(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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:

[tex]\frac{dE}{dt}[/tex] = [tex]\frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}}[/tex]

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[tex]^{2}[/tex] --> m/s[tex]^{2}[/tex]

[tex]\epsilon_{0}[/tex] --> [tex]\frac{C^{2}}{N*m^{2}}[/tex]--> [tex]\frac{A^{2}*s^{4}}{kg*m^{3}}[/tex]

c-->[tex]\frac{m}{s}[/tex]

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

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

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 [tex]\frac{dE}{dt}[/tex] and showing that it has equivalent units to what i found above.

I know the equation for electric field is:

[tex]\vec{E}[/tex] = [tex]\hat{r} k \frac{q}{r^{2}}[/tex]

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

Thanks for all the help!

-Ben

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# Homework Help: Time derivative of electric field? Electromagnetic radiation energy emitted

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