Problem 10 - University Physics with Modern Physics

Question

A point charge \(q_1 =\) 4.00 nC is located on the \(x\)-axis at \(x =\) 2.00 m, and a second point charge \(q_2 = -\)6.00 nC is on the \(y\)-axis at \(y =\) 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?

Step-by-Step Solution

Answer

(a) 0 Nm²/C, (b) -679 Nm²/C, (c) -226 Nm²/C.

Step 1: Understand the Problem

We need to calculate the total electric flux through a sphere for different radii due to two charges located at specific points. Use Gauss's Law which states that the total electric flux through a closed surface is equal to the charge enclosed divided by the electric constant.

Step 2: Analyze the Sphere with Radius 0.500 m

For a sphere with radius 0.500 m centered at the origin, neither of the charges, located at (2.00 m, 0) and (0, 1.00 m), lie within the sphere. Therefore, the enclosed charge is 0 C, resulting in zero electric flux.

Step 3: Calculate Flux for Radius 1.50 m

With a sphere radius of 1.50 m, check the positions of the charges relative to the sphere. Only charge \( q_2 = -6.00 \) nC at (0, 1.00 m) is inside the sphere, while \( q_1 = 4.00 \) nC at (2.00 m, 0) is outside. Thus, the flux is \( \Phi = \frac{q_{ ext{enclosed}}}{\varepsilon_0} = \frac{-6.00 \times 10^{-9}}{8.85 \times 10^{-12}} \approx -679 \text{ Nm}^2/\text{C} \).

Step 4: Calculate Flux for Radius 2.50 m

For a sphere with radius 2.50 m, both charges \( q_1 = 4.00 \) nC and \( q_2 = -6.00 \) nC are within the sphere. The net enclosed charge is \( 4.00 - 6.00 = -2.00 \) nC. The flux is \( \Phi = \frac{-2.00 \times 10^{-9}}{8.85 \times 10^{-12}} \approx -226 \text{ Nm}^2/\text{C} \).

Key Concepts

Gauss's Law

Gauss's Law is an essential principle in electromagnetism, providing a link between electric flux and the charge enclosed by a surface. This law is expressed using the formula \( \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \), where \( \Phi \) is the electric flux, \( Q_{\text{enclosed}} \) is the total charge inside the closed surface, and \( \varepsilon_0 \) is the electric constant. This concept implies that electric flux only depends on the enclosed charge within the surface and not on the specific distribution or the location of the surface outside the charge.

In our exercise, Gauss's Law helps us determine the electric flux through a sphere for various radii centered at the origin. The placement of point charges relative to the sphere determines if they are enclosed or not, directly affecting the calculation of electric flux.

Point Charges

Point charges, such as those in this problem, are particles with an electric charge concentrated at a single point in space. In our scenario, one charge is located on the \(x\)-axis, and another on the \(y\)-axis. The effect of these point charges on an external electric field diminishes with distance due to the spherical symmetry of the surface considered, which Gauss's Law accounts for effectively.

The electric field generated by a point charge decreases with the square of the distance from the charge. In terms of Gauss's Law, only the charges within the Gaussian surface contribute to the electric flux, aligning perfectly with the properties of point charges acting in space.

Enclosed Charge

The enclosed charge refers to the sum of electric charges within a closed surface under consideration. It plays a pivotal role in computing the electric flux using Gauss's Law. In each scenario of our exercise, the radius of the sphere determines which charges are enclosed and, consequently, contribute to the electric flux.

For a sphere with radius 0.500 m, neither charge \( q_1 \) nor \( q_2 \) is inside, resulting in zero enclosed charge and zero electric flux. At a radius of 1.50 m, only charge \( q_2 = -6.00 \) nC is enclosed, contributing to the electric flux of \( -679 \text{ Nm}^2/\text{C} \). Finally, at a radius of 2.50 m, both charges are enclosed, resulting in a total flux consistent with the net enclosed charge of \(-2.00\) nC.

Electric Constant

The electric constant, also known as the permittivity of free space \( \varepsilon_0 \), is a fundamental physical constant used in electrostatics. It represents the ability to permit electric field lines in a vacuum and is instrumental in the application of Gauss's Law.

The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^{2}/\text{N m}^{2} \). This constant is crucial in determining the magnitude of the electric flux as it appears in the denominator of the Gauss's Law formula \( \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \). Therefore, understanding \( \varepsilon_0 \) allows us to accurately calculate the flux through any given Gaussian surface, as demonstrated in the step-by-step solution of our exercise.