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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) \(3.955 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C}\), (b) \(-2.599 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C}\), (c) \(1.356 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C}\).

Step by step solution

01

Understanding Gauss's Law

Gauss's law states that the electric flux \( \Phi_E \) through a closed surface is equal to the charge \( Q \) enclosed by the surface divided by the permittivity of free space \( \varepsilon_0 \). The formula is given by: \[ \Phi_E = \frac{Q}{\varepsilon_0} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).
02

Calculating Flux for a Positive Charge

For the charge \( Q_1 = +3.5 \times 10^{-6} \, \text{C} \), apply Gauss's law: \[ \Phi_E = \frac{Q_1}{\varepsilon_0} = \frac{3.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 3.955 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \].
03

Calculating Flux for a Negative Charge

For the charge \( Q_2 = -2.3 \times 10^{-6} \, \text{C} \), apply Gauss's law: \[ \Phi_E = \frac{Q_2}{\varepsilon_0} = \frac{-2.3 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -2.599 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \].
04

Calculating Flux for Combined Charges

The total charge enclosed when both charges are present is \( Q_1 + Q_2 = +3.5 \times 10^{-6} + (-2.3 \times 10^{-6}) = 1.2 \times 10^{-6} \, \text{C} \). Using Gauss's law, the flux is: \[ \Phi_E = \frac{1.2 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.356 \times 10^{5} \, \text{N} \cdot \text{m}^2/\text{C} \].

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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 measure of the electric field passing through a surface. It helps us understand how much of the electric field is "flowing" through a given area. Imagine you have a net in a stream of water. The amount of water flowing through this net is analogous to electric flux, except with electric fields. When you calculate the electric flux, you are considering two main factors:
  • The strength and direction of the electric field.
  • The size and orientation of the surface through which the field lines pass.
The flux is represented by the symbol \( \Phi_E \). For a uniform electric field passing through a flat surface, the flux can be calculated via this formula:\[ \Phi_E = E \cdot A \cdot \cos(\theta) \]where \( E \) is the magnitude of the electric field, \( A \) is the area of the surface, and \( \theta \) is the angle between the field lines and the normal to the surface.
Closed Surface
A closed surface, in the context of Gauss's law, is any three-dimensional boundary that fully encloses a volume. Think of it like a balloon or a spherical shell, where there's a clear inside and outside. When we talk about calculating electric flux for a closed surface, we're interested in how the electric field interacts with the entire surrounding area. The significance of a closed surface comes into play with Gauss's law, which applies to these surfaces specifically.
  • The law tells us that the total electric flux through a closed surface relates directly to the charge inside the surface.
  • The shape of the surface doesn't matter as much as the charge enclosed by it.
This is why, even if your surface is oddly shaped, as long as it is closed, Gauss's law allows you to find the electric flux through it based on the enclosed charge alone.
Permittivity of Free Space
Permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It quantifies how an electric field is influenced in a vacuum and plays a critical role in Coulomb's law and Gauss's law.Here are some key points on permittivity of free space:
  • Its value is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).
  • It sets the scale for the force between two electric charges in a vacuum.
  • Permittivity affects how much electric field (flux) can be generated by charges in a vacuum.
In the context of Gauss's law, \( \varepsilon_0 \) appears in the denominator of the equation \( \Phi_E = \frac{Q}{\varepsilon_0} \). This equation links the electric flux through a closed surface to the charge \( Q \) inside, scaled by \( \varepsilon_0 \). As a universal constant, it ensures consistency in calculations involving electric fields.

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Most popular questions from this chapter

ssm Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{cm}\) and (b) \(x=+6.0 \mathrm{cm}\)

Conceptual Example 13 deals with the hollow spherical conductor in Figure \(18.30 .\) The conductor is initially electrically neutral, and then a charge \(+q\) is placed at the center of the hollow space. Suppose the conductor initially has a net charge of \(+2 q\) instead of being neutral. What is the total charge on the interior and on the exterior surface when the \(+q\) charge is placed at the center?

mmh A solid nonconducting sphere has a positive charge \(q\) spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is \(\frac{q}{\frac{4}{3} \pi R^{3}} .\) Use Gauss' law to show that the electric field at a point within the sphere at a radius \(r\) has a magnitude of \(\frac{q r}{4 \pi \epsilon_{0} R^{3}}\) (Hint: For a Gaussian surface, use a sphere of radius r centered within the solid sphere of radius. Note that the net charge within any volume is the charge density times the volume.)

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 \(\operatorname{sign}\) )?

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. (a) The object has a charge of \(+20.0 \mu \mathrm{C}\) and the instrument indicates that the electric force exerted on it is \(40.0 \mu \mathrm{N}\), due east. What are the magnitude and direction of the electric field? (b) What are the magnitude and direction of the electric field if the object has a charge of \(-10.0 \mu \mathrm{C}\) and the instrument indicates that the force is \(20.0 \mu \mathrm{N},\) due west?

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