Problem 66 - Fundamentals of Physics

Question

A charged particle causes an electric flux of \(-750 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\) to pass through a spherical Gaussian surface of \(10.0 \mathrm{~cm}\) radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the charge of the particle?

Step-by-Step Solution

Answer

(a) -750 N·m²/C; (b) -6.64 × 10⁻⁹ C

Step 1: Understand Gauss's Law

Before solving the exercise, we need to recall Gauss's Law, which states that the electric flux \( \Phi \) through a closed surface is equal to the charge \( Q \) enclosed divided by the permittivity of free space \( \varepsilon_0 \). Mathematically, this is expressed as \( \Phi = \frac{Q}{\varepsilon_0} \).

Step 2: Flux through the Original Surface

We are provided with an electric flux of \(-750 \, \mathrm{N \cdot m^2/C}\) passing through a spherical Gaussian surface. This flux is calculated using Gauss's Law, and since the flux is only dependent on the charge enclosed and not the size of the surface, it remains constant even if the surface changes, provided that no additional charges enter the Gaussian surface.

Step 3: Calculate Flux through the New Surface

Knowing that the flux through the Gaussian surface only depends on the charge enclosed, if we double the radius of the Gaussian surface, the flux remains unchanged. Therefore, the flux through the new surface after doubling the radius is still \(-750 \, \mathrm{N \cdot m^2/C}\).

Step 4: Find the Charge of the Particle

To find the charge \( Q \) that results in this flux through the Gaussian surface, use the equation of Gauss's Law: \( \Phi = \frac{Q}{\varepsilon_0} \). Rearrange this equation to solve for \( Q \): \( Q = \Phi \cdot \varepsilon_0 \). Substituting the values, \( Q = -750 \, \mathrm{N \cdot m^2/C} \times 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \), which gives \( Q \approx -6.64 \times 10^{-9} \, \mathrm{C} \).

Key Concepts

Electric Flux

Electric flux is a measure of how much electric field passes through a given surface. Imagine the electric field as invisible lines radiating from a charged object. Electric flux tells us how many of these lines pass through a certain area.

• If lots of lines pass through, the electric flux is high.
• If few lines pass through, the electric flux is low.

When we talk about electric flux mathematically, it's given by the formula: \( \Phi = \vec{E} \cdot \vec{A} \), where \( \vec{E} \) is the electric field, and \( \vec{A} \) is the area vector of the surface. But in Gauss's Law, we use a spherical surface where electric field lines radiate symmetrically. The formula becomes simpler, and directly relates to the charge enclosed and permittivity of free space: \( \Phi = \frac{Q}{\varepsilon_0} \). This formula tells us that, for a given charge, the electric flux doesn't depend on the size of the surface, but only on the amount of charge inside it.

Gaussian Surface

A Gaussian surface is an imaginary boundary we use to simplify complex electric field calculations, especially when using Gauss's Law.

• It's typically chosen to make math easier. For example, a sphere is often a good choice because it's symmetrical.
• This surface is just a boundary. It doesn't interact with electric fields physically.

Imagine a balloon that can resize but doesn't physically exist. For this exercise, the Gaussian surface is spherical and centered on the charge causing the electric field. This symmetry allows us to use Gauss's Law quite effectively as it simplifies the math to relate the charge enclosed directly to the electric flux through the surface. It's critical to remember that the electric flux only depends on charge inside the Gaussian surface, not its actual shape or size. As in the exercise, doubling the radius does not change the flux, demonstrating that Gauss's Law is highly reliant on charge rather than geometry.

Charge Enclosed

Charge enclosed refers to the total amount of electric charge found inside a Gaussian surface. This concept is vital when applying Gauss's Law.

• Gauss's Law only cares about the charge inside the surface. Any external charges are irrelevant to the flux calculation.
• This understanding helps in greatly simplifying complex field problems by just focusing on what’s inside, ignoring what's outside.

In our context, knowing the amount of charge enclosed allows us to compute the flux using the formula \( \Phi = \frac{Q}{\varepsilon_0} \). Here, \( Q \) is the charge inside the surface, and \( \varepsilon_0 \) is the permittivity of free space. Here's the beauty: once you determine the charge, the flux remains constant, even if the Gaussian surface grows or shrinks. In the exercise provided, the charge calculation—\(-6.64 \times 10^{-9} \, \mathrm{C}\)—helps us find the electric flux through various Gaussian surfaces enclosing the charge, staying consistent with the principle of Gauss's Law.