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6

Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to: $$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$ The solution to the last equation is: $$ \phi = ...


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 ...


5

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 ...


4

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 difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


2

You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't. In the case of a switched AC outlet, the switch will be ...


2

Basically, the system is simply that which is studied in a problem in physics. It refers to that which we want to know more about, in this case the moving electric charge in the presence of the electric field. Be cautious with the terms 'electric potential' and 'potential energy', since they're two different things. Electric potential is defined as ...


2

Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ...


2

Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small ...


2

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 ...


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

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

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

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 ...


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

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

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 ...


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

It is not that we 'need' to use an infinite wire or plane. For example, I am sure there must be problems in the N.C.E.R.T. textbooks where you use a finite charged wire to calculate the electric field at a given point. The point of using infinite wire or plane is that once you know the perpendicular distance of the point from the wire or the plane of the ...


1

If the charges are kept at a fixed distance R, the force will be given by: $$ F = \frac{kQ_1Q_2}{r^2} $$ The smallest possible charge that can exist freely is that of an electron or proton which is numerically equal to $1.6 \times 10^{-19}$ coulomb, put the values in the equation and you get the answer.


1

The charged sphere induces surface charge density on the conductor. This is necessary because field lines are always perpendicular to the surface of the conductor. This induced surface charge density modifies electric potential in the region. It will no longer be $V(d)=Q/4\pi\epsilon_0.d$. If you know the potential at the conductor(which I think is necessary ...


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.


1

The problem is translationally symmetric along one spatial dimension, which means that it is effectively a two-dimensional problem. You have a weird understanding of what "the dimension of a potential" is - there's really no such concept. But if you define the dimension $n$ by matching a radially symmetric $\phi$ to either $v(r)=b\log r +c$ when $n=2$ or ...


1

In this question the force F has to be calculated using coulomb's law.In coulomb's law it clearly states that both the 02 bodies should be charged.So in this question one will gain charge up to 6 uC and the other will be zero.So there will be no force between them. The question has not mentioned whether the two balls are insulators or conductors.It should ...


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

You are right that the charge gained potential energy. But this statement is only true because the charge is part of a system. We cannot talk about the electrical potential of a charge unless it is in an electric field - which means that there is "something else" that is essential for our definition of the potential. We say the "system" (the charge, plus ...



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