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Uggh. I hate this kind of problem because of all the different components you have to deal with. But I'll just start it, like you asked. The electric field of a charge is defined as $E=\frac{F}{q}$. Now we know that $F=k\frac{q_1q_2}{r^2}$, so $E=k\frac{q_1}{r^2}$. A test chargeanywhere in this space has to contend with four electric fields: ...


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You're reasoning from a faulty premise. The voltage difference depends on the strength of the electric field between the two points. If and only if the electric field strength increases proportionally with the charge will your reasoning be correct. But, in the two examples you give, this isn't the case. For example, when you connect the two identical ...


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Earth doesn't suck the electrons. The electrons in the negatively charged body repel each other and push electrons out of the body into Earth until both are at the same potential. It's like two balloons at different pressure connected with a tube and a valve. When you open the valve the lower pressure balloon doesn't want the higher pressure gas. It repels ...


1

Just for the sake of having an answer: In $\phi \nabla ^2 \phi$, consider the term $\phi \frac{\partial ^2 \phi}{\partial x^2}$. By the product rule, this is equal to $\frac{\partial}{\partial x} \left(\phi \frac{\partial \phi}{\partial x}\right) - \left(\frac{\partial \phi}{\partial x}\right)^2$. Combining all three terms, we get $\phi \nabla^2 \phi = ...


1

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 ...


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There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy Ep as an associated potential field such as $\mathbf{F}=-\frac{\partial Ep}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(Ep)$. That's no more than a definition of the potential energy. ...


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electric field strength is $$E=\frac Fq=\frac Vd$$ with $V$=voltage, $d$=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but $Fd$=energy $$\therefore {\rm energy}=qV$$


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The polarisation field can be imagined as due to lots of tiny dipoles within the medium, that are created in response to an applied electric field. The dipoles reduce the electric field within the medium (at least for materials with a dielectric susceptibility >0). Imagine the situation where you have a charged, plane-parallel capacitor, with a dielectric ...


2

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


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

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

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 ...


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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Take a infintesimally small element 'dl' ,now the potential due to this element is dv=1/4piεo λdl/r^2 and r =sqrt(z^2 +R^2) ,(which is const) And thus integrate 'dv' And since 'r' and all others in the equation are const except 'l' thus integrating 'dl' will give 2piR (problem solved)


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This question is using spherical coordinates, and in spherical coordinates the divergence of a vector (like $\nabla\cdot \textbf{E}$) is: $\nabla\cdot \textbf{A}=\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2 A_r \right) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta} \left( A_\theta\sin\theta \right) + \frac{1}{r\sin\theta}\frac{\partial ...


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The electric field given in the problem only depends on the radius $r$. Therefore, the divergence is most conveniently computed in spherical coordinates. The factors of $r^2$ and $1/r^2$ come from the transformation of $\nabla$ to spherical coordinates, see here. For a function $f(r)\hat{r}$ the divergence is $$\nabla\cdot ...


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Look at the expression for the divergence in the spherical coordinates.


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The ring can be parameterized by ${\bf r}' = R \left(\cos \theta \ \hat{\bf x} + \sin \theta \ \hat{\bf y}\right)$. Noting that $$ \left|d{\bf r}'\right| = R d\theta \left|\left(-\sin \theta \ \hat{\bf x} + \cos \theta \ \hat{\bf y}\right)\right| = R d\theta $$ and $$ {\bf r} - {\bf r'} = \left(x-R\cos\theta\right) \ \hat{\bf x} + \left(y-R\sin\theta\right) ...


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

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 ...


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

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 ...


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On the one hand, we let our reference voltage be at infinity. On the other, we say the sphere is grounded. "Grounded" could mean two things: The potential of the sphere is defined to be the reference. This contradicts the previous definition of the reference to be at infinity. The sphere is connected by an ideal conductor to an infinite source of charge ...


0

You are confusing work on a closed loop, with an integral on a closed surface. What is true is that for eletrostatics, we have $$\oint_C\mathbf{E}\cdot d\mathbf{l}=0,$$ where $C$ is a closed curve, which is Ampère's law. What Gauss law states is that the electric flux over a closed surface is equal to the charge enclosed by it.


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

$\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, ...


0

The geometry is not the same, they are complementary. The case where $\beta \approx0$ describes a very deep corner, so the space not occupied by the conductor is very narrow. The case where $\beta \approx 2 \pi$ is the case where the conductor is almost a thin sheet, that is, the space occupied by the conductor is very narrow. As expected, the charge ...


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 \\ ...


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 ...


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 ...


0

1.You are or more precisely the graph is considering the distance between the spheres ,where $ \vec E$ for both conducting and non conducting sphere is same.So the graph is same irrespective of whether its conducting or not.So the first answer is wrong. 2.As from the graph,at the midpoint the field is 0,it is clear that both have the same charge of same ...


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

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

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.


0

The answer to the main question is YES. Two electrons will "touch" each other when their centers are at a separation equal to one electron diameter. Since the diameter of an electron is not zero, an infinite amount of energy, is not required to make them "touch." With a (calculated) electron diameter = to $2.82 \times 10^{-15}$ m, the required energy can be ...


0

How does a parallel plate capacitor emit a constant electric field between its plates? Remember that the parallel plate capacitor is an idealization. The rules for it really only apply in the limit as the transverse dimensions of the plates go to infinity. in my book it is given that outside the parallel plate capacitor the electric field is 0, but ...


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 ...


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 ...


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.


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 ...


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


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 ...


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

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 ...


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 ...


0

The magnitude of the forces $q_1$ and $q_2$ exert on each other is equal. According to Coulomb's law, the magnitude of the force that a charge $q_1$ will experience due to a charge $q_2$ is $$|\mathbf F_{12}|=k_e{|q_1q_2|\over r^2}\ ,$$ where $k_e$ is Coulomb's constant and $r$ is the distance between the charges. But that equation is symmetrical in $q_1$ ...


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This equation will always give you a volume charge density. One way to see this is that surface charge density and volume charge density have different units - $\mathrm{C/m^2}$ and $\mathrm{C/m^3}$ respectively - and in order for the units to be consistent, $\rho$ has to be the latter. The fact that the equation is written with $\rho$ is a helpful reminder ...


0

As this is a conductor, the charges are on the surface. The statement that $d$ is small indicates that you are expected to model this as an infinite plane surface. You can compute the electric field just outside the conductor by taking $\frac {d\varphi}{dx}$ The electric field inside the conductor is zero because it is a conductor. You should have a ...


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 ...


0

From Griffith section 2.4.4 comments on Electrostatic Energy, you can get your answer. If you consider point charges, then actually, this integral is related with self-energy which is infinite at usual, to make finite we often introduce cutoff radius $\delta$. (In particle physics, we often use bare and renormalized terminology, renormalization is a some ...



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