The solar radiation incident on the Earth has a power density of approximately 900W/m2. Find the rms value of the electric field.
I know that Erms = Emax/squareroot(2)
But how can I pass from W/m^2 to electric field?
Update: I have tried to do:
Pd = E^2 / Zo = E^2 /120pi
E^2 = Pd * 120pi
E = squareroot(900 * 120pi) = 582.48 V
Would this be correct?
yes, that is correct.
Thanks.
To find the rms value of the electric field, we can use the relationship between power density (W/m^2) and the root mean square (rms) value of the electric field.
Given that the power density incident on the Earth is approximately 900W/m^2, we need to find the corresponding electric field.
The relationship between power density (P) and the electric field (E) is given by the equation:
P = (1/2) ε₀ c E²
Where:
- P is the power density (in W/m²)
- ε₀ is the vacuum permittivity constant (approximately equal to 8.85 x 10^-12 F/m)
- c is the speed of light (approximately equal to 3 x 10^8 m/s)
- E is the electric field (in V/m)
Rearranging the equation, we get:
E² = (2P) / (ε₀c)
Substituting the given value of power density (P = 900 W/m²) and the constants, we can calculate the square of the electric field:
E² = (2 * 900) / (8.85 x 10^-12 * 3 x 10^8)
E² ≈ 7.6277 x 10^15
To find the rms value (Erms) of the electric field, we divide the squared electric field by the square root of 2:
Erms = √(E² / 2)
Erms = √(7.6277 x 10^15 / 2)
Erms ≈ 1.381 x 10^8 V/m
Therefore, the rms value of the electric field is approximately 1.381 x 10^8 V/m.