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You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?

Short Answer

Expert verified
The electric flux is equal to the charge divided by the vacuum permittivity.

Step by step solution

01

Understanding the Electric Field and its Flux

To find the electric flux through the surface of a sphere, we start by understanding that the electric flux (\( \Phi \)) through a closed surface surrounding a point charge \( q \) is given by Gauss's Law: \[\Phi = \frac{q}{\varepsilon_0}\]where \( \varepsilon_0 \) is the vacuum permittivity, approximately \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \). The sphere has a radius of \( 0.150 \, m \), and the electric field, \( E \), is \( 1.25 \times 10^6 \, \mathrm{N/C} \).

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism that relates the distribution of electric charge to the resulting electric field. It states that the total electric flux through a closed surface is directly proportional to the charge enclosed by the surface. This relationship can be expressed with the equation:\[ \Phi = \frac{q}{\varepsilon_0} \]where
  • \(\Phi\) is the electric flux,
  • \(q\) is the total charge within the surface, and
  • \(\varepsilon_0\) is the vacuum permittivity, a constant.
Gauss's Law is particularly useful because it simplifies the calculation of electric fields when symmetry is present. By using a Gaussian surface, like a sphere around a point charge, calculations become more straightforward. This law helps us understand the strength and behavior of electric fields in spatial regions surrounding electrical charges.
Electric Field
An electric field ( \(E\) ) is a vector field that surrounds electric charges and exerts force on other charges within the field. The electric field due to a point charge can be calculated using the formula:\[ E = \frac{k \, q}{r^2} \]where
  • \(E\) is the electric field strength,
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2}\),
  • \(q\) is the point charge, and
  • \(r\) is the distance from the charge.
Electric fields represent how charges interact in space, and they can vary in strength and direction. In problems like the one surrounding a point charge, the electric field tells us how any test charge placed in this field would behave—attracting or repelling depending on the signs of the charges involved. Understanding the concept of an electric field is crucial in predicting the behavior of charged particles in various physical situations.
Point Charge
A point charge is an idealized model of a charged particle that is considered to have all its charge concentrated at a single point in space. While real charges have spatial dimensions, the point charge approximation simplifies calculations in physics, especially when determining electric fields and forces. Point charges are used because they eliminate complications arising from the geometry of a charged object. In the context of Gauss's Law, when considering the flux through a spherical surface with a point charge at its center, the electric field is uniform across the sphere's surface. This simplification allows us to calculate electric flux and fields without needing to worry about irregular charge distributions. The concept of a point charge is fundamental in electrostatics, making problems involving electric fields much more manageable.
Vacuum Permittivity
Vacuum permittivity ( \(\varepsilon_0\)), also known as the electric constant, is a measure of the ability of the vacuum to permit electric field lines. It appears in Coulomb's law as well as Gauss's Law and defines how much electric field "spreads out" in free space.This constant, approximately \(8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}\), plays a crucial role in determining the strength of the electric force between charges separated by a vacuum. It's part of the relation:\[ \Phi = \frac{q}{\varepsilon_0} \]without which we cannot compute electric flux or the influence of electric fields in a vacuum.Understanding vacuum permittivity is important for comprehending the behavior of electric fields in free space and calculating forces between charged particles with precision. It is one of the fundamental constants of nature and essential in the study of electromagnetism.

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Most popular questions from this chapter

A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 7.50 nC of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

A very long uniform line of charge has charge per unit length 4.80\(\mu \mathrm{C} / \mathrm{m}\) and lies along the \(x\) -axis. A second long uniform line of charge has charge per unit length \(-2.40 \mu \mathrm{C} / \mathrm{m}\) and is parallel to the \(x\) -axis at \(y=0.400 \mathrm{m} .\) What is the net electric field (magnitude and direction) at the following points on the \(y\) -axis: (a) \(y=0.200 \mathrm{m}\) and ( b \() y=0.600 \mathrm{m} ?\)

A hemispherical surface with radius \(r\) in a region of uniform electric field \(\vec{E}\) has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

A solid conducting sphere with radius \(R\) that carries positive charge \(Q\) is concentric with a very thin insulating shell of radius 2\(R\) that also carries charge \(Q .\) The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction in each of the regions \(0 < r < R, R < r < 2 R,\) and \(r > 2 R .(\) b) Graph the electric-field magnitude as a function of \(r .\)

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