Time derivative of electric field? Electromagnetic radiation energy emitted

[GreenLantern]

Homework Statement

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by:

$$\frac{dE}{dt}$$ = $$\frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}}$$

where c is the speed of light.
(a). verify that this equation is dimensionally correct

Relevant equations/ The attempt at a solution

q -> Coulombs-> Amp*Seconds -> A*s
a -> meters/second$$^{2}$$ --> m/s$$^{2}$$
$$\epsilon_{0}$$ --> $$\frac{C^{2}}{N*m^{2}}$$--> $$\frac{A^{2}*s^{4}}{kg*m^{3}}$$
c-->$$\frac{m}{s}$$


now i have no problem doing all of that and boiling it down to $$\frac{dE}{dt}$$ = $$\frac{kg*m^{2}}{s^{3}}$$

but my problem comes in when it comes to proving that $$\frac{dE}{dt}$$ is supposed to have those units. The book asks to prove it. I took what they gave me and worked it down to $$\frac{kg*m^{2}}{s^{3}}$$

now what I seem to need is help with figuring out how to go the other way around the same circle by taking the derivative of the electric field with respect to time $$\frac{dE}{dt}$$ and showing that it has equivalent units to what i found above.

I know the equation for electric field is:

$$\vec{E}$$ = $$\hat{r} k \frac{q}{r^{2}}$$

but how do I go about taking the d/dt of that?!?


Thanks for all the help!
-Ben

 

[collinsmark]

Hello GreenLantern,

I believe E here represents energy, not electric field. Here, dE/dt is the change in energy per unit time.

 

[collinsmark]

By the way, In case you haven't seen this:

http://www.xkcd.com/687/

 

[GreenLantern]

lol

thanks for the input on getting me away from electric field
okay so energy,

energy of...??

so that would energy flow per unit time which would be S (poynting vector magnitude) times area? (because S is the energy flow per unit time per unit area)

I was figureing something along theses lines after i posted and then got lost in a world of confusing thoughts trying to figure it all out my self using S = (1/A) dU/dt = $$\epsilon_{0}$$ * c * E$$^{2}$$ (in vacuum)

ugh still so confused. can anyone help me out some more with the solution to this?

 

[collinsmark]

I'm not sure if this will help. But I'm pretty sure the units you are looking for is the Watt. Power is defined as

$$P(t) = \frac{dE}{dt}$$

I'm pretty sure the relationship is more of a definition than a derivation. In the SI system, the unit of power is the Watt. I don't think you can prove/derive the relationship any further than that. The Watt unit can be broken down to J/s (the the Joule can be broken down further too, if you need to).

Do you need to actually prove that

$$\frac{dE}{dt} = \frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}}$$

or simply prove that the units are dimensionally correct? Because I think you've pretty much proven the units already (well, as long as you know that energy is measured in Joules).

 

[GreenLantern]

Ahhh okay i figured it out. Thanks so much for the help!
-GL