# Tag Info

0

The field of each quarter circle is evaluated separately about its symmetry axis: which is why the limits are taken the way the answer shows.

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The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...

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A conductor can have a zero net charge. One way to do this is to have an inner surface with a net charge of $-Q$ and an outer surface with a net charge of $+Q.$ So for instance you can have a surface charge density of $-Q/(4\pi R^2)$ on the surface $r=R$ and you can have a surface charge density of $+Q/(4\pi (2 R)^2)$ on the surface $r=2R$ and this has zero ...

2

You can attract metal with static electricity. Consider the text-book example of a conducting sphere vs. a dielectric sphere in an electric field. Let's assume the field is homogeneous. This field polarizes both spheres, but in different ways: Conducting sphere: The free electrons rearrange themselves on the surface until the total electric field is ...

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The fact is that metals do get polarized but in a different way. Since electrons are rather free, they distribute themselves along the frontier of the system, being attracted by the field. But they do so in such a way that the field becomes zero inside the metal, because the contribution from the external field and these electrons cancel each other. You ...

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The red lines represent equipotential lines, which have the additional property of having an electric field of zero. The electric field at a point on the red line can be defined as $$\mathbf E_{red}=\sum_{i=1}^{4}\mathbf E_i$$ $$\mathbf E_i=\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\cdot{\mathbf{\hat{r}_{0i}}}$$ Let's take the case where all charges ...

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Displacement Current actually does not exist, it is a theoretical misnomer. When we consider a Capacitor as a low Characteristic Impedance Transmission Line we can think of energy flow between the conductors (Parallel Plates) We see a TEM wave (ExH) moving at the speed of light for the medium. What Maxwell thought of as Displacement current is the ...

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If you have a two charge configuration like this one, every point in space is going to have an electric field vector with contribution from both the charges. There is only one electric field, and it wouldn't make sense to talk about the two electric fields as different entities (and consequentially, about ideas like 'compressibility'). You must remember that ...

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How quickly discharge will occur in the situation you sketch depends entirely on the surface properties of the negative electrode. For current to flow, electrons need to be released from the negative electrode; once they are free, they will accelerate unimpeded to the positive electrode. They will arrive there with 1.5 eV of energy, causing a small amount of ...

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Electrons (and other charge carriers, e.g., ions) in vacuum travel without resistance. However, as pointed out, correctly, in the other answers, there are no charge carriers in vacuum. Nevertheless, electrons can escape from the terminals if they have a kinetic energy which is bigger than the potential barrier of the terminal surface, i.e., the work ...

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Lines of electrostatic force exist between the positive and negative poles of the battery, even though they're separated by a vacuum. Vacuum permittivity is ε0 = 8.854 * 10^-12 farads per meter. By convention, this is called the dielectric constant of 1, a baseline against which the dielectric permittivities of other materials are compared. ...

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In vacuum there are no charge carriers like ions or electrons. With nothing to carry charge, i.e. current, such a battery would discharge much, much slower than when the battery poles are connected by something that can carry charge like a conductor or an imperfect insulator.

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Well for the three sides touching the particle, the vector $\bf E$ coming from the particle, is always contained in the surface. This means that there is no component in the direction normal to any of the surfaces. So this means $\bf E$$d \bf S and therefore the flux through them, \int \bf{E} d \bf S will result zero. 0 1) One eighth of the charge is contained in the cube because you can place eight cubes such that they have a common vertex, the one your charge is at. If your charge was on the middle of an edge, it would be shared by four cubes. If your charge was in the middle of one of the faces, it would be shared by two cubes and so on. 2) The three adjacent faces have ... 1 Suppose you have a charge in a plane. The electric field from the charge is spherically symmetric, and that means every field line from the charge that intersects the plane has an equal and opposite field line intersecting the plane. So when we integrate \mathbf{E}\cdot\text{d}\mathbf{A} the two field lines will cancel out and the net flux will be zero. ... 1 My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides. Not really. You see, the electrostatic field \vec{E} of the charge is always radially outwards. If the charge is situated at the exact corner of the cube, then the field is exactly coplanar with the ... 0 Hint: You can always construct imaginary cubes (7 more, in this case.) such that your charge is completely enclosed inside your multi-cube system. You can find the flux through this system via Gauss' Law. The total flux through each side (of the new configuration) would be equal via symmetry. 0 Without more detailed data, further experimentation, I cannot provide a definite answer. But with regards to your supply transformer that feeds the primary - could it possibly be a GFI protected transformer? Many of the later date neon light transformers (Franceformer for example) built GFI protectors into the transformers for safety. Using these ... 1 Let's look at the simplest case: two like charges a distance d away. If free to move, they move away from each other and away from the center. Now imagine three like charges on the corners of an equilateral triangle, if free to move they also move away from the center. Now imagine four like charges on the corners of a square, if free to move they also move ... 1 This is the electric field of a point charge: The electric field is radially outward from the point charge in all directions. The circles represent spherical equipotential surfaces. Notice the r in the denominator. For r=0, which is your question, the field becomes infinite. In classical electrodynamics a particle can never have zero dimensions ... 0 By Gauss's law, the scenario you describe can only happen when there are electric charges within this volume. If they are positive charges, you need to draw field lines coming out of them. If they are negative charges, you need to draw field lines ending at them. 1 You are right! The trick to remember here is that vector fields permeate all of space (literally all of it) and field lines are only a convenient representation of this. When a new field line is added due to the increased magnitude of the field at that point in space, the field vector 'arrow' that is introduced always existed there but was just small ... 0 I don't think this situation is possible.Consider finite field lines passing through a given area vector in vaccum.Now If you enclose the given field lines and charge by Gaussian surface where one of the plane is parallel to x-y plane.By Gauss's law the net electric flux is constant. While changing the magnitude of electric field with same enclosure i.e ... 8 Modulo constants, this cross product is called the Poynting vector,$$ \mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B. $$It represents the flow of energy in the electromagnetic field, so that the energy density u=\tfrac12\varepsilon_0\mathbf E^2+\tfrac1{2\mu_0}\mathbf B^2 obeys, in free space, the continuity equation$$ \frac{\partial u}{\partial ... 0 Here's the easy way to do it. Do Gauss's law for a single uniformly charged$yz$-plane with surface charge$\sigma$. You'll find that the electric field is distance-independent except for its sign; $$\vec E = \frac{\sigma}{2\epsilon_0} ~ \operatorname{sgn}(x) ~ \hat x.$$Here$\hat x$is the unit vector in the$x$-direction and$\operatorname{sgn}(x) = x / ...

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Imagine two opposite charges very very close together In the tiny region between the charges the field goes from the positive charge to the negative charge. However if you come towards the positive charge from the opposite direction as the negative charge the field points in the opposite direction. And if you come towards the negative charge from the ...

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The reason is simple: the convention is that the polarization vector goes from negative to positive, not the other way around.

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If you have some (static) charge distribution $\rho(\mathbf{x})$ the the electric field is given by: $$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$$ or it might be easier to calculate the potential using Poisson's equation: $$\nabla^2 V = -\frac{\rho}{\varepsilon_0}$$ and then calculate the field using: $$\mathbf{E} = -\nabla V$$ In your ...

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