On a clear day, the Earth’s atmospheric electric field near the ground has a magnitude of 100 N/C and points vertically down. Inside the ground, the electric field is zero, since the ground is a conductor. Consider a mathematical box of 1.0 m x 1.0 m x 1.0 m, half below the ground and half above. What is the electric flux through the sides of the box? What is the charge enclosed by the box?
Use Gauss' Law. Assume there is no flux in the downward direction or out the sides. The flux through the top (E* 1 m^2) will be proportional to the charge inside the box. Look up the law for the poprtionality constant.
That charge will reside at the surface, since the groud is a conductor.
To determine the electric flux through the sides of the box and the charge enclosed, we need to understand the concept of electric flux and Gauss's law.
Electric flux (Φ) is a measure of the total electric field passing through a given surface. It depends on the electric field strength and the area of the surface. Mathematically, it is defined as the dot product of the electric field (E) and the surface area vector (A) integrated over the surface.
Φ = ∫E · dA
Gauss's law relates the electric flux passing through a closed surface to the charge enclosed within that surface. It states that the electric flux through a closed surface is directly proportional to the total charge enclosed.
Φ = Q_enclosed / ε₀
where Q_enclosed is the charge enclosed within the closed surface, and ε₀ is the permittivity of free space (a constant).
Now, let's calculate the electric flux through the sides of the box:
Given:
Electric field near the ground (E) = 100 N/C (vertically downward)
Area of each side of the box (A) = 1.0 m x 1.0 m = 1.0 m²
Since the electric field points vertically down and the sides of the box are perpendicular to the field, the dot product of the electric field and the surface area vector will be E.A = E * A * cosθ, where θ is the angle between E and A. Since the angle is 0° (cos 0° = 1), the dot product simplifies to E * A.
Electric flux through each side of the box (Φ_side) = E * A = 100 N/C * 1.0 m² = 100 N.m²/C
Since there are four sides in total, the total electric flux through all the sides of the box is:
Total electric flux (Φ_total) = 4 * Φ_side = 4 * 100 N.m²/C = 400 N.m²/C
Next, let's determine the charge enclosed by the box using Gauss's law:
Using the equation Φ = Q_enclosed / ε₀, we can rearrange it to solve for Q_enclosed:
Q_enclosed = Φ * ε₀
Substituting the values, we get:
Q_enclosed = 400 N.m²/C * ε₀
The value of ε₀ (permittivity of free space) is approximately 8.85 x 10^-12 C²/N.m². Plugging in this value:
Q_enclosed = 400 N.m²/C * 8.85 x 10^-12 C²/N.m²
Simplifying the equation:
Q_enclosed = 3.54 x 10^-9 C
Therefore, the charge enclosed by the box is approximately 3.54 x 10^-9 C.