Time derivative of electric field? Electromagnetic radiation energy emitted

In summary, the rate at which energy is emitted from an accelerating charge is given by dE/dt = (q^2*a^2)/(6*pi*epsilon_0*c^3), where c is the speed of light. The equation is dimensionally correct and the unit of power is the Watt.
  • #1
GreenLantern
30
0

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:

[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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  • #2
Hello GreenLantern,

I believe E here represents energy, not electric field. Here, dE/dt is the change in energy per unit time.
 
  • #3
By the way, In case you haven't seen this:

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

:rofl:
 
  • #4
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 = [tex]\epsilon_{0}[/tex] * c * E[tex]^{2}[/tex] (in vacuum)

ugh still so confused. can anyone help me out some more with the solution to this?
 
  • #5
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

[tex] P(t) = \frac{dE}{dt} [/tex]

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

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

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).
 
  • #6
Ahhh okay i figured it out. Thanks so much for the help!
-GL
 

1. What is the definition of the time derivative of electric field?

The time derivative of electric field is the rate of change of electric field with respect to time. It measures how the electric field strength changes over time.

2. How is the time derivative of electric field related to electromagnetic radiation energy emitted?

The time derivative of electric field is directly related to the rate of change of electromagnetic radiation energy emitted. This means that if the electric field changes more rapidly, the amount of energy emitted as electromagnetic radiation will also increase.

3. How is the time derivative of electric field calculated?

The time derivative of electric field can be calculated by taking the derivative of the electric field function with respect to time. This can be done using mathematical techniques such as differentiation.

4. What is the significance of the time derivative of electric field in electromagnetic radiation?

The time derivative of electric field is important in the study of electromagnetic radiation because it helps us understand the behavior and properties of electromagnetic waves. It is also used in various applications such as antenna design and radar technology.

5. How does the time derivative of electric field affect the intensity of electromagnetic radiation?

The time derivative of electric field directly affects the intensity of electromagnetic radiation. A higher rate of change of electric field results in a stronger intensity of radiation. This is because a more rapidly changing electric field produces more energy, which is then emitted as electromagnetic radiation.

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