Time derivative of electric field? Electromagnetic radiation energy emitted

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Homework Help Overview

The discussion revolves around the dimensional analysis of the energy emitted by accelerating charges in the context of electromagnetic radiation. The original poster seeks to verify the dimensional correctness of a given equation relating charge, acceleration, and energy emission rate.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to verify the dimensional correctness of the equation for energy emission, expressing confusion about how to relate the derivative of the electric field to the units of energy. Participants discuss the interpretation of the symbols and the relationship between energy and power.

Discussion Status

Participants are exploring the dimensional analysis of the equation and clarifying the definitions of terms involved. Some guidance has been offered regarding the interpretation of energy and power, with a suggestion that the relationship may be more definitional than derivational. The original poster expresses confusion but also indicates progress in understanding.

Contextual Notes

There is a mention of the need to prove the dimensional correctness of the equation, with some participants questioning whether a deeper derivation is necessary or if the dimensional analysis suffices. The discussion reflects a mix of interpretations regarding the physical concepts involved.

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.
 
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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