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There are many many misconceptions tied into knots in your question. Firstly, the electric force between two charges doesn't depend on just the distance between them and their charges, it also depends on their velocity. When charge A moves then its electric field is different, so the electric force charge B feels is different. This means you can't reason ...


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What you're talking about is a part of the free electron theory in solid state physics. I believe you have to consult a solid state physics book, like Kittel's or Aschcroft's. A good start can be in Wikipedia at the Screening effect.


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When we bring a test charge let us say (+q) to a certain point, we exert some force. Our exerted force has to be equal or greater to the force exerted by the electric field in order to overcome it. Thus the forces being opposite and equal cancel each other and hence the direction is undetermined. Since it is a compulsion for vector to have both magnitude and ...


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It might help you to redraw the arrangement? Draw vertically a stack of 4 metal plates with dielectric in between but without any connections between the plates. Now redraw the diagram but change the middle metal plates into two plates connected by a wire like a capital $I$ so it now looks like that you have 3 capacitors in series. Now make the ...


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http://physics.stackexchange.com/a/65392/101895 Brilliantly explained there, relate the contracting effect with classical sound Doppler effect if you could. Again, only force acting on a particle due to another particle is solely coulomb's force.The most important thing is that concept of exchange of particles is supposed to create the force {basically, a ...


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Yes. If you have two positively charged plates of different charge magnitudes and fire a stream of electrons between them, the electrons will deflect toward the "more positive" plate. Though really the electron stream isn't exactly necessary, since the two plates will repel each other indicating a potential difference right off the bat.


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If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +9, then there will be a potential difference. If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +1, then there won't be a potential difference.


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There are no consequences. It's a convention that negative is for attraction forces and positive is for repulsion. This leads to forces tending to reduce the potential when attracting. You remove that sign, then attraction forces because positive, and forces will tend to increase the potential. The worst that would happen is a few inconsistencies within ...


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The gradient of a function $\nabla V$ is as you know a vector and this vector points to the (locally) steepest ascent. i.e. if you are on the hill the vector $\nabla V$ points to the tip of the hill and when you are on top of the hill the vector vanishes because there is no "steepest ascent" at that point. You can prove this fact mathematically but I'll omit ...


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The gradient of a scalar function always points in the direction of greatest ascent. That is, a scalar function has the most rapid increase in that direction. If you imagine walking around a mountain, the gradient of the height function of the mountain at each point will point towards the steepest direction. On the other hand, forces in a conservative field ...


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Take a simple 1D example. We'll plot gravitational potential energy as a function of horizontal position for the side of a hill: We normally define $F = -\nabla V$ and that means the force points downhill i.e. if we let the particle go it will roll downhill. If instead we defined $F = \nabla V$ the force would point uphill i.e. if we let the particle go ...


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One way of looking at the change is that you are changing the definition of potential from The potential at a point is the work done on a unit mass in going from an arbitrarily chosen zero of potential to the point to The potential at a point is the work done by a unit mass in going from an arbitrarily chosen zero of potential to the point So ...


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This the same as changing $V$ to $-V$. For example, if $V$ described a potential well then changing it to $-V$ would describe a bump.


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There is a potential difference due to an electric field that is (to a good approximation) contained only between the plates. The battery builds up this potential difference by driving a current. Because the capacitor does not allow charge to flow through it, as current flows, one plate is left with a deficit of electrons and the other with a surplus of ...


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The diagram is not well drawn. The charges reside on the other side of each plate. Thus as a reasonable approximation the E-field only resides between the plates. When a battery is connected across the capacitor charges flow and charge the capacitor until the voltage across the plates of the capacitor is equal to the voltage of the battery. When you connect ...


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Wear a tinfoil hat covered in conducting needles. These will dissipate any charge buildup on your body.


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There is no easy, obvious way to avoid building up a static charge. Carrying around an ion generator to get rid of excess charge is not exactly practical. A static shock hurts because the discharge current passes through a very small area of skin at the point of discharge. The classic way to avoid a painful shock is to hold a key by the flat area and touch ...


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It's your clothing. You generate the static as you walk, etc. Changing the materials, or using an anti-static treatment will reduce the static. Static is worse in very dry weather.


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The connection between electric potential and E-field strength can be conceptually difficult usually because some important ideas are forgotten. The first thing is the definition of potential. Potential is defined as the work done by an external force in taking unit positive charge from a arbitrarily chosen zero of potential to the point. For convenience I ...


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In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


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Does the electric field at a point on the surface change? Yes, it change. Where distance is decreased Electric field increase. Does the total flux through the Gaussian surface change? No, it remains same. Consider a spherical gaussian surface of radius $2m$ and a charge $q$ inside it. Let this charge emit $10$ field lines. Then, clearly, all the field ...


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-ve Sign which comes during integration is due to opposite directional motion of test charge. You can't consider it twice, once during integration and next during antiparallel motion. Hint: Why do you think - ve sign comes during integration? Since you know Electric field was positive, How integrating it could be -ve.


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Atoms are electrically neutral. Because of this they shouldn't attract or repel each other - but atoms do show a slight attraction, which is the reason most molecules form. This is called the residual electromagnetic interaction. In short, the positive parts of one atom attract the negative parts of the other, and vice versa. There is a good little diagram ...


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"So resistivity of a substance is defined as the voltage provided over the current flowing" Well you seem to be confusing Resistance with Resistivity. While the former depends upon the physical attributes of the substance, the later one is the intrinsic property of the substance. Resistance is defined by Ohm's law $V=IR$. $R$ is simply the required ...


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In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...


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Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct. First consider no cavity Potential at center of sphere due to uniformly charged complete sphere $ V = 3kq/2a$ Now, potential due to positive charged sphere $cavity$ at center. $$ V_{1}= \frac{4\rho π ...


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What are the factors that effect the strength of an electric field? Suppose I put 5 bulbs in a single circuit, the resistivity will be high, so less current flow, but what happens to the field? Will the charges not get constant energy due to the field? The factors vary depending on the source of non-EM electromotive force in the circuit. For example, ...


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First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


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At the beginning let us talk about water. Water consists of molecules. There are two buckets, one of them is on the level 0 meters and one is on the level 1 meter. I fill the high-level-bucket with water and connect the two buckets with a small pipe. It takes 10 sec and all the water is in the low-level-bucket. How I have to relate the molecules properties ...


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The other answers are ok in the sense that the nylon carpets often used in planes form a very good surface to get charged up with static electricity as you shuffle across it. However, they miss the absolutely crucial point that the air on an aeroplane at high altitude is partly recirculated but partly taken in (and heated) from the environment, which is ...


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Lets take your language.Charges lose their energy as they move in resistor. But they don't lose all their energy and have enough energy to come back to positive terminal. Even if they lose all their energy the electrons are attracted by positive terminal and gain energy again. If you will apply infinite bulbs their resistance would be so high that no ...


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The motion of the charges is totally unlike you setting off from home walking 10 km and feeling a little tired and slowing down and then walking another 10 km and feeling even more tired and slowing down and then walking a final 10 km and arriving home at a very slow pace, indeed just reaching your front door and stopping. Here is a simple model of what ...


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Assuming incandescent bulbs rather than, say, LEDs: they won't take "almost all" the energy, they'll take all of it. It doesn't matter if you use one bulb or five, it'll use up all of the energy that the battery is supplying. That's why multiple bulbs will be dimmer than one single bulb. Electrons collectively don't have a very good sense of how much energy ...


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The charge is on the "perpendicular bisector" of the dipole - that means the diagram looks something like this: Can you see it now?


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Ok, I think I figured it out. We are so used to assuming that a shell can be collapsed to a point charge, that it is easy to miss the subtlety that this is only true for the regular Coulomb law. As soon as we deform the Coulomb law as given above, we get corrections. If you do all the integrals over the charge distribution you will get the following fields ...


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Your general analysis is correct except the fact that you expect to exist a surface density just because there is a charge volume density. One way to see this is the following. In your uniform sphere consider a shell with small thickness. What is the amount of charge in this shell? If this shell has radius $r$ it is $\Delta Q=4\pi r^2 \Delta r \rho$ (volume ...


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No, atoms have the same number of protons and electrons so they have no net charge. On the other hand ions (cations and anions) would be repelled or attracted depending on their net charge. Atoms are bound together in a molecules by different means like covalent bonding, ionic bonding (which can be easily explained in terms of electrostatic forces) or ...


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Consider a more general case in which the three charges are $\lambda q, q$, and $q$, with $\lambda\geq 0$. Let $A$, $B$, and $C$ be the respective positions of the charges. By symmetry, we have $|AB|=|AC|$. Due to the conservation of momentum, the center of mass $O$ of the system is stationary. Let $\theta$ be the angle $OBC$, and $\alpha$ be the angle ...


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This is actually a particular case of the three-body problem (https://en.wikipedia.org/wiki/Three-body_problem), but with repulsion, rather than attraction. There are very few exactly solvable cases of the three-body problem, among them - the Lagrange's case, where the three bodies are at the vertexes of an equilateral triangle at each moment. Therefore, I ...


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It might help you to think about the symmetry of the situation. First in the application of conservation of momentum and then what the trajectories of the charges must be to keep the centre of mass $C$ at the same position. This will give you a connection between $v_y$ and $v_x$.


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You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what ...


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There many configurations that statisfy your assumptions. But you had forgot about many other constraints like potential difference, charge density, end effects etc. When you consider PD you come to know, ignoring end effects, that only one configuration is possible. Since 1st plate has charge Q its surface charge density of 1st side is Q/2A 2nd side again ...


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In classical electrodynamics, assuming a point charge to be having a finite charge, the net electrostatic self energy carried by it is given by $$ Self Energy = 1/2 \int E^2 dV$$ Upon performing the intergral in three dimensions, since the electric field of a point charge diverges at the origin, therefore the rest mass by the virtue of the electrostatic ...


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When we say proton flow it means H+ flow and by charge flow it may be H+ or any other ion.


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Current constituted by positive ions doesn't imply that the current is constituted by protons. The positive charge might also be constituted by ions, or holes in semiconductors. If you are referring specifically to a circuit made up of solid material, then protons are tightly bound and hence don't flow and positive charge if any, is due to holes. Protons ...


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Look if you substitute the boundary conditions in the given solution it will be identically satisfied, so you don't need them since those were used in the derivation of the given solution. So you have found the charge density and this should be satisfactory, though the final result should be: $$ \rho=-\frac{\rho_0}{\epsilon_0}\cos\beta x , $$ as pointed out ...


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The direction is correct, just note that you can write $$ \frac{\partial^2\phi}{\partial x^2} = - \beta^2\phi $$ and $$ \frac{\partial^2\phi}{\partial y^2} = \beta^2\cdot\left( \phi -\frac{\rho_0}{\epsilon_0\beta^2}\cos{\beta x}\right) $$ This can ease the calculations. The result you obtain is an oscillating potential in $x$, not depending on $y$. The ...


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In a copper wire there are copper ions with 28 bound electrons orbiting the nucleus. Approximately one electron per copper atom is free to roam throughout the metal and these electrons are called unbound or free electrons ans are responsible for electrical (and heat) conduction in copper. The positive copper ions are bound together and only vibrate about a ...


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Positive charges are massive in relation to negative charges, so, what actually flow are negative charge carriers such as electrons or holes. But due convention, we say that the current flows from positive to negative



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