Why is ${\rm d}\vec{A}_1$ pointing up and ${\rm d}\vec{A}_2$ pointing down?  Electric field ${\rm d}\vec{E}_1$ is pointing up so isn't it supposed to be perpendicular to the surface?
123561
Why does S3 equal to 0 ? for E3 dot dA = 0?  Could I also know why does dA1 and dA2 point up and down? while dA3 point to the right?
98190
specify two vector and two scalar (differential) equations. That implies 8 components in the equations. But between vector fields $\vec{E}=(E_x,E_y,E_z)$ and $\vec{B}=(B_x,B_y,B_z)$, there are only 6 unknowns. So we have 8 equations for 6 unknowns. Why isn't this a problem?  As far as I know, the answer is basically because the equations aren't actually independent but I've never found a clear explanation. Perhaps the right direction is in  on arXiv.  Apologies if this is a repost. I found some discussions on PhysicsForums but no similar question here.
89160
The  is a well-known and very useful tool for solving problems in electrostatics.  Unfortunately, when I was taught this method, it was presented simply as an algorithm. No real physical justification was given for its usage, and there was a complete absence of rigorous mathematics. The method has always been a little hazy to me since.  If someone could give a rigorous first-principles derivation of this method from Maxwell's equations (perhaps simply Coulomb's law?), it would be most appreciated.
45177
I'm having trouble understanding how a  (not the  that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I don't see how this alignment gives rise to "attraction" or force.
99893
Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.    My Attempt: Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}................(1)$ And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).......(2)$ And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).........(3)$. Here $\lambda,\mu,t$ are scalars. I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result. Please help me.Thanks.
100113
Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.  Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.  Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.  We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as  \begin{align*}  \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as  $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2}  = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k} \end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?
121403
Allow me to preface this by stating that I am a high school student interested in physics and self-studying using a variety of resources, both on- and off-line, primarily GSU's HyperPhysics website, Halliday &amp; Resnick's Fundamentals of Physics, Taylor's Classical Mechanics, and ultimately the Feynman lectures (mirrored by Caltech). Hopefully this gives somewhat of a feel of my level of physics understanding so as to avoid any answers that fly far above my head.  As I've understood from previous reading of electromagnetism (for example, in Halliday), a point charge is not affected by its own electromagnetic field. Unfortunately, as I recently read in the Feynman lecture on electromagnetism, :     For those purists who know more (the professors who happen to be reading this), we should add that when we say that () is a complete expression of the knowledge of electrodynamics, we are not being entirely accurate. There was a problem that was not quite solved at the end of the 19th century. When we try to calculate the field from all the charges including the charge itself that we want the field to act on, we get into trouble trying to find the distance, for example, of a charge from itself, and dividing something by that distance, which is zero. The problem of how to handle the part of this field which is generated by the very charge on which we want the field to act is not yet solved today. So we leave it there; we do not have a complete solution to that puzzle yet, and so we shall avoid the puzzle for as long as we can.   At first I figured I must've misunderstood, but upon rereading, it's clear Feynman states the that electromagnetic field due to a point charge does, in fact, influence said charge; I inferred this "self-force" must be somewhat negligible for Halliday to assert otherwise. What stood out to me was that Feynman states this problem had not yet been solved.  I suppose my first major question is simply, has this problem been solved yet? After a bit of research I came across the  which appears to refer exactly to this "problem of self-force". As the article states the formula is entirely in the domain of classical physics and a quick Google search indicates it was derived by Abraham and Lorentz in 1903-4, why is it that Feynman state the problem was still unsolved in 1963? Has it been solved in the classical case but not in QED?   Lastly, despite the Wikipedia article , is this problem of self-force present with other forces (e.g. gravity)? I believe it does state that standard renormalization methods fail in the case of GR and thus the problem is still present classically, though it does mention that non-classical theories of gravity purportedly solve the issue. Why is there not a similar Abraham-Lorentz-esque force possible in GR -- is there an underlying fundamental reason? Due to the relative weakness of gravity, can these self-force effects be ignored safely in practice?  I apologize for the long post size and appreciate any help I can receive. I only hope my post isn't too broad or vague!
97608
I'm trying to understand the derivation of the torque equation $\vec{r} \times \vec{F} = I \alpha$. My textbook derives this easily enough from Newton's 2nd Law for a single point with mass $m$ and  radial distance $r$, with the force applied at the same distance $r$, as (if we drop the vector notation for simplicity) $F=ma=m(r\alpha)$, so $rF=mr^2 \alpha = I\alpha$. (Note, there are both 'ay's and 'alpha's there; they look similar.)  The textbook stops there and concludes that the equation holds for any rotating body. But then I tried this derivation for two points positioned along a rigid, massless rod (which points perpendicular to an axis about which it rotates). If there's a point-mass $m_1$ a distance $r_1$ from the axis of rotation and another point-mass $m_2$ at a distance $r_2$, and the force is applied at (for example) distance $r_2$, I get $F=m_1 a_1 + m_2 a_2 = m_1 r_1 \alpha + m_2 r_2 \alpha$, so  $r_2 F= (m_1 r_1 r_2 + m_2 {r_2}^2 )\alpha$. But $m_1 r_1 r_2 + m_2 {r_2}^2$ isn't the correct value for $I$ here. What am I missing?
45913
I am currently wondering about the famous .  Where does it come from mathematically that when you point with your thumb in the direction of the current, your curved fingers will point in the direction of the $B$-field? In other words, which mathematical operation does it come from? Probably it has to do with some vector product or Stokes theorem application, but I am not quite sure about it.
2949
My professor said that the electric field is zero wherever two equipotential surfaces intersect. I can't come up with a reason why.  He also claimed two equipotential surfaces cannot intersect as that would give two different potentials at the same point. Why can't there just be two different equipotential surfaces with the same potential that intersect or touch?
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?
98357
I am currently learning electromagnetism and I am getting really annoyed using a physical hand to accomplish the right hand rule. I get it, i've seen diagrams, and i've solved problems with it but still somehow I mess up because I bend a finger or twist my wrist a strange way. I am tired of using my hand.  So my question is can I find an entirely mathematical basis for this?  Lets say for any given diagram I define the positive $x$ axis as going to the right, the positive $y$ axis as going up and the positive $z$ axis as going into the page. Then I just did unit vectors for everything to determine the missing component?  As I understand it $\vec{F} = q (\vec{V}\times\vec{B})$ and with that definition I am confident I could find out the direction of the electric field given the velocity of a particle of charge and the magnetic field but how can I re-arrange this so I can solve for other variables? For example what does $\vec{B}$ equal in terms of a cross product involving $\vec{V}$ and $\vec{F}$?
521
I'm stuck on problem 2.10 from Vector Analysis and Cartesian Tensors by Kendall:      Show that the four points with position vectors $\vec{r_1}$, $\vec{r_2}$, $\frac{r_2}{r_1}\vec{r_1}$, $\frac{r_1}{r_2}\vec{r_2}$ , where $r_1\neq0$ and $r_2\neq0$, lie on a circle.   I tried supposing that there exists some vector $\vec{d}$ which gives the position of the circle centre and then trying to prove that the distance from each point to this centre is equal. But I just arrive at the condition that $\hat{r_1}=\hat{r_2}$.   Any suggestions for another strategy, am I just messing something along the way?
13841
What was the improvement that Maxwell did to the  and why?  I understand that he combined the main equations so that you could get a wave equation for the vectors of $\bf E$ and $\bf B$ but is there anything else?
73420
Consider the following parallel plate capacitor made of two plates with equal area $A$ and equal surface charge density $\sigma$:    The electric field due to the positive plate is   $$\frac{\sigma}{\epsilon_0}$$  And the magnitude of the electric field due to the negative plate is the same. These fields will add in between the capacitor giving a net field of:  $$2\frac{\sigma}{\epsilon_0}$$  If we try getting the resultant field using Gauss's Law, enclosing the plate in a Gaussian surface as shown, there is flux only through the face parallel to the positive plate and outside it (since the other face is in the conductor and the electric field skims all other faces).   $$\Phi = \oint \vec{E}\cdot\vec{dA} = EA$$  where $E$ is the electric field between the capacitor plates. From Gauss's Law this is equal to the charge $Q$ on the plates divided by $\epsilon_0$  $$\frac{Q}{\epsilon_0}\implies E = \frac{Q}{A\epsilon_0} = \frac{\sigma}{\epsilon_0}$$  I know there is something fundamentally incorrect in my assumptions or understanding, because I frequently get conflicting results when calculating electric fields using Gauss's Law. I am, however, unsuccessful in identifying this.   Edit: Also, another problem I noticed was that even if we remove the negative plate from the capacitor and then apply Gauss's Law in the same manner, the field still comes out to be $\sigma/\epsilon_0$ which is clearly wrong since the negative plate contributes to the field. So, maybe the problem is in the application of Gauss's Law.
10882
The above question i'm trying for so long but failed. This is from my class $12$ book(india).First of all i'm confused with tangential component of electric field, I mean why this? So far I've studied that electric field(Whether its own or someone's else) is always perpendicular to a charged body. So, there shouldn't be any tangential component. Could someone please explain me visually what's happening?  Strict Note: A class $12$ answer required. No complex statements please. Only for school childs.
114089
I've been trying to derive this (which Feynman warns takes a lot of work) for a couple of days now, without success. My current best derivation which however doesn't give the right answer is:  First, realizing that to go from derivatives with respect to time, $t$, to ones with respect to retarded time, $t'=t - \frac{r}{c}$, we need:  $$\frac{d t'}{d t} = 1 - \frac{\dot{r}}{c} \tag{1}$$  Where $r=|\vec{r}|=|\vec{r}_{1}-\vec{r}_{2}(t')|$ Where $\vec{r}_{1}$ and $\vec{r}_{2}(t')$ are the fixed (time-independent) position vector of the observation point and the retarded position vector of the charge (at time $t'$), respectively. And the dot represents derivation with respect to $t$  The Lienard-Wiechert potentials are:  $$\phi(\vec{r}_{1}, t) = \frac{q}{4\pi\epsilon_{0}(r-\frac{\vec{v}\cdot\vec{r}}{c})}$$ $$\vec{A}(\vec{r}_{1}, t) = \frac{q\vec{v}}{4\pi\epsilon_{0}c^{2}(r-\frac{\vec{v}\cdot\vec{r}}{c})}$$  Where $\vec{v} = \frac{d \vec{r}_2}{d t'}|_{t'=t - \frac{r}{c}}$; that is, the standard retarded velocity.  Now, it is useful to note:  $$\frac{1}{1-\frac{\vec{v}\cdot\vec{r}}{rc}} = \frac{1}{1+\frac{\frac{d r}{ dt'}}{c}} = \frac{1}{1+\frac{\dot{r}}{c-\dot{r}}} = 1-\frac{\dot{r}}{c} \tag{2}$$  Where we have used $(1)$ to transform the time derivative.  Then I rewrite the LW potentials as:  $$\phi(\vec{r}_{1}, t) = \frac{q}{4\pi\epsilon_{0}r}\Big(1 - \frac{\dot{r}}{c}\Big)$$ $$\vec{A}(\vec{r}_{1}, t) = \frac{q\dot{\vec{r}}}{4\pi\epsilon_{0}c^{2}r}$$  Finally, I can work out the electric field:  $$\vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} = \frac{-q}{4\pi\epsilon_{0}}\bigg(\frac{-\vec{r}}{r^{3}}\Big(1 - \frac{\dot{r}}{c}\Big)-\frac{1}{rc}\vec{\nabla}\dot{r}+\Big(\big(1 - \frac{\dot{r}}{c}\big)\frac{1}{r^{2}}\frac{dr}{dt'}-\frac{1}{r}\frac{d}{dt'}\big(1 - \frac{\dot{r}}{c}\big)\Big)\frac{\vec{\nabla}r}{c}+\frac{\ddot{\vec{r}}}{c^{2}r} - \frac{\dot{\vec{r}}\dot{r}}{c^{2}r^{2}}\bigg)$$  Where the spatial gradient is with respect to $\vec{r}_{1}$, and where I've had to derive with respect to $\vec{r}_{1}$ directly and then with respect to $t'$ because it too depends on $\vec{r}_{1}$ through $r$. Now, $\vec{\nabla}\dot{r} = \frac{\partial}{\partial t}(\vec{\nabla}r)=\frac{\partial}{\partial t}(\frac{\vec{r}}{r})$ because these partial derivatives commute. Finally, I can again convert the time derivatives using $(1)$ so:  $$\vec{E} = \frac{q}{4\pi\epsilon_{0}}\bigg(\frac{\vec{r}}{r^{3}}-\frac{\vec{r}\dot{r}}{r^{3}c}+\frac{1}{rc}\frac{\partial}{\partial t}\Big(\frac{\vec{r}}{r}\Big)+\frac{\ddot{\vec{r}}}{c^{2}r} - \frac{\dot{\vec{r}}\dot{r}}{c^{2}r^{2}} - \frac{\vec{r}}{rc}\Big(\frac{\dot{r}}{r^{2}} + \frac{\ddot{r}}{r(c-\dot{r})}\Big)\bigg) = \frac{q}{4\pi\epsilon_{0}} \bigg(\frac{\vec{r}}{r^{3}} + \frac{r}{c} \frac{\partial}{\partial t}\Big(\frac{\vec{r}}{r^{3}}\Big)+\frac{1}{c^{2}}\Big(\frac{\ddot{\vec{r}}}{r} - \frac{\dot{\vec{r}}\dot{r}}{r^{2}}-\frac{\vec{r}\ddot{rc}}{r^{2}(c-\dot{r})}\Big)\bigg)$$  The first two terms are right but the third, although close, isn't right (specially annoying is that $c-\dot{r}$ in the denominator). The actual equation is found in . I've found a  (pages 22-23) that says that the Heaviside-Feynman formula can't actually be derived from the LW potentials, but I don't know, I think I trust Feynman more. Has anyone here done this derivation?