The answer would be center of mass, B
By definition,
q = 1.22y/D
Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna
At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance
But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m
Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
Answer:
0.0025H
Explanation:
I didn't come here to be part of this all I wanted is just information for my research
