Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
Explanation:
From the question we are told that
The time constant
The potential across the capacitor can be mathematically represented as
Where is the voltage of the capacitor when it is fully charged
So at
Generally energy stored in a capacitor is mathematically represented as
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at
The energy stored can be evaluated at as
Hence the fraction of the energy stored in an initially uncharged capacitor is
The equation for work (W) done by an electric field is:
W = qΔV
where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:
10 = 2ΔV
ΔV = 5
Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²