Tag Info

New answers tagged

0

But how can the circuits be of absolutely any shape? Circuits can be any shape because you can bend wires. But don't worry I'll explain how the charged react to the weird shape so as to have a nice flow of current. If I have an extremely long, convoluted wire, how can we analyze voltage as simply as we can? A lumped matter model means you treat the ...


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).


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 ...


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

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

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

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 ...


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 ...


0

In a plasma the particles are very hot. So hot that the atoms move so fast and collide so hard that they kick off electrons from each other and create ions. Molecules are hardly stable at these conditions. At these temperatures everything is no longer solid or liquid. So you can make plasma with everything like water, iron, tungsten, etc, but they will ...


0

A plasma is, by its nature a gas - it has no structure and the constituent particles are not strongly bound to each other. Well, at least it's approximately a gas, and it's certainly not a liquid or a solid, but has been described as "the fourth state of matter" http://pluto.space.swri.edu/image/glossary/plasma.html But it's easy (or at least ...


0

If you move q/2, both end up with charge q/2 and the force will scale with $\frac{q^2}{4}$. If you move 2Q, the product of the charges will be $-2Q^2$ the magnitude of which is obviously bigger (although in the first place the force will be repulsive and in the second case it will be attractive).


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.


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 ...


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 ...


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

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

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 ...


0

The reason being the charge imbalance in the frame of the moving electrons would produce a smaller charge imbalance force Since $$k = \frac{1}{4\pi \epsilon_0}$$ decreasing $k$ is equivalent to increasing permittivity $\epsilon_0$ of free space. Assuming the permeability $\mu_0$ of free space is held constant, decreasing $k$ decreases the invariant ...


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 ...


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.


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 ) $$\langle S \rangle = \frac12 \epsilon_0 c E^2$$ ...


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 ...


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

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 ...


-4

You have a moving charge and a circular moving that is an acceleration for what you need a force. Remember the Lorentz force $ \vec F = q \vec v \times \vec B $. This vector cross product can be rewritten to $$ \vec B = \dfrac{\vec F \times \vec{qv}}{\|\vec{qv}\|^2}$$. Edit: The equation was edited following ...


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 ...


0

The things that induce a magnetic field are a current or a changing electric field (with respect to time). In this situation you have charges moving,so you have a J.dS, which means you have a current. EDIT: If you prefer a different approach, then consider the cylinder to be an infinite amount of circular circuits (each with different radius because of ...


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 ...


0

How can E be constant in magnitude? E depends on F and since F follows an inverse square law therefore the magnitude of E should decrease as we move away? The 1/r² inverse square law only applies for a pointlike source. If you contrive a number of pointlike sources into a column (see the infinite line on Andrew Duffy's physics course), the force diminishes ...


0

So,the integral is equal to zero because the Q that is enclosed in the Gaussian surface is zero.That does not necessarily mean that the electric field is equal to zero.The only other way for the integral to be equal to zero is if the sum of the dot products inside the integral is equal to zero.It means that you have equal negative dot products as positive ...



Top 50 recent answers are included