17
Especially the hydrogen atom, with a proton in the nucleus and an electron revolving acting as a dipole
This is a problematic way of understanding the hydrogen atom ─ it basically tries to insist on treating it within classical mechanics, and this is doomed to fail. Instead, the hydrogen atom must be treated within quantum mechanics. This introduces a bunch ...
6
If the curl of a static electric field were nonzero, you could move a point charge around a closed circuit and it would have more energy when it got back to its starting point than it did when it was there last time. This would be a violation of conservation of energy.
3
The Ohm's law was concluded from what was found from experiments. It is true that "electric field cause charges to accelerate", but under conditions the Ohm's law apply (e.g., in conductors), there are a whole lot of other forces that affect the movement of electrons. The collisions of a moving electron to non-moving atoms serve as friction force which slows ...
2
To add to Dale's answer, $\vec{D}=\epsilon_0\vec{E}+\vec{P}$.
Here, $\vec{E}$ is the Electric Field, $\vec{P}$ is the Polarization field, and $\vec{D}$
is the Electric Displacement field.
The Polarization $\vec{P}$ is the electric dipole density.
Electric fields can induce electric dipoles in materials. This is often proportional to the strength of ...
2
the net charge of the wire is zero. So, no electric field will be produced.
The charge of the wire being zero means that there is no electrostatic E-field. However there are other ways to produce E fields. Faraday’s law says $$\nabla \times E=-\frac{\partial}{\partial t}B$$
Which means that you can also get a circulating E field by having a B field which ...
2
It isn't.
Consider, for example, an infinite horizontal sheet of positive charge with an infinite current-carrying wire, parallel to the sheet, suspended above it.
The electric field from the charged sheet points upward everywhere above it. The magnetic field curls around the wire. This means that, in the plane that contains the wire and is parallel to the ...
2
The graph depicts the field of a plane wave, for which the amplitude does not vary with distance.
Plane waves are infinite in transverse extent (and longitudinal extent, for that matter), so they are unphysical, but they do describe the behavior of waves in volumes that are small enough that spatial variations can be ignored. Waves whose origin is a point ...
1
By convention, the direction of an electric field is the direction of the force that a positive charge would experience if placed in the field. What does that tell you about the direction of the force that a negative charge would experience if placed in the field?
Hope this helps.
1
First of all electric field due to a system of charges and an electric field due to a single charge are two different concepts.
We know that The net electric field due to two equal and oppsite charges is 0.
Where do you saw the above statement?We can say that algebraic sum of equal and opposite charges is zero but it does not imply that the net electric ...
1
The net electric field due to two equal and oppsite charges is 0.
This is only true if the two charges are located in the exact same location. For example, a block of copper sitting on your lab bench contains an equal amount of electrons and protons, occupying the same volume of space, so the block of copper produces no net external electric field.
But if ...
1
The inverse-square law has very little to do with your graph. Like you said, it is a time snapshot of the EM wave showing the field strength and direction of a polarized wave at several points along the C axis. There is no depiction of what the E-field is along the red axis, i.e., the red axis is not a spatial dimension. Your graph is depicting only one ...
1
To explore the concept further, note that the overall charge density of the wire is in fact something that depends on your frame of reference.
To see this, imagine that you're on a starship next to an infinitely long, neutrally-charged but current-carrying wire. As you accelerate closer to the speed of light, you "catch up" to the electrons moving in the ...
1
Well, in short, there is no deeper understanding as to why the curl of the electric field is zero. The Maxwell equations are the most basic laws of electrodynamics and when you constrain yourself to the static case, that the curl of the electric field vanishes is one of the statements of the Maxwell equations.
However, you can try to better understand this ...
1
As you write, the expression for the charge density can be found by considering that the total charge within the sphere is $Q$. Indeed, if we define $\rho(r) = A r^n$ we have
$$
Q = 4 \pi A \int_0^{R_s} r^n r^2 dr = 4 \pi A \int_0^{R_s} r^{n+2} dr
$$
where $R_s$ is the radius of the sphere.
The Gauss law in this case takes the form
$$
4 \pi R^2 E(R) = \...
1
At the end of the day, flux is the number of field lines passing through a surface area. Imagine that the cylinder you mentioned is enclosed in a sphere, now the number of field lines passing through the cylinder should be equal to the filed lines passing through the sphere(assuming there doesn't exist any other electric filed). Therefore we can say that the ...
1
If we have a function $V$ that solves Poisson's equation, then it must be true that
$$\nabla^2V=\nabla\cdot(\nabla V)=-\frac{\rho}{\epsilon_0}$$
By definition of potential, we have $\nabla V=-\mathbf E$, so then
$$\nabla\cdot\mathbf E=\frac{\rho}{\epsilon_0}$$
Therefore, the answer to
does every solution to Poisson's equation correspond to a function ...
1
Due to the applied field, charges will be accumulated in a non-uniform
manner on the inner and outer surfaces of the shell
No, the charges will accumulate only on the outer surface of the shell, as the inner surface will not have bear any charges.
Now, remember that the whole conducting shell is at the same potential, which means that the inner surface ...
1
Hint: Apply guass law
The mistake you are doing is that the field E=kq/r^2 is valid at a distance r from the point charge(so spherical symmetry). All points on the face of the cube are not at a distance of x or x+a from the point charge.
Also as garyp mentioned in the comment below, all field lines do not exit from the opposite faces, so finding flux ...
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