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4

In this case, the image method can be used to calculate the potential (and hence the electric field) in the region $z>0$, with a negative charge $-q$ located at $(0,0,-d)$, since the potential would be $V(x,y,0)=0$, in this case. But for points in the region $z<0$, the potential is given by the solution of Lapalace equation $\nabla^2 V=0$, with ...


4

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...


3

I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up. The gravitational potential a distance $r$ from a mass $M$ is $$ \phi_g=-\frac{GM}{r}, $$ the gravitational field is $$ {\bf g} = - \nabla \phi_g, $$ and the gravitational potential energy (of two masses $M$ and $m$ separated by a distance ...


3

A charge radiates every time is accelerated. The power radiated is given by the Larmor formula. Putting this into the introductions to the motion of a charge in electromagnetic fields would be a meaningless complication, as much as considering air friction. But yes, a charge in a magnetic field would not spin indefinitely, even in vacuum.


3

$\oint E\cdot dS = \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV $ if $\frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV$ then $\frac{\rho}{\epsilon} = \nabla \cdot E$ if $\rho=0$ then $\frac{\rho}{\epsilon} = 0 = \nabla \cdot E$ Is this what your looking for? $\rho$ would be zero say, ...


3

Given that $\rho$ is the charge density, the integral, $$\frac{1}{\epsilon_0}\iiint_{V} \rho\, dV = \frac{Q}{\epsilon_0}$$ Now, Gauss' law states that, $$\iint_{\partial V} E \, dS = \frac{Q}{\epsilon_0}$$ Hence, we arrive at your 'global form' by simply equating: $$\iint_{\partial V} E \, dS = \frac{1}{\epsilon_0}\iiint_{V} \rho\, dV$$ By the ...


3

The definition of current density is $J = \frac{I}{A}$, or more precisely, $J = \frac{\mathrm{d}^2 I}{\mathrm{d}^2 A}$. It is always true, by definition. $J=\sigma E$ is a different equation: it's equivalent to Ohm's law, which you know better as $V = IR$. Ohm's law is not universal; it only works for certain materials, called ohmic materials. For an ohmic ...


2

That's probably for charged solid sphere, not a cylinder. In any case, setting the potential at infinity as zero, we have for $r>R$: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'\implies V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$ For $r<R$, we got: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'=-\int_\infty^RE(r')dr'-\int_R^rE(r')dr' \implies \\ ...


2

If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


2

The electron in the $n$ semiconductor and the hole in the $p$ type semiconductor are delocalised and not bound to any particular atom, so arguments based on completing octets aren't useful. This diagram shows roughly how the depletion layer forms: In the $n$ type semiconductor the doping creates donor states in the band gap, and electrons from these ...


2

We have $$\tan \alpha=\frac{a-x}{b}\implies\sec^2 \alpha \,d\alpha=-\frac{dx}{b}\implies\frac{r^2}{b^2}d\alpha=-\frac{dx}{b}\implies rd\alpha=-\cos\alpha\,dx,$$ which is the desired relation with a minus sign, that must be present, since an increase of $x$ decreases $\alpha$.


2

Perhaps looking at it like this will help clarify things: You have two triangles, one with angle $\alpha$ and one with angle $\alpha + \mathrm{d}\alpha$. The sides opposite those angles differ by $\mathrm{d}x$, but the hypotenuses are essentially the same, both equal to $r$. Write an equation expressing the fact that the triangles' opposite sides differ ...


2

Here is your picture, with a couple of additional angles and segments drawn: Do you see it now?


2

You probably did some wrong calculation, because your reasoning works. Take a circle of radius $r$ a distance $a$ above the charge. The increase $d\Phi$ of the flux by increasing the radius by $dr$ is given by $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{2\pi r \cos \theta dr}{a^2+r^2}=\frac{q}{2\epsilon_0}\frac{ar dr}{(a^2+r^2)^{3/2}},$$ where $\theta$ is the ...


2

Your reasoning is correct, it's just a lot harder to do with the surfaces you've chosen. Draw a small, elemental ring at some arbitrary height above the charge. A line from any point on the ring to the charge subtends the polar angle $\theta$ with the z-axis and is a radial distance $r$ from it. (i.e. $r$ and $\theta$ are the usual spherical polar ...


2

First, one of the implications of the electrostatic potential satisfying Laplace's equation is that the extremes are at the boundaries. If the potential of the surface of the sphere is zero and the potential at infinity is zero, the only solution for the potential outside the sphere is the trivial solution, i.e., the potential is zero everywhere outside the ...


2

Here's a simple way of looking at it: If you are close to an infinite plane, you may be feeling stronger attraction by every individual part of it, but "more" of those parts are pulling you at a significant angle. This way, a lot of the attraction is canceling out. As it happens (this is anything but coincidence though), these two opposite effects exactly ...


2

The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ...


2

It's not the case. In the second, the electron will radiate. This is how light species lose energy, and cool in Penning traps, and one of the factors that limit the energies of particles in circular particle accelerators. For a reference see: http://en.wikipedia.org/wiki/Cyclotron_radiation


1

In every negative acceleration electron loos energy, and this of course in the form of photons. This is not surprising because negative acceleration could be only after positive acceleration, the electron has to move befor he could be stopped or declined. And how the electron can be accelerated? By electric fields where the electron get the kinetic energy ...


1

Low voltage sources don't have enough potential to conduct through your skin or body so touching either the positive or negative doesn't make a difference. For you to feel it or get tissue damage, current must flow through you. This won't happen with very low voltage sources unless you're covered with something more conductive like wet salt water. But 12V ...


1

An electric dipole is some configuration of charge, namely two opposite charges separated by a distance. However, such a description is an effective, quite general description for various physical situations. For example, Carbon monoxide $CO$ can be described as a dipole. Relative to the oxygen-side of the bond there is slightly more negative charge located ...


1

Well, the electric field $\vec E$ is different from the force field $\vec F$ a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too $$ q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W$$ where I use the letter $W$ in order not to have confusing notation. The ...


1

Your body is a detector of electric current, in what you describe. Static electricity coming from clothes should not have such high current levels. Check with a voltmeter, it is safer . I suspect that even those small screw drivers for checking live wires will light up. There must be a leakage.


1

Please see the following diagram: This is the picture you need to draw - two spheres, slightly displaced. You can now see how the angle $\theta$ is defined (the dashed lines are supposed to go through the center of the system, midway between $C_1$ and $C_2$. It doesn't quite look like that...). When the distance between the centers is very small, you can ...


1

If the two plates were not connected together by a conducting material, like the metal strip, then this would not work as a capacitor. My guess is that the metal strip symbolizes the conducting wires / lanes in a curcuit. They are simply trying to make the question sound more non-curcuit-like by calling it "two plates connected by a metal strip".


1

Keeping it simple: You can think $J = I/A$ as one definition of the $J$. Since the current $I$ is related with the eletric field, then the $J$ must depend from the electric field as well. As a first approximation we guess a linear relationship between $J$ and $E$. Naturally it is not a linear relationship, but works well as approximation in some cases ...


1

One way of doing is parametrizating a ring. For instance, this is a ring: $\gamma(t) = R(\cos t, \sin t)$. Actually it is a circunference of radius $R$. It is charged with linear density $\lambda$. The potential in the point $P(0, 0, z)$ is $$dV = \frac{kdq}{r} = \frac{k\lambda d\gamma}{r}$$ where $r$ is the distance between an element of charge $dq$ and ...


1

A single Maxwell (for instance) BCAP0350 2.7v ultra capacitor that's about the size of a D cell has a capacity of 1300 Joules (1.3 x 10^3 J). It is extremely useful to use ultracaps to charge batteries if the nature of the power source is intermittent and high current (say, at 35 to 175 Amps, also within spec of the one I listed). Charge the ultracaps ...


1

First think of initially putting charges $\pm Q$ uniformly on the two plates and then letting the system arrive at the final configuration in isolation. If fringe effects were ignored, the initial uniform charge distribution on the plates will remain unchanged. In this case, the potential difference between the two plates will be $V_\infty=Q/C_\infty$, where ...



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