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) 0.0254 N·m²/C, (b) -0.0254 N·m²/C.

Step 1: Understanding the Gaussian surface

We have a hemisphere with a radius of \( R = 5.68 \text{ cm} = 0.0568 \text{ m} \). The uniform electric field \( E = 2.50 \text{ N/C} \) is perpendicular to the flat base and directed into the surface. The curved surface does not contribute to net charge.

Step 2: Calculating electric flux through the base

The electric flux \( \Phi \) through a surface is given by the formula \( \Phi = E \cdot A \cdot \cos(\theta) \), where \( A \) is the area of the surface and \( \theta \) is the angle between the electric field and the normal to the surface. For the base of the hemisphere, \( A = \pi R^2 \), and \( \theta = 0^\circ \), so \( \cos(\theta) = 1 \). Therefore, the flux through the base is \[ \Phi_{\text{base}} = E \cdot \pi R^2 = 2.50 \cdot \pi \cdot (0.0568)^2 \approx 0.0254 \text{ N·m}^2/\text{C}. \]

Step 3: Applying Gauss's Law for total flux

According to Gauss's Law, the total electric flux through a closed surface is zero if there is no net charge enclosed. Thus, the total flux \( \Phi_{\text{total}} = \Phi_{\text{base}} + \Phi_{\text{curved}} = 0 \), where \( \Phi_{\text{curved}} \) is the flux through the curved portion of the surface.

Step 4: Calculating electric flux through the curved portion

Since the total flux is zero, we have \( \Phi_{\text{curved}} = -\Phi_{\text{base}} \), meaning \[ \Phi_{\text{curved}} = -0.0254 \text{ N·m}^2/\text{C}. \]

Key Concepts

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in Gauss's Law to calculate electric flux or related properties, without affecting the electromagnetic field itself. It helps in visualizing and solving electrostatic problems. When you think of a Gaussian surface, imagine a bubble surrounding a charged region.
These surfaces are particularly useful because they simplify the application of Gauss's Law—especially in symmetric situations like spheres, cylinders, and planes. In the original exercise, the Gaussian surface is a hemisphere. This choice makes our calculations more straightforward due to the symmetry around the base and the curved surface.

• It allows for easy calculations of electric flux without complex integrations.
• Works well with symmetrical charge distributions where field components and surface normals align in a predictable manner.
• Does not interact with or alter the fields around it—it’s purely a geometrical convenience.

Understanding how to choose the appropriate Gaussian surface is crucial. For example, when dealing with spherical symmetry, one typically uses a spherical Gaussian surface. For the given problem, the hemisphere is optimal, especially since the field is uniform and perpendicular to the base.

Electric Flux

Electric flux represents the number of electric field lines passing through a given surface. It is a measure of the "flow" of the electric field through an area, akin to how water flows through a net. The concept of electric flux is essential when applying Gauss's Law, serving as the foundational piece in its calculation to find electric fields or charge distributions.
Mathematically, electric flux \( \Phi \) through a surface is defined as \( \Phi = E \cdot A \cdot \cos(\theta) \).

• \( E \) is the magnitude of the electric field.
• \( A \) represents the area of the surface.
• \( \theta \) is the angle between the electric field and the normal to the surface.

In our exercise, for the base of the hemisphere, the electric flux is quite straightforward to calculate since the electric field is perpendicular, meaning \( \theta = 0^\circ \). The resultant flux is then simply the product of the field and the area, considering the alignment parallel to the field lines.
This concept extends further when recognizing that if the surface doesn’t enclose a net electric charge, as in the given problem, the total electric flux through the surface is zero. That is derived from Gauss's Law.

Uniform Electric Field

A uniform electric field is characterized by having the same strength and direction at every point within the region it occupies. Picture it like standing inside a room where every corner and edge has the same breeze blowing in the same direction.
Uniform electric fields simplify many electrostatic calculations because you can expect continuity across the entire area of consideration. This is the reason calculations for electric flux, as in our exercise, become notably easier since no matter where on the surface you calculate, the field’s magnitude and direction are consistent.

• Such fields are typically created between two parallel plates (like in a capacitor) or exist far from any collection of charges where the field asymptotically approaches uniformity.
• An electric field line diagram would show parallel, equally spaced lines extending across the considered region.
• Leverages simple mathematical models for predictions and applications in physics problems.

For the hemisphere in the problem, the field's uniform nature ensures the dot product simplifies considerably, very much simplifying the integration process in the calculation of electric flux. This helps underline why the initial electric field perpendicular to the hemisphere base effortlessly leads to an accessible calculation of flux through the surface.