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Gauss's Law is a fundamental principle that relates the electric flux flowing out of a closed surface to the charge enclosed within it. Specifically, it states that the total electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space, \( \varepsilon_0 \). This can be mathematically expressed as:

• \( \Phi = \frac{Q_{ ext{enc}}}{\varepsilon_0} \)

Here, \( \Phi \) represents the electric flux, and \( Q_{ ext{enc}} \) is the enclosed charge. In scenarios where no net charge is enclosed, the total flux through the surface becomes zero, simplifying our calculations.
In the context of the exercise, the hemisphere encloses no net charge, leading to zero total flux through the hemisphere, hence illustrating how Gauss's Law is applied to find solutions in electrostatics.

When dealing with electrostatics problems, a hemisphere is a common shape used for Gaussian surfaces due to its symmetry. A hemisphere consists of two parts: a flat circular base and a curved surface. The area of the base is calculated using the formula for the area of a circle:

• \( A = \pi R^2 \)

Here, \( R \) is the radius of the hemisphere. For a hemisphere, understanding the distinction between the flat and curved surfaces is essential. It allows us to analyze the flux through different sections independently.
In the examined problem, the hemisphere's flat base plays a pivotal role because the electric field is perpendicular to it, simplifying the computation of electric flux. The symmetrical nature of the curved surface complements the requirement for symmetry when applying Gauss's Law.

The electric field magnitude \( E \) is a key parameter in determining the electric flux through a surface. The electric field represents the force that a charge would experience in the presence of electric energy. In the problem, the electric field has a magnitude of 2.50 N/C. Given the orientation and uniformity of the electric field, it impacts how we calculate the flux on different parts of the hemisphere.
Calculating the electric flux \( \Phi \) for a surface requires knowing the angle \( \theta \) between the electric field lines and the surface normal. In this problem, for the base of the hemisphere, this angle is 0 degrees, implying that \( \cos(\theta) = 1 \). Therefore, the formula for electric flux simplifies to \( \Phi = E \cdot A \), where \( A \) is the area of the surface.

Permittivity of free space, denoted as \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum of free space. It has a value of approximately \( 8.85 \times 10^{-12} \text{ C}^{2}/\text{N} \cdot \text{m}^{2} \). This constant appears in formulas that involve electric fields and forces, making it fundamental to Gauss's Law.
In the problem, although the hemisphere encloses no net charge, the permittivity of free space \( \varepsilon_0 \) is still crucial for explaining why the total flux through the closed surface equals zero. Understanding \( \varepsilon_0 \) provides insights into how electric fields are propagated in space and how they influence the calculations derived from Gauss's Law.