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## New answers tagged electric-fields

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You are talking about a p-n junction which forms a diode. Now a diode junction is very small compared to the rest of semiconductor. The diode junction can in a first approximation be taken as an insulator since it does not have any mobile charges compared to the rest of the region. When a voltage is applied at the ends the region outside the junction can be ...

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Then does all charges, no matter how much charged are they, ionize the material around them since distance can get smaller and therefore electric field gets bigger. Consequently, the electric field exerted by the charge would be higher than the electrical breakdown limit of the material around it. Let's say you magically stick a (positive) charge inside a ...

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Yes there is an electric field outside of a current carrying wire, in a direction along the wire axis (i.e. parallel to the wire). This is true in both the AC and DC case. There is also of course a magnetic field in the azimuthal direction. For a resistive wire oriented along the z-axis, the electric field inside the wire is given by Ohm's Law $E_y=\eta ... 1 Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where$r_a$is the initial position and$r_b$is the final position (and I have added$q$as the charge of the point charge). so, for example, if the point charge goes from$r$to$2r$we have two positive ... 1 The tangential component due to the locally flat piece of surface, is indeed zero, at the surface. But, the total electric field is the field due to the locally flat patch, plus the rest of the surface. Hence, the total tangential component need not be zero. 0 The explanation for this is partly Gauss' law and partly the nature of metal. Metals have free electrons that move to try to compensate for electric fields. The free electrons near the surface on the inside of the sphere will have an uneven distribution if the point charge is off-centre. These electrons will counteract the effect of the positive charge ... 0 I'm pretty rusty in electrostatics (and everything), so please let me know if I mess up the answer. The first step is to realize that the two conductors can be treated as separate problems. So, we will find the electric field for the two conductors separately, and them sum them together to find the total electric field from the conductors. The second step ... 0 You could try spraying them with an ion wind from a charged needle. If you need an interior rather than surface charge then you will probably need to hit them with a beam of charged particles of enough energy to penetrate to the interior. The easiest would be to charge them negatively using an electron beam or beta emitting radio isotope. How long they would ... 0 You ask how many cubes are required to fill the total volume about a point. Eight cubes. Hence the answer should be the total flux through whole space divided by 8. 2 The Rydberg electron - the electron in the high n level - is highly polarizable and very weakly held. The binding energy of the electron is very small. A small electric field will distort the wavefunction of the Rydberg electron so that it spends more time on one side of the atom than the other. The result is the formation of dipole. Rydberg electrons in ... 1 Take for example the$n = 13$line on the graph, which intercepts the$y$axis at$650\text{cm}^{-1}$. Converting$650\text{cm}^{-1}$to a wavelength by taking the reciprocal and dividing by$100$gives the corresponding wavelength as$\lambda = 15.38\mu\text{m}$, and converting to an energy using$E = hc/\lambda$gives$E = 1.292 \times 10^{-20}\text{J}$or ... 0 hv is the energy per photon, the power radiated is essentially this times the number of photons emitted per second, so a constant frequency antenna with variable power is emitting a variable number of photons per unit time. 1 Force per unit length would be given by $$F=\alpha E.$$ For an infinite line charge the electric field at a distance$d\$ is, by Gauss' Law, $$E=\frac{\alpha}{2\pi \epsilon_{0} d}.$$ The dielectric is made of dipoles, so you should be able to figure out why it makes no difference to the field outside the wire. And the two wires are far enough apart that we ...

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Divergence can be thought of as the flux of a vector field per unit volume. It is positive if there is a net flux out of a small volume and negative if there is a net flux inwards. When you say "its diagram" - of course there are different ways of plotting vector fields. Perhaps the most common way is using field lines. In which case it can be ...

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