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There is an alternative version of Ohm's law $\vec J = \sigma \vec E$ where $\vec J$ is the current density (current per unit area), $\sigma$ is the conductivity of the material and $\vec E$ the electric field strength. If $\vec E$ did become zero then no current would flow. If for a fixed $\vec E$ the conductivity increases (resistivity/resistance ...


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I am not a physicist but i am a curious person like you. So i seached for the same answer for so long, finally i got my answer from this Richard Feynman video in which he describes it perfectly for an uneducated mind like mine. http://youtu.be/3D2RaDVkylY


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Comments from @aquirdturtle have led me to rewrite my answer and to realise that it was a question worth asking. @ACuriousMind has likened the situation to a mass falling on the Earth. In that case the mass and Earth system loses gravitational potential energy and they both gain kinetic energy although almost all of it resides with the mass. Carrying on ...


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This is an answer to question 3 relating to simple motors and dynamos. A simple motor and a simple generator are one and the same thing When a current flows through the coil of a motor the coil rotates in a magnetic field and so the coil acts as a generator with the induced electromotive force (emf from Faraday’s Law) in the opposite direction to the ...


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The intensity of the total electric field will just be the sum of the two separate intensities, because the interference term will oscillate at a very fast frequency (namely $\omega_1 - \omega_2$) and its mean value will thus be zero. See this awnser for more details.


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You're correct. And you're not. Depending on how you read the question. There's more than one way to interpret the phrase "the force exerted by q3 on q1." You've interpreted it as just one in a bunch of terms in a net force equation, such that the net force on q1 is q1 times the sum of all the electric fields produced by all the charges in the problem. One ...


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Your approach using Gauss and the integration to find the total charge is correct. The method which just multiplies the charge density as though it was constant (independent of R) by the volume of a sphere is incorrect. The E-field lines will be radial and be in the outward direction if the charge is positive.


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The addition of two simple harmonic motions at right angles to one another produces what is called a Lissajous figure. If search for Lissajous Figures Simulation you will find a number of simulators. Make the x and y frequencies the same and the phase difference 90 degrees and you get your circle produced. I have not found a good one but here are two ...


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The field given by components $E_x = \cos(wt)$ and $E_y = \cos(wt-\pi/2) = \sin (wt)$ is always a rotating field, unless $w= 0$, and it will always have amplitude $$ \sqrt{E_x^2 +E_y^2} = \sqrt{\cos^2(wt)+\sin^2(wt)} = 1. $$ If $w>0$, the field will rotate counterclockwise, whereas if $w < 0$, the field will rotate clockwise. If $w = 0$, the field ...


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There shouldn't be any relationship if you do the modeling properly. Could it be that your simulation is not using fine enough time steps to see it? You need multiple steps per rotation - probably at least 20 - to see the rotation. It is possible that your choice of time step and $\omega$ makes it that the field appears to be "frozen" at a particular ...


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q1 and q2 induce charges of the same magnitude and of the opposite sign on the surface of the spherical cavities. This in turn means that on the outside surface of the sphere there are charges induced of the same magnitude and the same sign as charges q1 and q2. These charges are distributed on the outside surface of the spherical conductor. That is there ...


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In general you're right; if we know $V(0,0, z)$ we can only get its $z$-derivative. But the author is implicitly assuming the electric field is directed along the $z$ axis, because of the rotational symmetry of the problem. Therefore, the $z$ component of the gradient is all we care about.


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How can the position of that charge be determined without EM waves Accelerated charges and Cerenkof radiation can be useful in experiments, but if one really wants the location of charges one needs detectors, i.e. consecutive small interactions. These are mainly electromagnetic scatterings but so soft that they are included in the measurement error. To ...


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In your first sentence you mentioned charges. Electrons, protons as well as their antiparticles have permanent an electric field. This field seems not to be only an induced one during any measurement. The permanent state of the electric field of charges is a postulate because without any interaction a measurement is not possible (see the comments to your ...


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No, this is not true all the times for every region in space. Take this example: If you have two charges with the negative charge. At the middle of the straight line that connects them, the electric field(the force) will be equal to zero because the force from the electric field of each charge will be equal in magnitude but they will have opposite ...


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So how can the same virtual photons give rise to 2 different properties? The photon is an elementary particle, and in the quantum field theoretical framework, an electromagnetic field exists in all (x,y,z,t) which has zero vacuum expectation value unless a photon exists there, the excitation of the field. What is the vacuum expectation value? It is the ...


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Try to start from the end and recapitulate, what was unknown in Maxwell's time. Electrons, neutrons, protons as well as their antiparticles have spin. This spin carries an angular momentum. This we know since Einstein and de Haas carried out their experiment with electrons. Furthermore it is well known, that parallel to the rotational axis of this intrinsic ...


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Wires in a circuit are electrically neutral. They have as many positive charges as negative charges, so the net charge is zero. There is a spatial distribution of charges within a wire. The charge is not uniform across the cross section. But that variation in charge density is fairly small. Given the overall neutrality, you'd only have to move a short ...


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"Why?" questions are always a bit dubious in physics, but let me explain a few words here: "Voltage" is in itself a difference, and it is (by definition) the difference of electric potential energy. The difference in electric potential energy is by definition the work needed to move a test charge against the electric field. In other words: The electric ...


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in the problem that you describe the electric field is indeed conservative as you suggest. We have here a steady-state current, which means that $\text{div} \;\mathbf J = 0$ and therefore, according to the continuity equation $\frac{\partial \rho}{\partial t}+ \mathbf {\nabla} \cdot \mathbf{j}= \sigma$, the charge density is constant in time; this is an ...


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from Gauss' Law electric field inside a solid insulating sphere varies linearly not in conducting sphere. inside a conducting sphere it will be zero.


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The field inside a conducting sphere is zero, not linear in $r$. The field of a uniformly charged sphere goes linearly with $r$.


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The tangent line, in general, is the line that follows the gradient of a curve at a specific point - like in this diagram: The blue line is the tangent of the red line at the point where they touch.



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