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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 ...


3

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 ...


2

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 ...


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 ...


2

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 ...


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 ...


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

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 ...


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

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.


1

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. ...


1

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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