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Objects do not have potential energy. Systems of objects have potential energy due to forces of their interaction. We colloquially speak of the potential energy of an object in a gravitational field, but it's really the potential energy of a small mass and a planet. When the small mass falls, the majority of the potential energy changes into the kinetic ...


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If there were a way to remove/transfer the potential energy of a person or object, what would be left? I think you'd be left with gamma radiation and neutrinos. Like Acid Jazz said, you take potential energy out of an object by dropping it. What happens is that some of its mass- energy is converted into kinetic energy, and ends up getting radiated away. ...


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For a diode, no voltage means no current (and vice versa). That is, an operating point of a diode is (0 V, 0 A). So with $I_c=0$, there is no voltage dropped across the diode, so point $A$ is at the same potential as $V_{cc}$


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I think the electric field is zero on A, B, C, D and E, because otherwise there would be current, which would be odd And you are totally right for an electrostatic system (with no current). Instead of explaining it by, "this would be odd", let's have a look at what happens in the instant you add the wire to the battery pole.: Before the wire touches ...


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The potential, $\phi$, is defined as the the potential energy, $U$, of a system of charges, $Q$ and a unit charge. The potential energy of a system of charges is defined to be the negative of the work done by the electric field of one charge, $Q$, on another charge $q$, as $q$ is moved from infinitely far away to some distance $r$ from $Q$. If $q$ is a unit ...


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It is a convention that field lines point from positive to negative. Now because they point from positive to negative we get $$E = -{dV\over dr}$$ (in spherical symmetry). If you have a positive charge at the origin then the potential at the origin is positive and it decreases as you move away from the origin - thus $dV \over dr$ is negative because ...


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Indeed they can. This can even be seen in a classical set of equations if the interaction between the two polarizable objects is strong enough. Take a simple 1d problem with two polarizable point particles in a line. For low polarizations, we assume a linear relation: $$ p_1 = \alpha E(r_1) = \alpha (E_2 + E_{ext}) = \alpha (k p_2 + E_{ext})$$ $$ p_2 = ...


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Consider a plane of charge. For distances from the plane which are small compared to the size of the plane the electric field will be constant. In the capacitor, with two oppositely-charged plates and relatively small gap, the electric field is a constant, depending only on the overlapping plate area and the amount of charge, $E_{gap}=Q/(k\epsilon_o A)$, ...


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Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...


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I think your problem is that you're adding the electric fields like scalars, rather than breaking then into their vector components. You have: $ E_1 = \frac{q}{2\pi\varepsilon_0\rho_1} $ and $ E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2} $ What you should have is $ E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_{1x}}\hat{x} + \frac{1}{\rho_{1y}}\hat{y})$ and ...


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The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...


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Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...


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You have made two elementary errors. First, you forgot a negative sign and second, you forgot your limits (V at infinity is conventionally taken to be zero) $V = -\int_{\infty}^{r} E dr$ Since I believe you are taking the electric field at the x axis, instead of r you can use x, making that part of your work correct. These two things will fix your ...


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...why isn't the work done against the net force due to the system considered instead of simply adding up the work done against separate forces caused by individual charges? They're both equivalent, due to the principle of superposition. Basically, the net force is what you get when you add up the separate forces from the individual charges acting on ...


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If I understand correctly, your situation looks like this: The medium surrounding the electrodes is water - a conductor (but not a very good one). This means that a current will flow - more current will flow where the electric field is strongest. Now the conducting ring in the middle will make sure that the potential anywhere in the grey region is the ...


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One simple explanation is that the work you calculate does not depend on the path along which you move the charge. Or, in other words, if you move a charge in a closed path, the work done by the electric field is zero. See, $A=\oint q\vec E\cdot d\vec r=qQ(1/r - 1/r)=0$


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What the voltmeter is reading in the top circuit would be, if you used a Kirchhoff's loop rule in the loop containing 2 resistors and the voltmeter, the difference in the potential difference across the top-right resistor and the bottom-right resistor. The loop rule calculation would look something like this ($A,B,C,\&D$ are the top-left, top-right, ...


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I will presume that, as is standard and conventional, the electric potential at a place an infinite distance from the one in consideration is zero. Let the charge at point a be charge $a$, the charge at point b be charge $b$ and the charge at point c be charge $c$. With this, the formula for the electric potential is: $V = ...


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I'm assuming you mean a solid metallic sphere, not a shell. The analysis is slightly different for a shell but rests on the same principles. It is actually not correct that the total charge induced on the sphere in the case of grounding is $-q$. The basic concepts you need to know about electrostatics with conductors is that the entirety of the conductor ...


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You have a sign error. The potential is defined by see: http://en.wikipedia.org/wiki/Electric_potential


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The quantity that you can measure in the lab is the force experienced by a test charge (in reality sensor of some kind). This force is obtained via lorentz force that depends on the electric field. So the physically measurable quantity is the Electric field and not the potential. Remember that $ \vec{E} =-\nabla \phi $ so the electric field decreases with r, ...


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It is easiest if you consider a rectangular box in 3D, with length $L_x$ in the $x-y$ plane, and length $L_z \ll L_x$ in the $z$ direction. Now, $L_x$ is very large, and you are only interested in the physics deep within the bulk, far from the boundaries. Therefore it is of little importance exactly which boundary conditions you choose. It is often easier to ...


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On the one hand, you can't solve for the magnetic field without appropriate boundary conditions (e.g. there could always be an incoming electromagnetic wave that hasn't yet impinged on your cylinder). On the other hand if you have a fixed charge and current distribution you can always use Jefimenko's equations to find a solution to Maxwell's equations, and ...


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we know, $$E=V/d$$ (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html ) Where $V$,$E$ and $d$ are potential difference ,electric field and distance between the two plates respectively. Electric field between the plates only depends on charges of the plates and since charges must be conserved so when the plates are moved apart charges(the ...


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The potential at the surface of a charged sphere or radius $r$ is: $$ V = k\frac{Q}{r} \tag{1} $$ Since the area of the sphere is $4\pi r^2$, the charge density is: $$ \rho = \frac{Q}{4\pi r^2} $$ and rearranging gives: $$ Q = \rho 4\pi r^2 $$ and substituting for $Q$ in equation (1) we get:$$ V = 4\pi k\rho r $$ Can you take it from here?


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Because they should have the same electric potential and electric potentials of them depends on charge not charge density.


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If it's a converging series (which is in this case), then yes. According to uniqueness theorem if it satisfies the boundary then it's a complete field.


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I'm going to have to give an answer that's very different to Jimmy360's. Apologies. How does the potential and kinetic energy of a photon relate? They don't. The photon is all kinetic energy. Do they mean the same thing? No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic ...


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Quote from a webpage a bit over my head :-) greatians.com Photon has linear momentum. Photon travels in vacuum space at the ultimate speed of light. Photon has the quantized energy of hf as given by eq. WD.1.2. E = hf … eq. WD.1.1 where h = Plank’s constant ...


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Imagine a beam of light, going towards a massive object. It has potential energy in the gravitational field. Of course, the potential energy has to become kinetic energy. This is done by shifting frequency. The energy of a photon is given by $E = hf$ so to increase kinetic energy we must increase frequency. If the beam of light was red, it will be a higher ...


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Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ...


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A=qU Based on work-energy theorem: A=(mv^2)/2 (mv^2)/2 = qU mv^2=2qU mv=(2qU)/v and then you can find the de Broglie wavelength by :


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Basically, the system is simply that which is studied in a problem in physics. It refers to that which we want to know more about, in this case the moving electric charge in the presence of the electric field. Be cautious with the terms 'electric potential' and 'potential energy', since they're two different things. Electric potential is defined as ...


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You are right that the charge gained potential energy. But this statement is only true because the charge is part of a system. We cannot talk about the electrical potential of a charge unless it is in an electric field - which means that there is "something else" that is essential for our definition of the potential. We say the "system" (the charge, plus ...


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For a classical point charge, the field is divergent at $r=0$, and if you were to take the potential to be zero there, it would be infinite everywhere else. Meanwhile, you can approximate $r=\infty$ as the region with no interaction, so it's reasonably naturally to treat it in the way you would treat ground.


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You evidently understand that any constant can be added to a potential without affecting the physics -- or equivalently, any place can be taken to have zero potential. You also suggest, rightly, that there are really only two "natural" places to define the zero of the potential: either $r=\infty$ or $r=0$. For example, there's no particular reason to ...


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This is accurate, and it ultimately comes down to the fact that we can get arbitrarily close to an electric point charge in classical E&M. That means that the field right up next to the point charge could be arbitrarily large. So you get these huge, singular potentials close to point charges, which is really more-or-less fine. For instance, that huge ...


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You will get the potential outside from the method of images. However, the potential inside is obtained from the solution of the Laplace equation inside a cylinder with the given boundary condition. (Hint: the solution is immediate.)


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The key here is that E must exceed the minimum of $V(x)$. If one has a delta function well then the minimum is $V_0 = -\infty$ while for a delta function barrier the minimum is $V_0 = 0$ Hence for the delta function well, one can have bound states with $E<0$ or one can have scattering states with $E>0$. For the delta function barrier the necessary ...


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This is often referred to as the "step voltage". The image below illustrates the electric field. (it's caused by a wire and not by a lightning, but you get the idea) from here: http://commons.wikimedia.org/wiki/File:Potenzialtrichter_Freileitung.svg The problem of touching two points of the ground is that there can be a difference in the potential of the ...


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They are under the electric influence of the terminals. If they negatively charged, it will be attracted to the positive terminal.


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It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...


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Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...



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