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The charged sphere induces surface charge density on the conductor. This is necessary because field lines are always perpendicular to the surface of the conductor. This induced surface charge density modifies electric potential in the region. It will no longer be $V(d)=Q/4\pi\epsilon_0.d$. If you know the potential at the conductor(which I think is necessary ...


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If the belt is not insulating, any charge on the terminal will just flow back to ground, so you can't build up charge on the terminal and it will not rise in potential. The point of a Van de Graaff generator is to physically move charge against the electrical gradient, and you can't do that if the belt lets it slip away. Now, you can instead use the ...


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The formula for electric force is given by $F = k \dfrac{q_1q_2}{r^2}$ Where $q_1$ and $q_2$ are the charges, and $r$ is the distance. Therefore, there is no lower limit. It is always given by this formula.


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If the charges are kept at a fixed distance R, the force will be given by: $$ F = \frac{kQ_1Q_2}{r^2} $$ The smallest possible charge that can exist freely is that of an electron or proton which is numerically equal to $1.6 \times 10^{-19}$ coulomb, put the values in the equation and you get the answer.


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There is no range of electric force values, F = k*q/r^2 It will be fixed unless you having a mechanism that makes the charges to vary as required.


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Usually, applying Gauss's law to a problem with $$ \int_A \vec{E} \cdot d\vec{A} \propto Q$$ is only suitable, if one knows, that the electric field is perpendicular to the surface $A$ and is constant in magnitude over the whole surface. This leaves: $$ E \propto \frac{Q}{A}$$ On can conclude such statements if the problem is symmetric (e.g. spherical ...


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It is not that we 'need' to use an infinite wire or plane. For example, I am sure there must be problems in the N.C.E.R.T. textbooks where you use a finite charged wire to calculate the electric field at a given point. The point of using infinite wire or plane is that once you know the perpendicular distance of the point from the wire or the plane of the ...


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The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is: $$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$ When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging. And I ...


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This integral cannot be solved in terms of elementary functions. You can easily do an expansion in $\frac{1}{r}$ in the integrand after doing on of the integrations, then doing the second integral after expanding you get $$ \frac{ab}{r^2}\left(1 - \frac{a^2+b^2}{12 r^2} + \mathcal{O}\left( \frac{1}{r^4}\right)\right) $$ If you want to solve the poisson ...


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Something like that. In metal shel, charge distribute in the surface. So in this case, when you conect the center sphere with the wire, all the charges goes to the surface of the outer sphere. Then, the charge in the surface of the outer sphere is 2Q-Q=Q.


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If some charge is given to a conductor then its potential will be remain same through out the region, because work done on every charge is same.


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Now, the constants C1,C2,C3 appearing when we separate variables on Laplace's equation for electrostatic potential has some physical meaning? If they do, what is it? The constants are the related to the square of the spatial (angular) frequency or a spatial growth/decay constant. For an example of spatial frequency, let $$X(x) = A \sin (k_xx) + B ...


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In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two ...


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If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


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This is just the line integral which gives you the electric potential in two dimensions due to a charge distribution of one or more closed loops, i.e. closed charged wires. The logarithm is coming from the solution of laplace equation in two dimensions, replacing the $\frac{1}{|\vec{x} - \vec{x}'|}$ of three dimensions, and $f(w)$ is just the charge density ...


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You are confusing yourself. The statement $P - P_0$ would remain the same is false. Why would it remain the same? There is a certain amount of compressed gas inside the bubble, and there is a force that maintains it compressed. In the first case, this force is just the surface tension. In the second case, it is the surface tension reduced by the ...


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My textbook says: because the electric potential must be a continuous function. But why? Since, in the electrostatic case, $$\vec E(\vec r) = -\nabla V(\vec r)$$ then if the electric field is to be finite everywhere, $V(\vec r)$ must be continuous. Put less rigorously, the electric field would be 'infinite' wherever $V(\vec r)$ is discontinuous. ...


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Imagine you have a point charge inside the conducting sphere. Obviously, since the electric field inside the sphere is zero (as you state), there is no force on the charge, so no work done. Therefore the potential is constant. So far so good. Now as we approach the boundary, we can imagine moving an infinitesimal amount to go from $r = R - \delta r$ to $r = ...


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It should be seen more like: A stationary charge generates an electromagnetic field A moving charge generates an electromagnetic field An accelerating charge generates an electromagnetic field So the hierarchy/pattern you mentioned isn't really much of a hierarchy/pattern after all. But actually, you can show that only the second derivative enters the ...


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I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


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I don't think so. The whole point about the equivalence principle, is that gravity is indistinguishable from inertia. It is rooted in the fact that gravitational and inertial mass are the same. See this answer. This is not the case for the electromagnetic interaction. Two bodies with different charges but identical masses will not have the same acceleration ...


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What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+ $$ The force on the negatively charged particle is opposite the direction of the field from the positively ...


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The electric field lines show the direction of the electric force acting on a unit positive charge at a particular point in space. So, therefore, the force acting on a negative charge, as is in your question, will act in the opposite direction shown by the electric field lines. I believe that the fact that field lines are defined in terms of a unit positive ...


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Electron has less mass and therefore it start accelerating first therefore it is considered that dipole moment from negative to positive when placed in electric field


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A neutral object can be induced a non-zero charge when placed in an electric field. The charges or dipoles within that material will simply rearrange or rotate to aline slightly. An electric field will be generated, which will counteract the current field. Have a look at dielectrics. The gravitational constant $G$ is... A constant. Just like the ...


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The electron is relatively close to the uniformly charged sheet. You can simplify the problem to a electron in the homogeneous field of a capacitor consisting of two square plates with surface $A=0.25 m^2$. The force to a electron $e$ between the plates of the capacitor is $F=eE$, where $E = \frac{Q}{\varepsilon_0 \varepsilon_r A}$ is the strength of the ...


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Static comes from the same root as stasis, meaning stop, immovable, To create static electricity, you have to rub two different materials. At the moment you rub them, the electrons already moved Note the word "create", creation is not static, and yes there are transient fields and currents during creation of a static field. The static describes the ...


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Consider a sphere full of a pressurised gas. This is a spherically symmetric system, but if you cut the sphere along the equator the pressure of the gas would make the top half fly up and the bottom half fly down. So there is a net upwards force on the top half, just like in the charged sphere. The spherical symmetry just means the net force on the top half ...


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This is a very classical question in college physics, sometimes it can occur in high school physics level. If you use Faraday's law of induction, you should get the answer easily. Faraday's law is given as: $$ \oint_{\partial \Sigma} \vec{E} \cdot d\vec{l} = -\frac{d}{dt}\int\int_{\Sigma} \vec{B} \cdot d\vec{S} $$ First of all, choose the a circular contour ...


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Take for example, the case when the charge Q at the centre is negative, so the force is attractive. If it gets displaced even by a small value $\delta r$ towards either, the separation with this particular vertex would be $(\frac{l}{\sqrt 2}-\delta r)$, while it is slightly larger for the others. Due to the inverse dependence of the force on this ...


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Consider the square to be in the x-y plane. Then see what happens when you displace the central charge in the z-direction. Of course, the answer will depend upon whether Q is positive or negative and possibly also on the direction of displacement.


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Flux in two dimensions into a curve $\mathcal{C}$ should be $$ \Phi=\int_{\mathcal{C}} \vec{E}\cdot\hat{n}\, d\ell $$ Where $\hat{n}$ is the unit normal vector to the curve. This for instance is consistent with gauss's law $\Phi=\oint_{\mathcal{C}} \vec{E}\cdot\hat{n}\, d\ell=q_{in}/\epsilon$, when you notice that poisson equation in two dimensions for a ...


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When using Gauss's law to calculate the flux through a closed surface we take the field component in the normal direction of the surface. The normal direction always points outward for the closed surface. Going by the math, it seems like you're making a Gaussian cylinder that encloses one of the plates. The difference between making a cylinder that ...


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To integrate the expression over the area you need to write the area of the surface element (the ring of charge that is a distance $r$ away). If we write the position of a point on that surface in spherical coordinates (rather than (x',y',z')) then a little element of surface becomes $$dA = R \sin\theta d\theta R d\phi$$ which they stated explicitly in the ...


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The "direct" formula is $$V(r)=\frac{1}{4\pi\epsilon_0}\int\frac{dQ}{\lvert \vec{r}-\vec{R} \rvert}=\frac{1}{4\pi\epsilon_0}\iint_{sphere}\frac{\sigma(\vec{R})dS}{\lvert \vec{r}-\vec{R} \rvert}.$$ Now, think carefully about what the $\frac{1}{\lvert \vec{r}-\vec{R} \rvert}$ means---it is the reciprocal of the distance from an arbitrary point on the surface ...


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I suspect that the Earnshaw's theorem might make such motion unstable. You may consider a similar problem: a positive charge between two positive charges. All your reasoning seems applicable there, but it is intuitively obvious that the motion will be unstable.


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The metal plate, being a good conductor, will have its electrons rearrange in such a way as to neutralise the electric field inside the plate. The electrons would tend to bunch up in the plate at the point(s) closest to each of the two charges, alterring the electric field that they're exposed to and changing the electrostatic force on them. Strictly ...


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The charge distribution when the charge is only on the surface is given by: $$ \varrho(r,\varphi,\theta) = \frac{Q}{4\cdot\pi\cdot r^2} \cdot \delta(R - r)$$ where $Q$ is the total amount of charge on the sphere with radius $R$ and $\delta(R - r)$ is the Dirac-Delta. Now use $$ \varphi(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\varrho(\vec{r}\ ...


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The area element for a sphere of radius $R$ in spherical coordinates is $$dA = R^2 \sin\theta\ d\theta \ d\phi$$ and $dq = \sigma\ dA$. where $\sigma$ is the surface charge density. Your differential E-field element is $$d\vec{E} = \frac{k\ dq\ \hat{r}}{r^2}$$ where $\hat{r}=\vec{r}/r$ is a function of the angles. You will need to determine the vector ...


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will this cause a change in the electrostatic force acting on one charge due to another? ans is no. The two charges will induce some charges on the metal plate which will ofcourse change the electric field but according to principle of superposition force on one charge due to another charge is not affected by the presence some others charges near.(I meant ...


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The wikipedia article on Paschen's law should answer your question quite well. http://en.wikipedia.org/wiki/Paschen%27s_law Basically the breakdown field strength increases at smaller distances. That's why, when you "glue" two pieces of plastic foil together with static charge, you can easily get around 30 MV/m or more in the microscopic air gap that's in ...


3

The magnetic field is different, and is not the same result. The result using concentric circles actually makes use of the rotational symmetry of the system, but the same symmetry does not hold for the rectangular system. Don't expect a nice formula - these things can be difficult to calculate explicitly! Ampere's law and the Biot-Savart law still hold, but ...


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You neither have arbitrarily small distances or arbitrarily point-like charges: Usually, even "localized" charges are distributed over an entire ion, and any other molecule cannot come arbitrarily close (without becoming absorbed and hence part of the surface, acting as a buffer to other surrounding material). What matters in practice is not if an ...


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The electric potential can be negative. Both in difference and absolutely if you have chosen a gauge. To see that this must be so, just replace the charge distribution (not the test charge, all the others...) with one that has the opposite sign.


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Strictly talking about electrostatic conditions(not referring to induction): Coulomb's law has a constant in it Ke(k electrostatic) which is equal to 1/4πε where ε is the electric permittivity of the medium that the electric field is in.More electric flux exists in a medium with a low permittivity. So,this is the difference in your equation between the two ...


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I believe you are confusing the work done by the electric field with the work done on the particle. By the electric field: The radial force of the electric field is always pointing outwards, and the displacement of the charge in this case is going inward. Thus, the integral you've specified will be negative. That is, the work done by the electric field is ...


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The relationship between electric field and electric potential is just that the electric field is (minus) the gradient of the potential. Thus in the case of a uniform field extending from a uniformly charged plate (let's call it along the z-axis, with the late in the x,y plane) $$ E_z = - \frac{dV}{dz}$$. This of course means that $$ V(z) = - \int E_z\ dz = ...


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You can change the sign of your angle, or you can swap a and b, but you can't do both. When you swap a and b the angle gets increased by 180 degrees (which is the same as changing the sign of your angle), if you do this and change the sign of the angle (again) the two cancel each other out.


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Every differential equation problem has two parts: the differential equation itself, and the boundary condition. Neither can tell you the other in and of itself. Consider $y'' = -k^2 y$. Is the solution sine or cosine? The answer depends on boundary conditions, and I have to specify them. I can't get them by specifying anything in the equation. So, say your ...



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