91Ó°ÊÓ

Using Gauss's law, we know that the electric field related to the charge enclosed is \(E = \frac{Q_{encl}}{4\pi \varepsilon_0 r^2}\), where \(E\) is the electric field, \(Q_{encl}\) is the total charge enclosed by the Gaussian surface, \(\varepsilon_0\) is the vacuum permittivity, and \(r\) is the distance from the center of the sphere (radius of Earth). Given the electric field \(E = 150 \frac{N}{C}\), we'll first find the total charge on Earth's surface by solving for \(Q_{encl}\): \(Q_{encl} = 4\pi \varepsilon_0 r^2 E\) We know \(E = 150 \frac{N}{C}\), \(\varepsilon_0 = 8.85 \times 10^{-12} \frac{F}{m}\), and the radius of Earth \(r \approx 6.37 \times 10^{6} m\). Plug the values into the equation: \(Q_{encl} = 4\pi (8.85 \times 10^{-12} \frac{F}{m})(6.37 \times 10^6 m)^2 (150 \frac{N}{C})\)

After plugging the values into the equation and performing the calculation, we obtain the total charge on Earth's surface: \(Q_{encl} \approx 5.58 \times 10^5 C\) This is the total charge on Earth's surface.

To find the charge per unit area on Earth's surface, we will divide the total charge, \(Q_{encl}\), by the Earth's surface area. The surface area of a sphere is given by \(A = 4\pi r^2\), where \(r\) is the radius of the sphere. In this case, the radius of Earth is \(r \approx 6.37 \times 10^{6} m\). Thus, the Earth's surface area is: \(A \approx 4\pi ( 6.37 \times 10^6 m)^2\) Now, we will divide the total charge, \(Q_{encl}\), by the Earth's surface area to find the charge per unit area: \(\frac{Q_{encl}}{A} \approx \frac{5.58 \times 10^5 C}{4\pi(6.37 \times 10^6 m)^2}\)

After performing the charge per unit area calculation above, we obtain: \(\frac{Q_{encl}}{A} \approx 6.92 \times 10^{-9} \frac{C}{m^2}\) This is the charge per unit area on Earth's surface.

In part (b), we are given that the electric field at an altitude of 250m above Earth's surface is \(120 \frac{N}{C}\). We will apply Gauss's law to find the charge density in the air. We will consider a cylindrical Gaussian surface extending from Earth's surface to an altitude of 250m with a radius \(R\). The volume enclosed between the two surfaces will contain the charge of the air. Let the charge density be \(\rho\). The total charge enclosed by the Gaussian surface can be written as: \(Q_{air} = \rho \cdot V\) Where, \(V = A\cdot h\) is the volume of the cylinder and \(h= 250m\). The electric field at the outer surface of the cylinder can be written as: \(E = \frac{Q_{air}}{2\pi\varepsilon_0 R h}\) Substitute the total charge as \(Q_{air} = \rho \cdot A \cdot h\) and the given electric field value \(E = 120 \frac{N}{C}\): \(120 \frac{N}{C} = \frac{\rho \cdot A \cdot h}{2\pi\varepsilon_0 R h}\)

Rearrange the equation above and solve for the charge density \(\rho\): \(\rho = \frac{120 \frac{N}{C} \cdot 2\pi\varepsilon_0 R h}{A \cdot h}\) We are given the altitude \(h = 250m\), vacuum permittivity \(\varepsilon_0 = 8.85 \times 10^{-12}\frac{F}{m}\), and we can choose \(R \gg h\). Hence, the charge density \(\rho\) can be calculated: \(\rho \approx \frac{120 \frac{N}{C} \cdot (2\pi)(8.85 \times 10^{-12}\frac{F}{m})(250m)}{A}\)

After performing the calculation above, we obtain the charge density of the air at an altitude of 250m: \(\rho \approx 1.71 \times 10^{-8} \frac{C}{m^3}\) This value is the charge density (charge per unit volume) of the air assumed constant at an altitude of 250m above Earth's surface.