If an electron in a universe constantly generates an electric field why does it not get annihilated ? I am confused because I read that an accelerating charge radiates and loses energy. So, why won't a stationary electron radiate and lose energy?
[Physics] Why doesn’t a stationary electron lose energy by radiating electric field (as per coulomb’s law)
electromagnetic-radiationelectronsenergyquantum-electrodynamics
Related Solutions
You are right. An electron in a uniform magnetic field will travel in circles (or in a helix, up to a change in frame of reference), but this means that it is an accelerated charge and it must therefore radiate and lose energy. This radiation is known as synchrotron radiation, and it is a major design issue for particle accelerators. (In fact, it is the reason for a recent trend to go back to linear accelerators, which are less efficient as each accelerating stage only works once per particle, but are not subject to this.) It can also be harnessed to make synchrotron light sources, and with some extra work one can build a free-electron laser using that principle.
In short, then, the electron will spiral down to the centre and lose all its kinetic energy as electromagnetic radiation.
(For the more quantum-mechanical minded, now that Landau eigenstates have joined the fray, this means that all excited Landau states will have to decay through radiative coupling to the ground state with zero angular momentum. Once there, though, the uncertainty principle kicks in and stops the electron getting localized to radii smaller than the characteristic harmonic oscillator length $$x_0=\sqrt\frac{\hbar\omega_c}{m}=\frac{\sqrt{\hbar eB}}{m}$$ corresponding to the cyclotron frequency $\omega_c=eB/m$.)
Now I am left wondering why does the heat become lost as if travels slightly.
It is not lost. It is spread more out.
If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100.
The heat radiation sent from the source like the sun spreads out with distance. The same amount of energy every second is spread out at a larger and larger sphere as it travels away from the source.
Surface area of such a sphere is $A=4 \pi r^2$. Doubling your distance to the campfire means:
$$A_2=4 \pi r_2^2=4 \pi (2 r_1)^2=4 *4 \pi r_1^2=4 A_1$$
The area that the radiation is spread over is four times as large for just the double distance.
This of course regards the sun with only space surrounding it. Since the campfire is place on ground, downwards radiation in that case will be smaller.
Solar constant
For your interest, the Suns' intensity on our planet Earth is called the solar constant (or solar coefficient) $S$. In units $\mathrm{W/m^2}$. This tells us how much radiation that reaches a square meter on Earth (or any other planet at the same distance from the Sun) every second. Mercury which is closer will have another solar constant.
To find the Suns' intensity at any distance, you will need to know how much energy is generated per second within the Sun, that is the power of the Sun $P$. This energy will be spread out while travelling:
$$P=S*A=4 S \pi r^2$$
where both intensity $S$ and area $A$ are for a specific sphere at a specific distance. The Suns' own intensity at its surface is then found by insertings the Suns' own radius and isolate S.
Also, our Sun can be viewed as a socalled blackbody. That is, it emits radiation very efficiently. The Stefan–Boltzmann law of blackbody radiation then gives us the emissive power from the Suns' surface:
$$P=\sigma T^4 A= \sigma T^4 4 \pi r^2 $$
with $T$ being the surface temperature of the Sun (around $5800^\mathrm{o C}$ if I remember correctly).
Campfire
Regarding the campfire as pointed out in the comments, convection might be considerable if you are standing very close to the campfire or maybe even reaching over it.
By natural convection, heated air will flow upwards, and this will carry a lot of energy that way. The radiation itself is negligible at that position.
Walking slightly further away might remove the convection effect entirely. This will feel like a huge decrease in heating. Adding a little windy weather, you might not feel any wind while being in "equilibrium" in the convection zone near the campfire. But walking two steps away can have a large cooling effect now that forced convective cooling from the wind acts as well.
Best Answer
You can think of it this way -
The field isn't being created 'constantly'. That's like saying our hands and legs are being 'created constantly'. The field is coupled to an electric charge, and that's how it is.
When a charge is being accelerated, it is gaining energy (since work is being done on the charge) and it also looses energy (through the radiation). The electrostatic field doesn't have anything to do with this picture. When a charge is static, the field is also static.
In a universe where there is exactly one charge sitting in a vacuum, the charge will have an electric field. But the field does no work, and no work is done on the field. So there is no loss of energy.