# Tag Info

## Hot answers tagged electric-fields

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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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Yes, in the sense that the charge carriers in the superconductor will experience a force when there's a changing flux.

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Physicists have a slightly different (and in my opinion incorrect) way of deriving nodal circuit diagrams than engineers, which would make his statement correct because the circuit diagram would imply that the wires form loops of a finite area, but this way of thinking breaks the usefulness of circuit diagrams in fundamentally undesirable ways. If you ...

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OK, in this video you've kindly provided me, Lewin essentially talks about this circuit: The easy rule of thumb that's common to all electrical engineers is to say: a current $I$ goes through this loop, causing a voltage drop $R I$ across the resistor, a voltage drop $\int_0^t dt~I(t)/C$ across the capacitor (assuming it is uncharged at $t=0$), and a ...

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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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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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"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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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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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 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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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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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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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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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 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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