# Tag Info

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You are right, the vector decomposition doesn't care about the medium. You decompose your vector into unit length basis vectors in a particular coordinate system, e.g in the Cartesian system, $$\mathbf{v} = c_1\mathbf{i} + c_2\mathbf{j} + c_3\mathbf{k}$$ This decomposition was done with no reference to any medium or its transformations. In this way you ...

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First, when you found your volume it looks like you missed a sign. In my last integration over $\phi$, I had $$\frac{4\pi}{3}\int^{\phi}_{0}d\phi R^3\sin\phi-\frac{d^3\sin\phi}{8\cos^3\phi}\Rightarrow$$ $$\frac{4\pi}{3}\left[ -R^3\cos\phi - \frac{d^3}{16\cos^2\phi}\right]_0^{\cos\phi=\frac{d}{2R}}.$$ If you switch the sign on the second term, it looks ...

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It will help if you study this diagram of what a vacuum tube is If a cathode is heated, it is found that electrons from the cathode become increasingly active and as the temperature increases they can actually leave the cathode and enter the surrounding space. When an electron leaves the cathode it leaves behind a positive charge, equal but ...

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The field lines either has to be attracted if chargers are opposite or repulsed if chargers are same. Hence can never cross

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If you take a point in the horizontal line in the middle,then the potential caused by the upper left charge will be the same as the lower left charge but with opposite signs.The same goes for the two charges on the right.Just take a random point on that line and let a be the distance from upper or lower left charges to the point and b the distance from the ...

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Technically the resistance of a wire is never 0. Typical wires of copper have electrical conductivity of $10^{7} S/m$ at room temperature. So there is indeed a very slight potential drop but highly negligible compared to the one that would be experience at a proper resistor. As for the problem of electrostatic, indeed from a single charged particle the ...

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A single charged object is sufficient to produce an electric field. Following Coulomb's law: $\textbf{E} = {Q \over 4\pi \epsilon_0\textbf{r}^2}\hat{\textbf{r}}$ where $\textbf{E}$ is the vector electric field, Q the charge of the object in question, $\epsilon_0$ the permittivity of vacuum or the electric constant, $\textbf{r}$ the vector position of the ...

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There are also electric fields outside of a real capacitor as well, any capacitor with finite-sized plates. The energy in a capacitor is stored in the electric field, and since some of the electric field is outside the plates, some of the energy is also already outside the plates already. Imagine a bunch of surfaces everywhere in space that are orthogonal ...

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At equilibrium, the field inside an ideal conductor is zero. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c2 A charge moving through such a conductor neither gains nor loses energy. We can't attach an ideal conductor to an ideal voltage source. Something has to give. There will be a voltage drop along a real wire due to non-zero ...

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In a parallel plate capacitor there's accumulation of electrons on one side and lack of them on the other. Since one plate is in front of another, the fields on each are equal in magnitude and opposite and therefore the field lines are straight (away from the boundaries) and cancel. In a battery the field is chemically produced inside the structure of the ...

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An electric charge has an electrostatic potential associated with it. This potential is a scalar, so it has no direction associated with it, and its value at any distance $r$ from the charge is simply: $$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$ If we graph the potential created by the charge as a 2D plot then we get something like this: (picture ...

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The question asks for direction of force "at this instant" so we need to look at the force on the particle when it has the velocity as given (no need to figure out what happens after that). Now the electrical force is $$\vec{F_e} = q\vec{E}$$ so the electrical force is always in the X direction (along $\vec{E}$). The magnetic force is perpendicular to ...

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First of all, I think you might have written the equation for the net force incorrectly: $$\vec{F}_{net} = q\vec{E} + q(\vec{v} \times \vec{B})$$ The second term is $q(\vec{v} \times \vec{B})$ and not $q(\vec{E} \times \vec{B})$. From the first term of the force equation ($q\vec{E}$), we can see that the electric field will try push the proton parallel to ...

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Maybe in a first very rough approach you could consider the metal sphere as being an oscillating charge hence radiating fields like a Hertzian dipole or an antenna. It will indeed induce currents inside the sphere in addition to the already existing from the sinusoidal field generated by the capacitor. By the way the field created by this kind of hertzian ...

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You're close: we can write the "operator" $p.\nabla$ as the "scalar" $$p.\nabla = p_x{\partial\over\partial x}+p_y{\partial\over\partial y}+p_z{\partial\over\partial z}$$ and $$E = E_x i + E_y j + E_z k = (E_x,E_y,E_z)^T$$ Then the result will be the vector $$\left(p_x{\partial E_x\over\partial x}+p_y{\partial E_x\over\partial y}+p_z{\partial ... 1 It depends on how you define the potential. In mechanics one has the convention$$ F = -\nabla U$$Since the electric field exerts a force via F=qE it is only natural to apply this convention in electrostatics too. In this way the electric potential V can be directly interpreted as mechanical potential energy U=qV. Option number (2) is therefor the ... 2 Coulomb's law says$$ \vec{E(\vec{r})} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\vec{r}') (\vec{r} - \vec{r}') \, d^3 r'}{|\vec{r} - \vec{r}'|^3} .$$The charge density is a constant \lambda and is linear. We'll fix that the wire runs along the Y axis and we're calculating the field along the X axis. This being the case, let us fixe ... 0 If your question is "why is the electric force of a static electric field conservative", it might help you to ask yourself the question "what would happen if the work required to move a charged particle from r_1 to r_2 would depend on the particluar path the particle travels?". If this would be true, there would be (at least) two different paths between ... -1 You can use the superposition principle, in general this is more difficult as that requires you to know the charge distribution on all the objects. Note that this is not always as simple as it may seem. E.g. you say "uncharged due to attraction between two plates", but this is not the correct reason, attraction alone would not yield this, it's the inverse ... 0 Note that the Gauss law comes from the the Maxwell Eq. DivE=constant X charge density. When you place a charge in a dielectric the polarization so developed gives a bound charge opposite to that of that embedded. The total charge therefore decreases. The electric flux will therefore decrease since the total charge will now give lesser E. In other words the E ... 0 Electric flux is the surface integral of the electric field along or over a given surface . By Gauss's law , it depends upon the charge enclosed inside the surface . However, I guess what you mean by electric flux is not the total flux through the surface but just the flux through one of it's surfaces . Let us consider a slab made up of a material having a ... 1 For a Gaussian surface between the two plates, the total flux through the surface is zero. For the particular surface you give, all of the electric field lines crossing one of the circular faces cross the other face but in an opposite sense. This is because the outward normal vector for one face of the cylinder is opposite in direction to the outward ... 1 I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at:$$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$I'm not sure if you then tried a u-substitution or just found an ... 0 Why is that? Where am I getting wrong? The proper approach is to note that the integrand for the x component is an odd function of x. But first, recall that$$\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = \int_a^0 f(-x)dx + \int_{0}^a f(x)dx$$Now, for an odd function,$$f(-x) = -f(x)$$thus$$\int_a^0 f(-x)dx + \int_{0}^a ...

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