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?
^^^^drwls
How did you get 167 - 151 = 46???
To calculate the volume charge density of the atmosphere, you can use Gauss's law in integral form. The equation is given by:
∮ E · dA = Q_enclosed / ε₀
Where:
∮ represents the closed surface integral,
E is the electric field,
dA is an elemental area vector on the closed surface,
Q_enclosed is the total charge enclosed by the surface, and
ε₀ is the permittivity of free space.
Since the electric field is known at two different altitudes (280 m and 430 m), we can consider a cylindrical Gaussian surface between these two altitudes. Let's assume the cross-sectional area of the Gaussian surface is A, and the height between the altitudes is h = (430 m - 280 m) = 150 m.
Since the problem specifies that the atmospheric charge density is uniform, the electric field is constant within the region between the altitudes. Therefore, the electric field in this region is given by the average of the two known electric fields:
E_avg = (151 N/C + 167 N/C) / 2 = 159 N/C (downward)
Now, let's calculate the charge enclosed within the Gaussian surface. As the problem assumes a uniform charge density, the charge within the Gaussian surface will be the product of the volume charge density (ρ) and the volume of the region between the altitudes (V).
The volume of the region between the altitudes is given by:
V = A * h
Considering the electric field points downward, the total charge enclosed within the Gaussian surface can be calculated using:
Q_enclosed = ρ * V
Substituting the value of V, we have:
Q_enclosed = ρ * (A * h)
Rearranging the equation, we find:
ρ = (Q_enclosed) / (A * h)
Now, let's calculate the area A. Since we are neglecting the curvature of the Earth, the Gaussian surface is cylindrical, and the area A is the circumference of the circular base multiplied by its height (h).
A = 2πr * h
To calculate r, we will consider the Earth's radius r₀ (approximately 6,371 km) and use the altitude values provided. The distance between sea level and 280 m altitude is:
d₁ = r₀ + 280 m
Similarly, the distance between sea level and 430 m altitude is:
d₂ = r₀ + 430 m
Using these distances, the effective radius r of the Gaussian surface is given by:
r = (d₁ + d₂) / 2
Finally, we substitute these values into the equation to obtain the volume charge density:
ρ = (Q_enclosed) / (A * h)
= (Q_enclosed) / (2πr * h)
Now that the approach is clear, you can calculate the volume charge density using the provided altitudes and electric field values. Ensure that you also consider the permittivity of free space (ε₀ = 8.85 × 10⁻¹² C²/(N·m²)) in your calculations.
If you encounter any issues or have further questions, feel free to ask.
Treat the atmosphere as planar and use Gauss' Law. The net flux from an volume of cross sectional area A and thickness t is:
(167 - 151) *A = 46 Volt/C * A
That equals the charge inside the volume, divided by ε.
46 V/C*A = (charge density)*A*(thickness)/ε
charge density = ε * 46 /t
You should have subtracted the two field strengths, not added them, since E enters the volume on one side and leaves ion the other. The flux is in opposite directions.