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Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


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Classical electrodynamics generally makes a continuity approximation for bulk materials. We're not interested on variation in the fields at the scale of the distances between atoms, so we just average them away.1 With that approximation, the conduction electrons are acted on by the mean field. They also contribute to the mean field but we treat the two bits ...


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Wouldn't the field line end at the charge +q, and so the field lines that enter don't necessarily leave, making +2q contribute a negative flux? The quantity of field lines that terminate on the charge -q is unchanged by the presence of charge elsewhere. Imagine -q field lines terminate on the -q charge. Then, the net electric flux outward through ...


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What we know is that the electrostatic force does not work nearby the nucleus. The interaction between the positive and the negative charged particles stops at some distance, where Quantum Mechanics describes it. These distances for small orbits are well studied. To calculate them is much more difficult, for the hydrogen atom you have the formula, for all ...


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Before we talk about the term "spherical shell" and "thin sphere", let us talk about the possible cases of the sphere itself. Generally speaking, it depends on what is given to you. If you have been given $\rho _s$, then probably it will be hallow. If you have been given $\rho _v$, then definitely it will be solid. Keep in mind that whether it was ...


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I have personally also contemplated this issue, and have come up with a simple solution that is satisfactory, to me at least. I'm sure this can also be found in many textbooks. In general, we have $$\tag{$\star$} Q=\int \rho\ d\tau$$ because we are considering a three dimensional space. Intuitively, we feel it should be possible to talk about a three ...


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As we know [...] in any frame of reference, its radius is a finite number Er ... do we know that? Relativity--as we understand it--does not allow for fundamental particles of finite size, and there is no experimental hint at all of any size associated with the electron. Pointedly the "classical electron radius" is about $2.8 \times 10^{-15} ...


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Given a volume with finite charge density $\rho$, the surface charge density $\sigma$ of an embedded surface will be infinitesimal. The charge of the volume is the integral of the infinitesimal charges of the embedded surfaces. Conversely, a finite surface charge density would give you an infinite charge density there - specifically a delta function which, ...


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I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...


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The net flux crossing the Gaussian surface (according to Gauss's law) is irrespective of those charges outside it. In some detail about your case: The total flux leaving the surface drawn is -q (so it is actually entering the surface not leaving it). So although the 2q charge outside the surface do contribute to the flux crossing the surface, the net ...


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Wikipedia> Electric charge: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric flux: In electromagnetism, electric flux is the rate of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a ...


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In a conducting sphere, like charges repel each other - so when there is a net charge, it will all appear on the surface (they try to get as far away from each other as possible). In a non-conducting material, charge will stay wherever you put it - so if you have a solid material with a net charge per unit mass (not sure how you achieved that), it will not ...



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