Time varying electric and magnetic field due to oscillating or moving charge Why does an oscillating or moving electric charge produces a time varying electric field is it time varying to some extent even in the state of rest.
 A: You are asking whether a charge at rest can produce an electric field. Yes, because an electric charge, like an electron, even at rest, would be time varying. Now what happens, is that it is a common misconception to think in only space.
You are asking whether a charge at rest (you think about rest in space) is creating an electric field? In reality, the charge moves in spacetime, not in space. An electric charge in spacetime, even if it is rest in space, it is moving in spacetime. It is because objects that are at rest in space, are moving in the time dimension at speed c. Only objects that do not have rest mass, that is they are moving in space at speed c, are at rest in the time dimension, their speed in the time dimension is 0, like a photon. But the photon has no electric charge.
Now in the case of an electron, it does have rest mass, and is moving in the time dimension with speed c, even if it is at rest in space. So an electron is never at rest in spacetime. That is why it is able to create an electric field, even when it is rest in space. 
Now in reality, an electron is never at rest in space, it is always moving in space at a speed less then c. So it is still moving in the time dimension with speed c. 
Now if this movement (in space) of the electric charge is constant (no acceleration), the electric charge will create an electric field and a magnetic field. The reason for this is relativity. And length contraction, when the charge moves, the distance between the charge and the observer seems shorter, and the electric force seems stronger.
If this electron moves in space with acceleration, it will create light. But the electric and magnetic field are not separate, but different views of the same EM field, so it is observer dependent. When the electron accelerates, it moves faster then the near field around it, and stirs it up, and this is the effect that creates light. 
A: Everyone knows that a point charge $q$ at rest at position $\mathbf{r}_0$ has a static, radial, spherically-symmetric electric field that decreases as the inverse square of the distance from the charge,
$$\mathbf{E}(\mathbf{r})=q\frac{\mathbf{r}-\mathbf{r}_0}{|\mathbf{r}-\mathbf{r}_0|^3},$$
and that it has no magnetic field,
$$\mathbf{B}(\mathbf{r})=0.$$
What about a point charge in uniform motion with velocity $\mathbf{v}$?
In this case, it has electric and magnetic fields which just "move along" with the charge at velocity $\mathbf{v}$.
If one takes the position of the charge at time $t$ to be $\mathbf{v}t$, its electric field turns out to be
$$\mathbf{E}(\mathbf{r},t)=q\frac{1-\beta^2}{(1-\beta^2\sin^2{\psi})^{3/2}}\frac{\mathbf{r}-\mathbf{v}t}{|\mathbf{r}-\mathbf{v}t|^3},$$
where $\beta=v/c$ and $\psi$ is the angle between $\mathbf{r}-\mathbf{v}t$ and $\mathbf{v}$,
and its magnetic field is
$$\mathbf{B}(\mathbf{r},t)=\frac{\mathbf{v}}{c}\times\mathbf{E}(\mathbf{r},t).$$
The vector $\mathbf{r}-\mathbf{v}t$ is just the vector from the position of the particle to the where the field is being calculated.
You can see that the electric field still drops off as the inverse square of the distance, and that it is still perfectly radial. Even arbitrarily far from the particle, it points directly away from the particle (assuming $q$ is positive).
But it is no longer spherically symmetric. The factor involving the angle $\psi$ (measured at the particle, between its direction of motion and the direction in which the field is being calculated) makes the field more intense in directions perpendicular to the direction of motion than in the direction of motion or opposite to it. In fact, when the speed of the charge approaches the speed of light, this factor smooshes the field into a pancake!
You can also see that the field depends on time. It cannot possibly be static. The charge is moving, so its field has to "move" so that the field is strong near the charge. Of course fields don't really move like an object moves; what happens is that the value of the field at each point simply changes with time. However, the field changes in such a way that the pattern of field lines simply moves along with the particle.
The magnetic field circles around the charge's direction of motion, drops off as the inverse square of the distance from the charge, and is more intense perpendicular to the direction of motion than in the direction of motion or opposite to it.
What about an accelerating point charge?
I'm not going to go through the math, because there are lots of different ways to accelerate and the fields depend on the kinematic details. But the most important thing that happens is that the fields no longer simply "follow" the charge. Far away, the fields don't "know" what the charge is going to do. For example, in the case where a charge at rest suddenly starts moving, the fields within distance $ct$ "know" that the charge is moving, but the fields farther away "think" it is still at rest. So a "kink" in the fields develops. The electric field is no longer purely radial everywhere (that is, it no longer simply points away from the charge) and there is a region where both fields now drop off as the inverse, rather than the inverse square, of the distance. If you have learned about the Poynting vector, which is proportional to $\mathbf{E}\times\mathbf{B}$, you know that this means that the fields are carrying energy (and also momentum and angular momentum) off to infinity, in an electromagnetic pulse!
When the charge undergoes periodic motion (back and forth like on a spring, or around in a circle), the field is an electromagnetic wave, again transporting away energy, momentum, and angular momentum because the radiation field drops off like $1/r$, not $1/r^2$.
Note: The above formulas are in Gaussian units.
