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Amplitude of an oscillating electric field

  • Thread starter DottZakapa
  • Start date
65
4
Homework Statement
The amplitude of the oscillating electric field at your cell phone is 4.0 μV/m when you are 10 km east of the broadcast antenna. What is the electric field amplitude when you are 20 km east of the antenna
Homework Equations
electric field
Homework Statement: The amplitude of the oscillating electric field at your cell phone is 4.0 μV/m when you are 10 km east of the broadcast antenna. What is the electric field amplitude when you are 20 km east of the antenna
Homework Equations: electric field

i've done

E=##\frac A {r^\left(2\right)}##

that in this case is ##4 \times 10^\left(-6 \right)=\frac A {10000^2} ##
i solve for A
##A= 4 \times 10^\left(-6\right) \times 10000^2##
then i divide ##A## by ##20000^2##
but i do not get the right result, why?
 

Doc Al

Mentor
44,803
1,061
The intensity drops off as ##1/r^2##, not the field.
 
65
4
The intensity drops off as ##1/r^2##, not the field.
Do i have to look in electromagnetic waves formulas?
##E\left(x,t\right)=E_0sin\left(kx-\omega t \right)##

in the data given there is not power so i cannot divide it by the half sphere surface, in order to get the intensity end from there the amplitude.
 
65
4
ok i got it
I start from this

##I=\frac 1 2 \times \epsilon_0 \times E^2##

##I= \frac P {half \space sphere \space area}##

## P=\frac {4 \times \pi \times r^2} 2 \times I##

in such way i get the power of the wave
dividing it by the half sphere area with radius 20000 meters i get the new average intensity
then

##E=\sqrt { \frac {I \times 2} {\epsilon_0 \times c}}\space## with ##\space c=3\times 10^8##
 

Doc Al

Mentor
44,803
1,061
No need for any calculation. Since the distance doubles, the intensity drops to 1/4. Since the amplitude of the field is proportional to the square root of the intensity, you can immediately find the new amplitude.
 

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