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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 ...


4

Now, the constants C1,C2,C3 appearing when we separate variables on Laplace's equation for electrostatic potential has some physical meaning? If they do, what is it? The constants are the related to the square of the spatial (angular) frequency or a spatial growth/decay constant. For an example of spatial frequency, let $$X(x) = A \sin (k_xx) + B ...


3

I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


3

If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


3

The magnetic field is different, and is not the same result. The result using concentric circles actually makes use of the rotational symmetry of the system, but the same symmetry does not hold for the rectangular system. Don't expect a nice formula - these things can be difficult to calculate explicitly! Ampere's law and the Biot-Savart law still hold, but ...


2

The electric potential can be negative. Both in difference and absolutely if you have chosen a gauge. To see that this must be so, just replace the charge distribution (not the test charge, all the others...) with one that has the opposite sign.


2

The "direct" formula is $$V(r)=\frac{1}{4\pi\epsilon_0}\int\frac{dQ}{\lvert \vec{r}-\vec{R} \rvert}=\frac{1}{4\pi\epsilon_0}\iint_{sphere}\frac{\sigma(\vec{R})dS}{\lvert \vec{r}-\vec{R} \rvert}.$$ Now, think carefully about what the $\frac{1}{\lvert \vec{r}-\vec{R} \rvert}$ means---it is the reciprocal of the distance from an arbitrary point on the surface ...


2

To integrate the expression over the area you need to write the area of the surface element (the ring of charge that is a distance $r$ away). If we write the position of a point on that surface in spherical coordinates (rather than (x',y',z')) then a little element of surface becomes $$dA = R \sin\theta d\theta R d\phi$$ which they stated explicitly in the ...


2

Without equations: The ideal dipole is made up of two oppositely charged particles infinitely close to each other. So we can immediately deduce that if the electric field does not change along the direction of the dipole, it exerts no force (because the force it exerts at the positive particle will identically cancel that at the negative one). The only way ...


2

My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...


2

The problem is about the energy needed to assemble the charges, assuming that they already exist. So, imagine that the charges are separated from each other by a very large distance. Then your third expression is effectively zero. As you bring the charges together, the first two expressions don't change at all, and the third expression does. So the work ...


2

There is an intuitive reason why the electric field cancels at the center. Notice that the charge distribution is a cosine function, which means that if you stay at the center and watch at any specific direction, the charge at the opposite direction will have the opposite sign, and will cancel out the net force. This happens for any arbitrary direction of ...


2

The paper strips were ironed to make them flat and easy to stick to the CRT screen. Old CRT color screens used (IIRC) around 25kV to accelerate the electrons to excite the phosphors on the screen itself. This resulted in the buildup of a static charge on the inside of the screen, and a corresponding charge on the outer, with the glass acting as a dielectric. ...


2

You are using the repulsive force as the force acting to move the charge from B to A(which is not actually moving the charge). We need an external force to move the charge from B to A, which will be taken into consideration(to calculate workdone).


2

I don't think so. The whole point about the equivalence principle, is that gravity is indistinguishable from inertia. It is rooted in the fact that gravitational and inertial mass are the same. See this answer. This is not the case for the electromagnetic interaction. Two bodies with different charges but identical masses will not have the same acceleration ...


2

You are confusing yourself. The statement $P - P_0$ would remain the same is false. Why would it remain the same? There is a certain amount of compressed gas inside the bubble, and there is a force that maintains it compressed. In the first case, this force is just the surface tension. In the second case, it is the surface tension reduced by the ...


2

Static comes from the same root as stasis, meaning stop, immovable, To create static electricity, you have to rub two different materials. At the moment you rub them, the electrons already moved Note the word "create", creation is not static, and yes there are transient fields and currents during creation of a static field. The static describes the ...


2

A neutral object can be induced a non-zero charge when placed in an electric field. The charges or dipoles within that material will simply rearrange or rotate to aline slightly. An electric field will be generated, which will counteract the current field. Have a look at dielectrics. The gravitational constant $G$ is... A constant. Just like the ...


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

If the belt is not insulating, any charge on the terminal will just flow back to ground, so you can't build up charge on the terminal and it will not rise in potential. The point of a Van de Graaff generator is to physically move charge against the electrical gradient, and you can't do that if the belt lets it slip away. Now, you can instead use the ...


2

You've stumbled on an interesting idea: how do classical systems that dissipate heat or energy via frictiom arise from quantum systems that perfectly conserve energy in their interactions? Particles in the collision kind of scenario you described don't really exhibit friction. One convenient point is that temperature and heat transfer in quantum physics is ...


2

No, this is not possible. Consider a field which always points in the same direction, and put your $z$ axis in that direction. Your field can then be described as $$\mathbf E=E_z(x,y,z)\hat{\mathbf z}.$$ As an electrostatic field, this must satisfy Gauss's law, which in vacuum reads $$ \nabla\cdot\mathbf E=\frac{\partial E_z}{\partial z}=0, $$ and means ...


1

I don't agree with Ben Crowell. I think that the reason that the moment is directed from negative to positive is because of the definition of moment: $$ \mathbf{p} = \sum\nolimits q*\mathbf{d} $$ q: charge, d:distance from the origin of coordinates to the carge If you have a negative and positive charge, this relation gives you the direction from negative ...


1

http://arxiv.org/abs/1001.3702 : "We give a rigorous computer-assisted proof that the triangular bi-pyramid is the unique configuration of 5 points on the 2-sphere that globally minimizes the Coulomb (1/r) potential. We also prove the same result for the (1/r^2) potential. The main mathematical contribution of the paper is a fairly efficient energy estimate ...


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

Here is an analogy. Let's consider dropping a ball on the earth's surface. We all know that Earth's gravitational field is a vector towards its centre. Only if the ball is dropped from rest, or was a given a downward shove, it's path will be straight(at least till it hits the ground). On the other hand, if you lob the ball horizontally, it's path is curved. ...


1

Your statement implies that the Electric field $\boldsymbol E$ will be parallel, during the whole motion, to the instant velocity $\boldsymbol u(t)$, i.e: $$\boldsymbol E = \boldsymbol E_\parallel + \boldsymbol E_\perp= \boldsymbol E_\parallel.$$ If you have a curved trajectory there must be a component of the force which is perpendicular to the velocity 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

The Earth is negatively charged. As a result anything electrically connected to Earth sharges this negative charge. That includes plants, but also you and I should we be walking barefoot.



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