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A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if the collection consists of (a) a single \(+3.5 \times 10^{-6} \mathrm{C}\) charge, (b) a single \(-2.3 \times 10^{-6} \mathrm{C}\) charge, and (c) both of the charges in (a) and (b).

Short Answer

Expert verified
(a) \(3.95 \times 10^{5} \mathrm{Nm}^2/\mathrm{C}\), (b) \(-2.60 \times 10^{5} \mathrm{Nm}^2/\mathrm{C}\), (c) \(1.36 \times 10^{5} \mathrm{Nm}^2/\mathrm{C}\).

Step by step solution

01

Understand Gauss's Law

Gauss's Law states that the electric flux \( \Phi_E \) through a closed surface is proportional to the charge enclosed, \( Q_{enclosed} \). Mathematically, it is given by \( \Phi_E = \frac{Q_{enclosed}}{\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \).
02

Calculate the Flux for +3.5 µC

For the charge \( +3.5 \times 10^{-6} \mathrm{C} \), the electric flux through the spherical surface is calculated using Gauss's Law: \( \Phi_E = \frac{3.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 3.95 \times 10^{5} \mathrm{Nm}^2/\mathrm{C} \).
03

Calculate the Flux for -2.3 µC

For the charge \( -2.3 \times 10^{-6} \mathrm{C} \), apply Gauss's Law: \( \Phi_E = \frac{-2.3 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -2.60 \times 10^{5} \mathrm{Nm}^2/\mathrm{C} \).
04

Calculate the Net Flux for Both Charges

When both charges are enclosed, calculate total charge: \( (3.5 - 2.3) \times 10^{-6} = 1.2 \times 10^{-6} \mathrm{C} \). Then, find the electric flux: \( \Phi_E = \frac{1.2 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.36 \times 10^{5} \mathrm{Nm}^2/\mathrm{C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Flux
Electric flux is a fundamental concept in electromagnetism and forms the cornerstone of Gauss's Law. It essentially measures the number of electric field lines passing through a given surface. This is crucial because it helps us understand how electric fields interact with surfaces.

The electric flux \( \Phi_E \) through a closed surface is mathematically defined as:
  • \( \Phi_E = \int \mathbf{E} \cdot d\mathbf{A} \)
where \( \mathbf{E} \) is the electric field, and \( d\mathbf{A} \) is a differential area on the surface with an outward normal vector.

For a uniform field and a flat surface perpendicular to the field lines, the dot product simplifies to multiplication. However, for a curved surface like a sphere, the integration evaluates the total flux. Importantly, Gauss's Law tells us that the total electric flux through any closed surface is directly proportional to the net charge enclosed within that surface.
Using the formula \( \Phi_E = \frac{Q_{enclosed}}{\varepsilon_0} \), where \( Q_{enclosed} \) is the total charge within the surface, allows calculation of flux effectively.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a key constant in electromagnetism. It characterizes how electric fields propagate in a vacuum, and is critical in calculations involving electric fields and flux.

The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \). This value appears in Coulomb's law and in the expression for electric flux, thereby linking it to the equilibrium interaction between electric charges.
  • The smaller the value of \( \varepsilon_0 \), the stronger the electric field generated by a given charge, and vice versa.
Understanding \( \varepsilon_0 \) is crucial for appreciating how effective electric forces operate within different media, compared to free space. In practical terms, it helps us recognize the baseline against which materials' permittivity is measured.
Spherical Surface
A spherical surface is particularly interesting in the context of Gauss's Law because of its symmetry. When dealing with electric fields and charges, symmetry significantly simplifies calculations.

For a charged spherical surface, the electric field is radially symmetric. This means that at every point on the surface, the field lines are perpendicular to the surface and have the same magnitude. Such symmetry allows us to use Gauss's Law effectively to calculate the electric flux.
  • Since every point on a spherical surface is equidistant from the center of a sphere, calculations become surprisingly straightforward.
Using spheres also often simplifies integration. Since the electric field remains constant across the sphere (assuming uniform charge distribution), the complex integral for flux across open surfaces reduces considerably.
This simplification benefits from the geometric property of spheres, making them ideal for theoretical problems and real-life applications in physics that involve electric flux and field calculations.

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Most popular questions from this chapter

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol},\) and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2}\), where \(q_{1}=\) \(+3.5 \mu \mathrm{C} .\) They are separated by a distance of \(0.26 \mathrm{m},\) and particle 1 experiences an attractive force of \(3.4 \mathrm{N} .\) What is \(q_{2}\) (magnitude and \(\operatorname{sign}\) )?

Iron atoms have been detected in the sun's outer atmosphere, some with many of their electrons stripped away. What is the net electric charge (in coulombs) of an iron atom with 26 protons and 7 electrons? Be sure to include the algebraic sign ( \(+\) or \(-\) ) in your answer.

Four identical metallic objects carry the following charges: +1.6 \(+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into \(\mathrm{con}-\) tact, so that each touches the others. Then they are separated. (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

Two particles, with identical positive charges and a separation of \(2.60 \times 10^{-2} \mathrm{m},\) are released from rest. Immediately after the release, particle 1 has an acceleration \(\overrightarrow{\mathbf{a}}_{1}\) whose magnitude is \(4.60 \times 10^{3} \mathrm{m} / \mathrm{s}^{2},\) while particle 2 has an acceleration \(\overrightarrow{\mathbf{a}}_{2}\) whose magnitude is \(8.50 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\). Particle 1 has a mass of \(6.00 \times 10^{-6} \mathrm{kg} .\) Find \((\mathrm{a})\) the charge on each particle and \((\mathrm{b})\) the mass of particle 2.

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