For all of those in the future who come seeking a solution, this is it.
keep going from where i left off, plug and chug by setting energy flux of the transmitter (P_tran / 2 Pi d^2 ) equal to (E_0^2 / 2 mu_0 c).
Then solve for d!
Technically mu_0 * c gives you 377 Ohms, i thought wolfram alpha would deal with it, but it did not.
I see what i did, I didn't square V/m. Ok, back to business then, is my 4th step correct, or no?
Sorry about it not being clear.
But I'm obviously lost then!
That's what I mean, I don't know WHY the units are wrong, I just know that when I plugged them into Wolfram Alpha to double check, they were very wrong!
Where would you recommend I start? Did I start on the right path and veer off...
.5 gives the average of the sinusoidal function.
E_0 = (V*lambda)/(Pi^2*r_ant^2) (according the my calculation of using the wave equation to derivate E_0).
The size of the antenna is used in E_0, because you're calculating the induced voltage through the loop.
The units are consistent...
Homework Statement
Assume that the power radiated by the television transmitter uniformly fills the upper hemisphere. A UHF television with a single-turn circular loop antenna of radius 8 cm requires a maximum induced voltage above 24 mV for operation.
The speed of light is 2.99792 × 108 m/s...
As you probably well know, the Electric field due to an infinite sheet is σ/(2ε(naught)). But this gives something about 2.43 times smaller than what was given. Does this proportion have anything to do with how the field distributes itself?
Homework Statement
Here is the question, which itself is rather confusing.
A nonuniform surface charge lies in the yz plane. At the origin, the surface charge den- sity is 3.5 μC/m^2. Other charged objects are present as well. Just to the right of the origin, the electric field has only an x...