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Chapter 22: Problem 10

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 these two point charges through a spherical surface centered at the origin and with radius (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

Short Answer

Expert verified
(a) 0, (b) 0, (c) -2.26 脳 10虏 Nm虏/C.

Step by step solution

01

Understand Gauss's Law

Gauss's Law states that the electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{enc} \) divided by the permittivity of free space \( \varepsilon_0 \):\[ \Phi = \frac{Q_{enc}}{\varepsilon_0} \]. To solve the problem, we need to find out how many charges are enclosed by the spherical surface for each given radius and then apply Gauss's Law.
02

Evaluate Spherical Surface of Radius 0.500 m

For a radius of 0.500 m, the sphere is centered at the origin and does not enclose any of the point charges, since both are located beyond this radius. Thus, the enclosed charge \( Q_{enc} = 0 \), and the electric flux \( \Phi = \frac{0}{\varepsilon_0} = 0 \).
03

Evaluate Spherical Surface of Radius 1.50 m

With a radius of 1.50 m, the sphere still does not enclose either charge, as \( q_1 \) is at \( x = 2.00 \text{ m} \) and \( q_2 \) is at \( y = 1.00 \text{ m} \), both lying outside the sphere. Therefore, \( Q_{enc} = 0 \), leading to an electric flux of \( \Phi = 0 \).
04

Evaluate Spherical Surface of Radius 2.50 m

For a radius of 2.50 m, the sphere encloses both point charges \( q_1 \) and \( q_2 \). Hence, we calculate the enclosed charge: \( Q_{enc} = q_1 + q_2 = 4.00 \text{ nC} - 6.00 \text{ nC} = -2.00 \text{ nC} \). Applying Gauss's Law, the flux is: \[ \Phi = \frac{-2.00 \times 10^{-9} \text{ C}}{\varepsilon_0} \], where \( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \), resulting in \( \Phi \approx -2.26 \times 10^2 \text{ Nm}^2/\text{C} \).

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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 crucial concept in electromagnetism, helping us understand how electric fields interact with surfaces. Consider it as the quantity of electric field lines passing through a given surface. This idea is pivotal in solving problems using Gauss's Law.
To visualize electric flux, imagine a light beam hitting a surface. The more perpendicular the light hits the surface, the stronger the intensity or flux. Similarly, the more electric field lines passing through a surface, the higher the electric flux.
Gauss's Law uses electric flux to relate closed surfaces and enclosed charges. In the context of a spherical surface, if no charge is enclosed, as in our example with spheres of radius 0.500 m and 1.50 m, the flux is zero. Conversely, a non-zero flux indicates the presence of an enclosed charge.
Point Charge
A point charge is an idealized charge located at a single point in space. It's like assuming the charge is concentrated at a tiny spot, allowing for simplification in calculations. Think of it like pinpointing the exact source of an electric field. Point charges are significant in theoretical problems because they allow limiting calculations to simpler forms without complex shapes involved.
In the problem provided, the point charges are located at summarily points on the x-axis and y-axis. By looking at their positions, we determine whether the sphere of interest encloses them, which defines if electric flux through the sphere is non-zero when using Gauss's Law.
  • Charge at \(x = 2.00 \text{m}\)
  • Charge at \(y = 1.00 \text{m}\)
These charges affect the calculation of electric flux depending on the radius of the sphere centered at the origin.
Enclosed Charge
The concept of an enclosed charge refers to any charge that lies within a surface when evaluating electric flux through that surface. Gauss's Law heavily relies on this concept, as it states that the total electric flux through a closed surface depends solely on the charge enclosed.
In our problem, we are dealing with spherical surfaces of different radii centered at the origin. The key step is determining which charges, if any, these surfaces enclose:
  • For a sphere with radius 0.500 m, no charge is enclosed, resulting in zero electric flux.
  • For a radius of 1.50 m, again, no charge falls within the sphere, making the electric flux zero.
  • At a 2.50 m radius, both point charges, one positive and one negative, are enclosed. This results in a net enclosed charge of \( Q_{enc} = -2.00 \text{nC}\)
By understanding the charges that lie within any surface, we can determine the necessary values when applying Gauss's Law to find electric flux.

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Most popular questions from this chapter

A very small object with mass \(8.20 \times 10^{-9} \mathrm{kg}\) and positive charge \(6.50 \times 10^{-9} \mathrm{C}\) is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density \(5.90 \times 10^{-8} \mathrm{C} / \mathrm{m}^{2} .\) The object is initially 0.400 \(\mathrm{m}\) from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 \(\mathrm{m}\) ?

A Uniformly Charged Slab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) -and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho .\) (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law,find the electric field due to the slab (magnitude and direction) at all points in space.

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.

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere with radius 0.355 m is 1750 N/C.(a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? (b) Calculate the electric field inside the sphere at a distance of 0.200 m from the center.

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 \(-35.0 \mu \mathrm{C} .\) 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.
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