/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 22.15. A point charge of \(+5.00... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

22.15. A point charge of \(+5.00 \mu C\) is located on the \(x\) -axis at \(x=4.00 \mathrm{m},\) next to a spherical surface of radius 3.00 \(\mathrm{m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{m} .\) (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{m}\) ) than on the far side (at \(x=-3.00 \mathrm{m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.

Short Answer

Expert verified
(a) 44950 N/C, (b) 918.3 N/C, (c) Flux is zero due to area and field density balance.

Step by step solution

01

Understand the Problem

We have a point charge of \(+5.00 \mu C\) located at \(x = 4.00 \, \text{m}\) and want to find the electric field at \(x = 3.00 \, \text{m}\) and at \(x = -3.00 \, \text{m}\) on a spherical surface centered at the origin with a radius of \(3.00 \, \text{m}\). We must then explain how the net flux through the sphere is zero even with field strength differences.
02

Calculate Electric Field at x=3.00 m

Using Coulomb's law, the electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by:\[E = \frac{k \cdot q}{r^2}\]where \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\). The distance \(r\) from the charge to \(x=3.00 \, \text{m}\) is \(4.00 \, \text{m} - 3.00 \, \text{m} = 1.00 \, \text{m}\). Thus, the electric field is:\[E = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-6})}{1.00^2} = 4.495 \times 10^4 \, \text{N/C}\]
03

Calculate Electric Field at x=-3.00 m

For \(x=-3.00 \, \text{m}\), the distance \(r\) from the charge is \(4.00 \, \text{m} + 3.00 \, \text{m} = 7.00 \, \text{m}\). Using Coulomb's law:\[E = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-6})}{7.00^2} = 9.183 \times 10^2 \, \text{N/C}\]
04

Explain Gauss's Law and Flux

Gauss's law states that the net electric flux through a closed surface is zero if there is no charge enclosed. Although the field is stronger at \(x=3.00 \, \text{m}\) than at \(x=-3.00 \, \text{m}\), the surface areas through which these fields penetrate differ. The electric field lines enter the near side more densely but exit the far side less densely over a larger area, keeping total flux balanced. The integral of \(E \, dA\) over the whole sphere, taking into account the angles and area, results in zero net flux.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 the study of electric fields and flux. It states that the net electric flux through any closed surface, also known as a Gaussian surface, is directly proportional to the net charge enclosed within that surface. Mathematically, it is expressed as: \[ \Phi_{E} = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( \Phi_{E} \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.
Gauss's Law is particularly useful for calculating electric fields when dealing with symmetrical charge distributions, such as spherical or cylindrical distributions. In the context of this exercise, even though the electric field strength differs at different points on the sphere, the net electric flux is zero because the sphere does not enclose any charge. This concept underscores how the distribution of electric field lines over the surface maintains balance, compensating for areas with higher intensity by covering wider regions at lower intensity.
Coulomb's Law
Coulomb's Law describes the electrostatic interaction between electric charges. This law states that the magnitude of the electric force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The equation is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the magnitude of the force between the charges, \( k \) is Coulomb's constant \((8.99 \times 10^9 \text{ Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
In the provided exercise, this law helps us determine the electric field magnitude at different locations relative to a point charge. By applying Coulomb's Law, we calculate that a point at \( x = 3.00 \text{ m} \) experiences a stronger electric field compared to a point at \( x = -3.00 \text{ m} \), due to the closer proximity to the point charge. This difference illustrates how electric fields diminish rapidly with distance from a source charge.
Net Electric Flux
Net electric flux refers to the sum of electric field lines passing through a closed surface, which provides insights into the presence and distribution of electric charges. It is computed by the surface integral of the electric field over a closed surface: \[ \Phi_{E} = \oint \mathbf{E} \cdot d\mathbf{A} \] The net electric flux through a closed surface is zero if there are no charges enclosed by that surface.
In the exercise, the sphere centered at the origin does not enclose the external point charge, even though electric field lines are present across the sphere's surface. On the side closer to the charge (at \( x = 3.00 \text{ m} \)), the field lines are denser, indicating a stronger electric field. However, these lines spread out as they pass through the sphere, balancing out the flux. The total flux remains zero because the field lines that enter the sphere eventually exit, maintaining equality between entering and exiting flux.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

22.12. (a) In a certain region of space, the volume charge density \(\rho\) has a uniform positive value. Can \(\overrightarrow{\boldsymbol{E}}\) be uniform in this region? Explain. (b) Suppose that in this region of uniform positive \(\rho\) there is a "bubble" within which \(\rho=0 .\) Can \(\vec{E}\) be uniform within this bubble? Explain.

22.43. A solid conducting sphere with radius \(R\) that carries poositive 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 \(r .\)

22\. 14. Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?(\mathrm{c})\) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

22.63. Positive charge \(Q\) is distributed uniformly over each of two spherical volumes with radius \(R\) . One sphere of charge is centered at the origin and the other at \(x=2 R\) (Fig. 2244 ). Find the magnitude and direction of the net electric field due to these two distributions of charge at the following points on the \(x\) -axis: (a) \(x=0 ;\) (b) \(x=R / 2 ;(c) x=R ;\) (d) \(x=3 R\) .

22.11. In a certain region of space, the electric field \(\overrightarrow{\boldsymbol{E}}\) is uniform. (a) Use Gauss's law to prove that this region of space must be electrically neutral; that is, the volume charge density \(\rho\) must be zero. (b) Is the converse true? That is, in a region of space where there is no charge, must \(\overrightarrow{\boldsymbol{E}}\) be uniform? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.