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A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{~m}\). What is the total electric flux due to the two point charges through a spherical surface centered at the origin and with ra\(\operatorname{dius}(\mathrm{a}) 0.500 \mathrm{~m},(\mathrm{~b}) 1.50 \mathrm{~m},(\mathrm{c}) 2.50 \mathrm{~m} ?\)

Short Answer

Expert verified
The total electric flux through spherical surfaces with radii \(a = 0.50 m\), \(b = 1.50 m\), and \(c = 2.50 m\) are \(0 N*m^2/C\), \(-677.96 N*m^2/C\), and \(-225.99 N*m^2/C\) respectively.

Step by step solution

01

Identify charges enclosed for different spherical surfaces

For \(a = 0.50 m\), no charges are enclosed within the sphere. For \(b = 1.50 m\), only the charge \(q_2 = -6.00 nC\) is enclosed. For \(c = 2.50 m\), both the charges are enclosed within the sphere.
02

Apply Gauss's law

According to Gauss's law, the electric flux, \(\Phi = Q/\epsilon_0\) where \(Q\) is the charge enclosed and \(\epsilon_0\) is the permittivity of vacuum with a value of \(8.85 × 10^-12 C^2/N∙m^2\).
03

Calculate the total electric flux for each case

For the sphere \(a = 0.50 m\), since no charge is enclosed, the total electric flux is 0. For the sphere \(b = 1.50 m\), the total electric flux is \((-6.00×10^-9 C) / (8.85×10^-12 C^2/N∙m^2) = -677.96 N∙m^2/C\). For the sphere \(c = 2.50 m\), the total charge enclosed is \((4.00 nC - 6.00 nC) = -2.00 nC\), hence the total electric flux is \((-2.00×10^-9 C) / (8.85×10^-12 C^2/N∙m^2) = -225.99 N∙m^2/C\).

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Key Concepts

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

Gauss's Law
Imagine wrapping a bubble around a charge; how would the charge interact with the bubble's surface? This is where Gauss's Law becomes a boon to physicists. It simplifies the work needed to determine the electric field around symmetrical charge distributions by relating the electric flux flowing out of a closed surface to the charge enclosed by that surface. In essence, for any closed surface, the net flux (\( \text{Phi} \)) puncturing through is proportional to the net charge (\( Q \) inside the surface.

Gauss's Law is mathematically expressed as \(\text{Phi} = Q/\epsilon_{0}\), where \(\text{Phi}\) represents the electric flux, \(\text{Q}\) stands for the total charge enclosed within the 'Gaussian surface', and \(\text{eps}ilon_{0}\) is the permittivity of the vacuum. It essentially tells us that whether we have a single point charge or a complex distribution of charges, as long as we can define a 'Gaussian surface' that encapsulates all the charges, the flux calculation remains fairly straightforward.

This law is particularly valuable because it lets us calculate electric fields and fluxes with relative ease—provided the charge distribution has a high degree of symmetry, such as spherical, cylindrical, or planar symmetry.
Point Charge
Zoom in on a charge until it looks like a speck with no dimensions—that's what physicists refer to as a point charge. This tiny dot of charge is central to many electrostatic concepts because it allows us to simplify complex problems. In reality, charges have spatial dimensions, but in many cases, considering them as point charges does not significantly affect the accuracy of calculations while drastically reducing the complexity.

The electric field \(\text{E}\) produced by a point charge is radially outward (repulsive) for positive charges and radially inward (attractive) for negative charges. The strength of this electric field is determined by Coulomb's Law, and it diminishes with the square of the distance from the charge. Because a point charge is so simple, it's an ideal candidate around which to apply Gauss's Law, helping us to calculate electric flux through closed surfaces surrounding it. The exercise provided is an excellent example of how point charges can be evaluated by applying Gauss's Law to different spherical surfaces centered at the origin.
Permittivity of Vacuum
Floating in space, away from any matter, the permittivity of vacuum (\( \text{eps}_{0} \) is a constant that characterizes how an electric field permeates the vacuum of free space. Permittivity in general measures how much electric field is 'allowed' through a material—and in a vacuum, it has a value of approximately \(8.85 × 10^{-12} C^{2}/N∙m^{2}\).Its role in Gauss's Law and electric flux is crucial: it serves as the proportionality constant that relates the amount of charge to the resulting electric flux. This universal value assures that the electric field in a vacuum behaves predictably and consistently, forming the foundational bedrock upon which many electrostatic calculations, including the ones in our practice problem, are grounded. The permittivity of vacuum is a gateway to understanding more complex materials, where the permittivity will vary, highlighting the influence of medium on electric properties.

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Most popular questions from this chapter

It was shown in Example 21.10 (Section 21.5 ) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E=\lambda / 2 \pi \epsilon_{0} r .\) Consider an imaginary cylinder with radius \(r=0.250 \mathrm{~m}\) and length \(l=0.400 \mathrm{~m}\) that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda=3.00 \mu \mathrm{C} / \mathrm{m} .\) (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r=0.500 \mathrm{~m} ?\) (c) What is the flux through the cylinder if its length is increased to \(l=0.800 \mathrm{~m} ?\)

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{~cm}\), giving it a charge of \(-49.0 \mu \mathrm{C}\). Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) \(5.00 \mathrm{~cm}\) outside the surface of the paint layer.

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 \(02 R\). (b) Graph the electric-field magnitude as a function of \(r\)

A long line carrying a uniform linear charge density \(+50.0 \mu \mathrm{C} / \mathrm{m}\) runs parallel to and \(10.0 \mathrm{~cm}\) from the surface of \(\mathrm{a}\) large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

(a) A conducting sphere has charge \(Q\) and radius \(R .\) If the electric field of the sphere at a distance \(r=2 R\) from the center of the sphere is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field of the sphere at \(r=4 R ?\) (b) A very long conducting cylinder of radius \(R\) has charge per unit length \(\lambda\). Let \(r\) be the perpendicular distance from the axis of the cylinder. If the electric field of the cylinder at \(r=2 R\) is \(1400 \mathrm{~N} / \mathrm{C},\) what is the electric field at \(r=4 R ?\) (c) A very large uniform sheet of charge has surface charge density \(\sigma .\) If the electric field of the sheet has a value of \(1400 \mathrm{~N} / \mathrm{C}\) at a perpendicular distance \(d\) from the sheet, what is the electric field of the sheet at a distance of \(2 d\) from the sheet?

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