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A point charge of \(9.20 \mathrm{nC}\) is located at the origin and a second charge of \(-5.00 \mathrm{nC}\) is located on the \(x\) axis at \(x=2.75 \mathrm{~cm} .\) Calculate the electric flux through a sphere centered at the origin with radius \(1.00 \mathrm{~m} .\) Repeat the calculation for a sphere of radius \(2.00 \mathrm{~m}\).

Short Answer

Expert verified
For the 1.00 m sphere, the flux is \( \approx 1.04 \times 10^3\, \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C} \) and for the 2.00 m sphere, \( \approx 4.75 \times 10^2\, \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C} \).

Step by step solution

01

Understanding Electric Flux and Gauss's Law

The electric flux \( \Phi \) through a closed surface is determined by Gauss's Law, which states \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \), where \( Q_{enc} \) is the net charge enclosed within the surface and \( \varepsilon_0 \) is the permittivity of free space.
02

Calculate Enclosed Charge for 1.00 m Sphere

The sphere with a radius of 1.00 m is centered at the origin. It only encloses the charge at the origin, which is \(9.20\, \mathrm{nC}\). The charge \(-5.00 \mathrm{nC}\) at \(x=2.75\, \mathrm{cm} \) is outside this sphere.
03

Calculate Electric Flux for 1.00 m Sphere

Using Gauss's Law, the electric flux through the surface is \( \Phi_{1.00} = \frac{9.20\times10^{-9}\, \mathrm{C}}{8.85\times10^{-12}\, \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2} \). Calculating this value gives \( \Phi_{1.00} \approx 1.04 \times 10^3\, \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C} \).
04

Calculate Enclosed Charge for 2.00 m Sphere

The sphere with a radius of 2.00 m encloses both charges, as they are both within this sphere. Thus, \( Q_{enc} = 9.20 \, \mathrm{nC} + (-5.00 \, \mathrm{nC}) = 4.20 \, \mathrm{nC} \).
05

Calculate Electric Flux for 2.00 m Sphere

Applying Gauss's Law again, the electric flux through the surface is \( \Phi_{2.00} = \frac{4.20\times10^{-9}\, \mathrm{C}}{8.85\times10^{-12}\, \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2} \). Calculating this value gives \( \Phi_{2.00} \approx 4.75 \times 10^2\, \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C} \).

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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. It relates the distribution of electric charge to the resulting electric field. The law states that the electric flux, \( \Phi \), through a closed surface is equal to the total charge, \( Q_{enc} \), enclosed by that surface divided by the permittivity of free space, \( \varepsilon_0 \). Mathematically, it is expressed as \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \).
This equation helps simplify the calculation of electric fields in situations where there is a high degree of symmetry, like spherical or cylindrical shapes. Overall, Gauss's Law is incredibly useful when learning about electrostatics, as it helps visualize how charges interact within different mediums and structures.
Point Charges
A point charge is an idealized model of a particle of charge that occupies a single point in space. It is a way to simplify complex problems by considering the charge, regardless of its actual size or shape, as a point to make calculations feasible.
Point charges are used to represent elementary particles, such as protons or electrons, or any smallest unit of charge in a larger system. Calculating interactions between point charges allows you to understand the fundamental forces at play in many electrical systems.
Enclosed Charge
The enclosed charge, \( Q_{enc} \), refers to the total electric charge within a given closed surface. When using Gauss's Law, it's crucial to accurately determine which charges lie within the closed surface (like a sphere) to find the electric flux.
In the scenario provided, for a sphere centered at the origin with a radius of \(1.00\, \mathrm{m}\), only the charge at the origin is enclosed. For the sphere with a radius of \(2.00\, \mathrm{m}\), both charges are enclosed.
  • For smaller spheres, only some charges may be enclosed.
  • For larger, encompassing spheres, nearly all nearby charges can be enclosed.
Understanding which charges are included is vital for correctly applying Gauss's Law.
Permittivity of Free Space
The permittivity of free space, represented by \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum of space. It is a crucial component in the equations governing electrostatics, including Coulomb's Law and Gauss's Law.
Its value is approximately \(8.85 \times 10^{-12}\, \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2\). This constant provides the link between electric charge and electric field in a vacuum.
Understanding the role of \( \varepsilon_0 \) helps with calculating electric flux, as it appears in the denominator of the Gauss's Law formula: \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \). It essentially scales how much electric field is produced by a given charge in free space.

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Most popular questions from this chapter

(II) A 15.0-cm-diameter nonconducting sphere carries a total charge of \(2.25 \mu \mathrm{C}\) distributed uniformly throughout its volume. Graph the electric field \(E\) as a function of the distance \(r\) from the center of the sphere from \(r=0\) to \(r=30.0 \mathrm{~cm}\)

(II) A long cylindrical shell of radius \(R_{0}\) and length \(\ell\) \(\left(R_{0} \ll \ell\right)\) possesses a uniform surface charge density (charge per unit area) \(\sigma\) (Fig. \(22-33\) ). Determine the electric field at points \((a)\) outside the cylinder \(\left(R>R_{0}\right)\) and \((b)\) inside the cylinder \(\left(0

(1) A uniform electric field of magnitude \(5.8 \times 10^{2} \mathrm{N} / \mathrm{Cpasses}\) through a circle of radius 13 \(\mathrm{cm} .\) What is the electric flux through the circle when its face is \((a)\) perpendicular to the field lines, \((b)\) at \(45^{\circ}\) to the field lines, and (c) parallel to the field lines?

(II) Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities \(\pm \sigma .\) Ignore edge effects and use Gauss's law to show \((a)\) that for points far from the edges, the electric field between the plates is \(E=\sigma / \epsilon_{0}\) and (b) that outside the plates on either side the field is zero. (c) How would your results be altered if the two plates were nonconductors?

(II) A point charge \(Q\) rests at the center of an uncharged thin spherical conducting shell. What is the electric field \(E\) as a function of \(r(a)\) for \(r\) less the radius of the shell, (b) inside the shell, and \((c)\) beyond the shell? \((d)\) Does the shell affect the field due to \(Q\) alone? Does the charge \(Q\) affect the shell?

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