In a particular region of Earth's atmosphere, the electric field above Earth's surface has been measured to be 151 N/C downward at an altitude of 280 m and 167 N/C downward at an altitude of 430 m. Calculate the volume charge density of the atmosphere, assuming it to be uniform between 280 and 430 m. (Hint: You may neglect the curvature of Earth. Why?)
I tried using E=pz/ε, where E was 151+167 and z was (430-280)/2, half the distance between the ends, and got, 3.75E-11 C/m^3, but that was wrong as the online software told me. What should I do?
To calculate the volume charge density of the atmosphere, you can use Gauss's law. Gauss's law relates the electric field to the enclosed charge within a Gaussian surface. In this case, we can consider a Gaussian surface between points at altitudes of 280 m and 430 m.
The formula for Gauss's law in integral form is:
∮ E · dA = (1/ε₀) ∫ ρ dV
where ∮ E · dA represents the closed surface integral of the electric field, ρ is the volume charge density, ε₀ is the permittivity of free space, and ∫ ρ dV represents the integral of the volume charge density over the volume enclosed by the Gaussian surface.
In this problem, the electric field values are given, so we can calculate the difference in electric field between the two altitudes:
ΔE = |E2 - E1| = |167 N/C - 151 N/C| = 16 N/C
The distance between the two altitudes is:
Δz = |430 m - 280 m| = 150 m
Now, we can rewrite Gauss's law as:
ΔE · A = (1/ε₀) Δρ ΔV
where A is the area of the Gaussian surface and ΔV is the volume between the two altitudes.
The area, A, is the cross-sectional area of the region between the two altitudes. However, since the problem assumes a uniform charge density, we can choose any cross-sectional area, as long as it remains constant. For simplicity, we can choose a unit area, which cancels out in the final answer. Thus, A = 1 m².
Now, the volume between the two altitudes is:
ΔV = AΔz = 1 m² * 150 m = 150 m³
Finally, we can solve for Δρ, which represents the difference in charge density between the two altitudes:
Δρ = (ΔE · A) ε₀ / ΔV
Substituting the known values:
Δρ = (16 N/C * 1 m²) / (150 m³ * ε₀)
Now, we need to use the value of permittivity of free space, ε₀, which is approximately 8.85 × 10^-12 C²/(N·m²).
Plugging in the values:
Δρ = (16 N/C * 1 m²) / (150 m³ * 8.85 × 10^-12 C²/(N·m²))
Calculating this expression gives us:
Δρ ≈ 1.203 × 10^-10 C/m³
Therefore, the volume charge density of the atmosphere is approximately 1.203 × 10^-10 C/m³.