Problem 71 - Fundamentals of Physics

Question

A Gaussian surface in the form of a hemisphere of radius \(R=\) \(5.68 \mathrm{~cm}\) lies in a uniform electric field of magnitude \(E=2.50 \mathrm{~N} / \mathrm{C}\). The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through (a) the base and (b) the curved portion of the surface?

Step-by-Step Solution

Answer

(a) The flux through the base is approximately 0.0253 N·m²/C. (b) The flux through the curved surface is approximately -0.0253 N·m²/C.

Step 1: Calculate the flux through the base

The electric flux \( \Phi \) through a surface is given by \( \Phi = E \cdot A \cdot \cos(\theta) \), where \( E \) is the electric field magnitude, \( A \) is the area of the surface, and \( \theta \) is the angle between the electric field and the normal to the surface. For the base, \( \theta = 0^\circ \) because the field lines are perpendicular to the base and directed into the surface. Therefore, \( \cos(\theta) = 1 \).The area of the base of the hemisphere is given by \( A = \pi R^2 \).Calculate the flux:\[\Phi_{\text{base}} = E \cdot A = E \cdot \pi R^2 = 2.50 \, \text{N/C} \cdot \pi \cdot (0.0568 \, \text{m})^2 = 2.50 \cdot \pi \cdot 0.00322624 \, \text{m}^2\]\[\Phi_{\text{base}} \approx 0.0253 \, \text{N} \cdot \text{m}^2/\text{C}\]

Step 2: Apply Gauss's Law for the entire closed surface

According to Gauss's Law, the net electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, \( \varepsilon_0 \). Since the hemisphere encloses no net charge, the total flux through the hemisphere is zero:\[\Phi_{\text{total}} = \Phi_{\text{base}} + \Phi_{\text{curved}} = 0\]

Step 3: Calculate the flux through the curved surface

Using the result from Gauss's Law that the total flux is zero, we find the flux through the curved surface by solving for \( \Phi_{\text{curved}} \):\[\Phi_{\text{curved}} = -\Phi_{\text{base}}\]Substitute the calculated value of \( \Phi_{\text{base}} \):\[\Phi_{\text{curved}} = -0.0253 \, \text{N} \cdot \text{m}^2/\text{C}\]This negative sign indicates the flux through the curved surface is directed out of the surface.

Key Concepts

Gauss's Law

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.

Hemisphere Surface

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.

Electric Field Magnitude

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

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.