Time derivative of electric field? Electromagnetic radiation energy emitted

AI Thread Summary
Electromagnetic radiation is emitted by accelerating charges, with the energy emission rate given by the formula dE/dt = (q²a²)/(6πε₀c³). The discussion revolves around verifying the dimensional correctness of this equation and understanding that dE/dt represents the change in energy per unit time, not the electric field. Participants clarify that the units of power (Watt) can be derived from energy (Joules) divided by time (seconds), confirming that dE/dt has units of kg·m²/s³. The conversation highlights the confusion between energy and electric field, ultimately leading to a resolution on the dimensional analysis. The thread concludes with a participant expressing gratitude for the clarification.
GreenLantern
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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
 
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Hello GreenLantern,

I believe E here represents energy, not electric field. Here, dE/dt is the change in energy per unit time.
 
By the way, In case you haven't seen this:

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

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