Problem 10 - University Physics with Modern Physics

Question

22.10. A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m}\) . What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radins (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

Step-by-Step Solution

Answer

(a) 0, (b) -0.678 x 10^3 Nm^2/C, (c) -0.226 x 10^3 Nm^2/C

Step 1: Understanding the concept of electric flux

Electric flux, denoted by \( \Phi \), through a closed surface in a uniform electric field is calculated using Gauss's Law: \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \), where \( Q_{enc} \) is the net charge enclosed by the surface, and \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \).

Step 2: Analyzing (a): Radius = 0.500 m

A spherical surface of radius 0.500 m is smaller than the distance of both charges \( q_1 \) and \( q_2 \) from the origin. Therefore, no charge is enclosed within this sphere. Substituting \( Q_{enc} = 0 \), the electric flux \( \Phi = \frac{0}{\varepsilon_0} = 0 \).

Step 3: Analyzing (b): Radius = 1.50 m

The spherical surface of radius 1.50 m encloses the charge \( q_2 = -6.00 \, \mathrm{nC} \) at \( y = 1.00 \, \mathrm{m} \), since it is inside while \( q_1 = 4.00 \, \mathrm{nC} \) at \( x = 2.00 \, \mathrm{m} \) is outside. Thus, \( Q_{enc} = q_2 = -6.00 \, \mathrm{nC} \). The electric flux is \( \Phi = \frac{-6.00 \times 10^{-9}}{8.85 \times 10^{-12}} \approx -0.678 \times 10^{3} \, \mathrm{N \cdot m^2/C} \).

Step 4: Analyzing (c): Radius = 2.50 m

The spherical surface with a radius of 2.50 m encloses both charges \( q_1 \) and \( q_2 \) since both are located within 2.50 m from the origin. Thus, \( Q_{enc} = q_1 + q_2 = 4.00 \, \mathrm{nC} - 6.00 \, \mathrm{nC} = -2.00 \, \mathrm{nC} \). The electric flux is \( \Phi = \frac{-2.00 \times 10^{-9}}{8.85 \times 10^{-12}} \approx -0.226 \times 10^{3} \, \mathrm{N \cdot m^2/C} \).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It makes use of the concept of electric flux through closed surfaces. In essence, Gauss's Law links electric fields to the charges that create them. The law states that the total electric flux through a closed surface is directly proportional to the enclosed electric charge. Mathematically, Gauss's law is expressed as: \[ \Phi = \frac{Q_{enc}}{\varepsilon_0} \]Here, \(\Phi\) represents the electric flux through the surface, \(Q_{enc}\) is the net charge enclosed, and \(\varepsilon_0\) is the permittivity of free space. Through employing this law, you can efficiently calculate the electric field created by geometrically symmetric charge distributions, like point charges inside spherical surfaces.

Point Charge

A point charge is an idealized model of a charged particle. It is a single charge with a negligible size that simplifies calculations in electrostatics. The electric field around a point charge diminishes with distance, adhering to a simple inverse square law. This means that the farther you move away from a point charge, the weaker the electric field becomes.In the exercise, we have two point charges:

• The first charge, \(q_1 = 4.00\, \mathrm{nC}\), is placed at \(x = 2.00\, \mathrm{m}\).
• The second charge, \(q_2 = -6.00\, \mathrm{nC}\), is positioned at \(y = 1.00\, \mathrm{m}\).

These charges generate electric fields, which we calculate using Coulomb's law, at various points in space, including through surfaces such as spheres when using Gauss's Law.

Spherical Surface

A spherical surface is a three-dimensional surface where all points are equidistant from a central point. It is a circular symmetric shape, which simplifies calculations for electric flux when applying Gauss's Law. This symmetry allows us to easily deduce if point charges are enclosed or not based on their position relative to the sphere.Let's break down the different scenarios:

• For a sphere with a radius of 0.500 m, neither charge \(q_1\) nor \(q_2\) is enclosed, leading to zero electric flux.
• At a radius of 1.50 m, the sphere covers only the charge at \(y = 1.00\, \mathrm{m}\) (\(q_2 = -6.00\, \mathrm{nC}\)), resulting in a calculated negative flux.
• With a larger sphere radius of 2.50 m, both charges are enclosed. Here, the total charge is the algebraic sum of both charges, affecting the net electric flux across the spherical surface.

Permittivity of Free Space

Permittivity of free space, often symbolized as \(\varepsilon_0\), is a constant that quantifies the ability of a vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}\). This fundamental constant is used in calculating electric fields in free space and plays a central role in Gauss's Law.This value is crucial when computing electric flux through a surface. In the context of our exercise, \(\varepsilon_0\) is used for determining the flux generated by point charges through spherical surfaces. It's an integral part of the formula:\[ \Phi = \frac{Q_{enc}}{\varepsilon_0} \]By using permittivity of free space in these calculations, we are able to determine how effectively the vacuum of space supports forming an electric field.