/*! 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 9 Field from two charges ** A ch... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 9

Field from two charges ** A charge \(2 q\) is at the origin, and a charge \(-q\) is at \(x=a\) on the \(x\) axis. (a) Find the point on the \(x\) axis where the electric field is zero. (b) Consider the vertical line passing through the charge \(-q\), that is, the line given by \(x=a\). Locate, at least approximately, a point on this line where the electric field is parallel to the \(x\) axis.

Short Answer

Expert verified
The point on the x-axis where the electric field is zero is \(x=2a\). Moreover, the point on the line \(x=a\) where the electric field is parallel to the x-axis is approximately at \((a,卤\frac{a}{\sqrt{3}})\).

Step by step solution

01

Identify Variables

Let E1 be the field from the charge \(2q\) and E2 from \(-q\). Because the charges are on the x-axis, the direction of the electric fields will be along the x-axis.
02

Determine the Electric Field of Each Charge

The electric field from a point charge is given by \(E=k\frac{|q|}{r^2}\) where r is the distance from the charge. For \(E1\), \(q=2q\) and \(r=x\), resulting in \(E1=k\frac{2q}{x^2}\). For \(E2\), \(q=-q\) and \(r=x-a\), resulting in \(E2=-k\frac{q}{(x-a)^2}\). The negative sign comes from the negative charge.
03

Determine the Position Where the Electric Field is Zero

Set \(E1 + E2 = 0\) and solve for \(x\). This gives: \(k\frac{2q}{x^2} = k\frac{q}{(x-a)^2}\). Simplifying this equation gives two roots, \(x=0, 2a\). However, \(x=0\) is the position of the \(2q\) charge and should be discarded. Thus, the electric field is zero at \(x=2a\) on the \(x\)-axis.
04

Locate point on line \(x=a\) for part (b)

Here, the electric field is parallel to the x-axis when the y-component cancels out. At any point in the plane, the electric field has two components: one along the x-axis and one along the y-axis. The field from a charge always points directly away from that charge if positive or towards it if negative. Now, let鈥檚 consider a position \(P(a, y)\) on line \(x=a\). At this position, the x-component from \(2q\) cancels with the y-component from \(鈭抭\), leaving only the y-component from \(2q\) and the x-component from \(-q\). For these components to cancel, they must be equal. So, \(k\frac{2q}{y^2} = k\frac{q}{y^2 + a^2}\). This results in \(y =卤\frac{a}{\sqrt{3}}\), at least approximately. Note that the location along the \(y\)-direction should be positive because the electric field is parallel to the \(x\)-axis in the \(y>0\) region.

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 91影视!

Key Concepts

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

Point Charges
Point charges are fundamental units in the study of electric fields. They are theoretical constructs representing charged particles concentrated at a specific point in space. Each point charge has a magnitude, which quantifies the amount of electrical charge, and a sign, either positive or negative.
Understanding point charges is crucial to analyze electric fields because they serve as the basis for calculating how charges interact with one another. The influence, or field due to a point charge, extends radially outward and diminishes with distance.
  • A positive point charge repels other positive charges and attracts negative charges.
  • A negative point charge attracts positive charges and repels other negative charges.
The fields created by point charges can combine if there are multiple charges, and this combination principle is key to solving problems involving several charges.
Electric Field Direction
The direction of an electric field is determined by the nature of the charge that creates it. For a positive point charge, the field vectors point radially outward from the charge, while for a negative charge, they point inward towards the charge. This directional property follows from the definition of electric fields as being the force per unit positive charge.
When solving problems involving multiple charges, it's essential to consider the direction of the electric field contributed by each charge. Since electric fields are vectors, they have both magnitude and direction. To find the net electric field at a point, you should consider both components:
  • Combine the fields by adding vectorially, meaning you must pay attention to both the magnitude and direction.
  • For charges aligned along an axis, such as the x-axis, the direction can be simplified to positive or negative along that axis, making calculations straightforward.
Hence, understanding how direction works helps in determining the resultant field at any given point.
Electric Field Equations
Electric field equations provide the basis for calculating the field at a point due to a point charge. The basic formula for the electric field (E) generated by a point charge (q) is given by:\[E = k\frac{|q|}{r^2}\]where \(k\)is Coulomb's constant, \(q\) is the charge magnitude, and \(r\) is the distance from the charge to the point of interest.
This equation reveals that the strength of an electric field decreases with the square of the distance from the charge, a principle known as the inverse square law.
  • For a charge arrangement, the electric field at a point is the vector sum of fields due to each charge.
  • Significance of charge sign is crucial as it dictates whether the field contributes positively or negatively. Negative charges reverse the direction of the field vectors.
  • In specific problems, like the one described, solving for positions of zero net field involves setting up and solving equations where the sum of field contributions equals zero.
This enables understanding interactions among charges in terms of the resultant field.
Charge Distribution on X-Axis
Understanding charge distribution on the x-axis allows us to solve more complex problems by simplifying them into one-dimensional analysis. In our problem, charges are placed along the x-axis, which significantly reduces the complexity of electric field computations because directions align along a common line.
The benefit of this setup is that it simplifies the analysis of field interactions due to their linear arrangement. Steps to remember include:
  • Identifying each charge's contribution and how distance affects these contributions due to inverse square dependence.
  • Recognizing the significance of charge positioning, such as one charge at origin and another at a distinct x-coordinate, helping in calculating point of zero field.
  • Utilizing symmetry, if present, in charge distribution makes the location of zero-field points easier to determine, as seen in the given problem where symmetry helped find the field cancellation point.
This approach helps students visualize and solve electric field problems efficiently in instances of simplified geometrical charge distributions.

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

Energy of concentric shells * (a) Concentric spherical shells of radius \(a\) and \(b\), with \(a

Deriving the energy density ** Consider the electric field of two protons a distance \(b\) apart. According to Eq. (1.53) (which we stated but did not prove), the potential energy of the system ought to be given by $$ \begin{aligned} U &=\frac{\epsilon_{0}}{2} \int \mathbf{E}^{2} d v=\frac{\epsilon_{0}}{2} \int\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right)^{2} d v \\ &=\frac{\epsilon_{0}}{2} \int \mathbf{E}_{1}^{2} d v+\frac{\epsilon_{0}}{2} \int \mathbf{E}_{2}^{2} d v+\epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} d v \end{aligned} $$ where \(\mathbf{E}_{1}\) is the field of one particle alone and \(\mathbf{E}_{2}\) that of the other. The first of the three integrals on the right might be called the "electrical self-energy" of one proton; an intrinsic property of the particle, it depends on the proton's size and structure. We have always disregarded it in reckoning the potential energy of a system of charges, on the assumption that it remains constant; the same goes for the second integral. The third integral involves the distance between the charges. Evaluate this integral. This is most easily done if you set it up in spherical polar coordinates with one of the protons at the origin and the other on the polar axis, and perform the integration over \(r\) before the integration over \(\theta\). Thus, by direct calculation, you can show that the third integral has the value \(e^{2} / 4 \pi \epsilon_{0} b\), which we already know to be the work required to bring the two protons in from an infinite distance to positions a distance \(b\) apart. So you will have proved the correctness of Eq. (1.53) for this case, and by invoking superposition you can argue that Eq. (1.53) must then give the energy required to assemble any system of charges.

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Fluids

Read Explanation

Kinematics Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Magnetism

Read Explanation
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.