# Tag Info

## Hot answers tagged electric-fields

11

What you call "discernibly" moving, is called a non-stationary current-density. Consider a wire with constant charge-flow at every cross section. Then this current is in the same sense not "discernibly" moving as your charge on the sphere. Still there is a magnetic field around the wire. You can think of the charged sphere as infinitely many circular wires ...

5

The reason is the samen as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...

2

Quantum mechanic predicts, that the allowed directions of the spins are quantized. This is one of the main findings of the Stern–Gerlach experiment. In a thermal beam I suppose the the spins to be equally in up and down. (There is no reason why they should not.) But "up" and "down" only correspond to a specific direction in space if there is an external ...

2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...

2

The electron is an elementary particle in the underlying building blocks of matter organized in the elementary particles table of the standard model of particle physics. Elementary particles are point particles. The standard model is a precis of a very large number of measurements (data) fitted by mathematical models of theoretical physics. A point ...

2

Now, from Coulomb's Law we can find a vector for the electric field due to this electron at all points in this space. When you read about Coulomb's law, you can see that it describes the force between two charged particles. When you set them down such that they have no initial relative motion, they move together or apart in a linear fashion. But note that ...

2

Why is this? By convention. If you put the negative (black) lead at GND (or, e.g., battery minus) and you put the positive (red) lead at VDD (or, e.g., battery plus) the reading on the meter is positive. It's telling you how much higher in voltage the red lead is than the black lead. It's a convention.

2

The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is: $$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$ When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging. And I ...

2

Plasma can be created from many substances. More importantly, plasma can have the same density as liquids or solids, or even much more dense. One case widely-studied nowadays is the laser-produced plasma. By focusing an intense laser on matter, the electric field of the laser light can be so strong to ionize atoms very quickly (less than a femtosecond) thus ...

1

Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...

1

In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...

1

You can find the expression for the electric field of a finite line element at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html - which gives for the Z component of the field of a line that extends from z=a to z=b $$E_z = \frac{k\lambda}{z}\left[\frac{b}{\sqrt{b^2+z^2}} + \frac{a}{\sqrt{a^2+z^2}}\right]$$ You can follow the approach in that ...

1

Prove I have taken that line charge is placed Vertically and one test charge is placed. Now the electric field experienced by test charge dude to finite line positive charge. $$Ex = \int dx cos \alpha$$ $Ey$ will be cancel out as they will be opposite to each other. $$Ex = \int k \frac{dq}{x^2+y^2}cos\alpha$$ $$Ex = \int k \frac{\lambda ... 1 Suposse that -Q=ne, where e is the electron charge and n a natural number. So, you can "add" positive charge to an object removing n electrons. Now imagine you have a object with charge 2Q, if you add 3n electrons, the final charge would be -2ne+3n3=(-2+3)ne=ne=-Q. Remember that electrons has negative charge. 1 Yet a voltmeter will provide a positive reading if you put the positive lead at the location with higher potential But this is precisely what one should expect given the quote in your question. Consider a conductor with resistance R, oriented vertically, and with a constant downward electric (conventional) current through. In this case, there is a ... 1 This integral cannot be solved in terms of elementary functions. You can easily do an expansion in \frac{1}{r} in the integrand after doing on of the integrations, then doing the second integral after expanding you get$$ \frac{ab}{r^2}\left(1 - \frac{a^2+b^2}{12 r^2} + \mathcal{O}\left( \frac{1}{r^4}\right)\right) $$If you want to solve the poisson ... 1 Since both you and your teachers are stumped I will give some pointers. Pointer 1 - light is an electromagnetic wave. The energy flow is given by the Poynting vector. In vacuum (or air) this is$$\vec S = \vec E \times \vec H$$Conveniently, for plane waves the time averaged Pointing vector is (see wiki )$$\langle S \rangle = \frac12 \epsilon_0 c E^2$$... 1 You neither have arbitrarily small distances or arbitrarily point-like charges: Usually, even "localized" charges are distributed over an entire ion, and any other molecule cannot come arbitrarily close (without becoming absorbed and hence part of the surface, acting as a buffer to other surrounding material). What matters in practice is not if an ... 1 The wikipedia article on Paschen's law should answer your question quite well. http://en.wikipedia.org/wiki/Paschen%27s_law Basically the breakdown field strength increases at smaller distances. That's why, when you "glue" two pieces of plastic foil together with static charge, you can easily get around 30 MV/m or more in the microscopic air gap that's in ... 1 The surface of the conductor has constant electrostatic potential V and the electric field is proportional to the gradient of the potential: \nabla V. By definition the gradient of a scalar quantity is always perpendicular to the level curves (surfaces). 1 Maxwell's equation tells us that for the general case$$ \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}  This vanishes not only when charges are stationary, but also when they are moving in a uniform continuous manner such as to produce a constant magnetic field (which describes your example). Another general case is when the charges ...

Only top voted, non community-wiki answers of a minimum length are eligible