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Hot answers tagged electric-field

14

1) yes, it basically will find a non-optimal solution. At every point, the top of the ray looks for the bigger potential gradient, the charge in the surrounding volume grows, polarizing surrounding material (air, in this case) until a bigger gradient shows up and the ray continues over that direction. This is why the lightining path looks like a jigsaw; its ...

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Electric current, by definition, is a flow of charged particles. When someone says it is the propagation of the electric field, usually he means the following: The velocity of the electrons in the wires is very slow (few cm/s if I remember it right), but when one turn on the light he doesn't see any delay. The lamp starts lighting when the electrons start ...

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James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential $\Phi$ relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is: $$\Phi_{inner} = \Phi \,q\, ... 9 Here are from wikipedia drawings of the field lines of two magnets in two orientations, like-like, like-unlike . North pole to north pole North pole to south pole. The lines distort but do not intersect. These field lines are solutions of the formal Maxwell differential equations. Differential equations do not give discontinuous solutions, as ... 8 Be careful here. Gauss's law tells you that the flux through the (whole) closed surface is proportional to the enclosed charge and therefore zero in this case. That's one fact. The second fact is that you have a constant electric field in this region of space, and that means that the flux through the circular end-cap (which is not a closed surface) is ... 7 Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the ... 7 There's an old and clichéd but actually pretty good analogy for understanding electric circuits, and that's to think of the circuit as water plowing through pipes. If you have a pipe full of water, and you turn on the tap at one end, water immediately starts flowing out of the other end. Well not quite immediately: when you turn on the tap you raise the ... 7 Seriously? Electromagnetic waves are neutral; electrons are charged. Electromagnetic waves have spin one; electrons have spin one half. Electromagnetic waves have lepton number zero; electrons have lepton number one. Electromagnetic waves have invariant mass zero; electrons have non-zero invariant mass. All of these things are observables. 6 No, the statement is false even in the electric case. At the very beginning, the acceleration is \vec a \sim \vec E so they have the same direction at t=0: the tangents agree. However, as soon as the particle reaches some nonzero velocity \vec v \neq 0, its acceleration is still \vec a\sim \vec E, in the direction of the field lines, however its ... 6 Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates. First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the z-direction. I will follow the notation of the Wiki article:$$ x=a\cosh\mu\cos\nu\cos\phi\\ y=a\cosh\mu\cos\nu\sin\phi\\ z=a\sinh\mu\sin\nu $$where ... 5 I'll give you the derivation from my book which includes a nice way to see how the delta functions arise: .............................................................................................................................................................. We can derive the potential field \vec{A} and the electromagnetic fields \vec{E} and ... 5 You tell us that one surface of the box is at B, but you're a little vague on where the opposite face is. You do say that your surface is "between the two sheets", so I think you may mean that the surface is entirely contained in the space between the two sheets. The box does not intersect any charged surface. With that, and a uniform electric field in ... 5 Gauss' law is applicable for a finite wire. But, it's useless in this case. In the infinite example, you assume some things due to symmetry, namely: It's pretty obvious why these things can be assumed--moving up and down the wire should not change \vec E, so we take it constant. Also, there should be no direction bias, so \vec E has no component ... 5 Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force \mathbf{F}=q\mathbf{E} on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. ... 5 If you followed the arguments carefully and checked what is demonstrably right and what is not, you would agree that what the argument actually does is to prove that a uniform electric charge density cannot have a uniform electric field. Your original task was to solve Maxwell's equations (well, Gauss's law), so if you find out that the equations aren't ... 5 The wikipedia article is quite good on this subject. For any discharge in the air the molecules of the air must be ionized. This ionization happens during thunderstorms because of the high static electric fields carried by the clouds which generate "streamers", i.e. paths for the electrons to flow downwards. Corresponding streamers are formed by conductors ... 5 In Classic Electrodynamics it's well known that an accelerated charge will radiate energy and the radiated power is given by the Larmor Formula$$P=\frac{\mu_0q^2a^2}{6\pi c},$$in SI units, where c is the speed of light, q is the charge of the particle and \mu_0 the magnetic constant. Well, to incorporate the Larmor formula into Newtonian mechanics ... 5 The exact solution is$${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$Clearly the field inside the conductor (that is, for r<R) vanishes. Here Q is the total charge on the disk. The field, for large values of r, looks ... 5 It's not intuition.It's a problem which can be solved. First we identify the sign of the charges. By seeing the direction of field lines we can see that the sign of charges. Field lines originate from +ve and end at -ve charges. Next by Definition of Flux, The number of field lines cutting per unit surface surface . And Gauss' Law The flux ... 5 The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire). This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have ... 4 The OP wants an intuitive answer to an intuitive obstacle to seeing its truth. Well, the intensity of the flux is like how many lines we draw per unit area. No one line « loses strength » so to speak. (There is no dissipation, no friction.) If it is a point source, the lines are not parallel, they diverge, and the greater distance between the lines leads ... 4 It is indeed correct that only the difference between two potential energies is physically meaningful. An in-depth explanation follows. For the rest of this answer, forget everything you know about potential energy. I suppose you know that when you have a conservative force \vec{F} acting on an object to move it from an initial point \vec{x}_i to a ... 4 No, there is nothing special about your right hand as compared to your left one. (Well, there might be, if you're a baseball player or a fiance, but there's nothing in classical electromagnetism that makes it special.) If you set up two wires next to each other and run current through them, they will attract if the current runs the same way and repel if ... 4 Conductors are defined by the freedom of some of the charges inside to more with little resistance. So, if there were a non-zero field, what happens? Answer: some of the free charges move until the field is again zero. You might be wondering if there are limits to this claim, but a introductory book of that sort is not worrying about extreme situations. ... 4 Yes, your friend is right. Within electrostatics, an electric field \vec{E} should be curl-free \vec{\nabla} \times\vec{E}= \vec{0}. The drawn electric field lines looks like the electric field is of the form$$ E_x=E_x(y), \qquad E_y=0, \qquad E_z=0, $$cf. the rule that to depict the magnitude |\vec{E}|, a selection of field lines is drawn such ... 4 (as per Chris White's suggestion) The diagram is confusing. It is drawing two sets of field lines: one set due to plate A (as if plate B didn't exist) and another due to plate B (as if plate A didn't exist). It is not showing the total field. This doesn't represent the total field if both plates are present! The electric field is a vector field \vec{E}: ... 4 The field strength is the negative first derivative of the potential. For example, in Cartesian coordinates, with electric potential V = V(x,y,z), the electric field is:$$ \boldsymbol E =-\nabla V= -\frac{\partial V}{\partial x}\boldsymbol i - \frac{\partial V}{\partial y}\boldsymbol j - \frac{\partial V}{\partial z}\boldsymbol k  Note that you can ...

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