c$ ($c$ constant) at the point $(x,y,z)$.  I would really appreciate if anyone can explain how, in this way, can divergence, curl and Laplacian be interpreted geometrically. (e.g. For a given vector field $\textbf{F}: \mathbb{R}^3\to \mathbb{R}^3$, the relation between $\nabla\times\textbf{F}$ and $F$ and questions like that.) I looked around Google a bit, but couldn't find what I was looking for.  Thanks in advance.
6787
I can understand how the line integral of a scalar function has pure-mathematical significance, representing an area beneath a curve. But is there a pure significance to the line integral of a vector field? The only understanding I have of this is the physical example of work done on a moving particle. Thanks for any insight.
51264
I'm learning about work in my dynamics class right now. We have defined the work on a particle due to the force field from point A to point B as the curve Integral over the force field from point A to B. From math I know that if a vector field has a potential, we only need to evaluate the potential at point B minus the potential at point A to get the result of the curve Integral. In the text that I'm reading, it's explained that if the integral over a force-field is path-independent, then the force field $F = -{\rm grad}(V)$, where $V$ is the potential. Why is it defined as the negative gradient? Doesn't one determine the potential from $F$ mathematically. Why do we impose the sign on the potential?