/*! 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 25 A positive charge \(q\) is fixed... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 25

A positive charge \(q\) is fixed at the point \(x=0, y=0,\) and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\) . (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\) -axis as a function of the coordinate \(x\) . Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\) -axis is \(V=0 ?(\text { d) Graph }\) \(V\) at points on the \(x\) -axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a\) (e) What does the answer to part \((b)\) become when \(x \gg a ?\) Explain why this result is obtained.

Short Answer

Expert verified
The potential is zero at \( x = -a \) and \( x = \frac{3a}{2} \) on the x-axis.

Step by step solution

01

Diagram the charges

To show the positions of the charges in a diagram, plot the positive charge \( q \) at the origin \((x=0, y=0)\) and the negative charge \(-2q\) at the point \((x=a, y=0)\) on a coordinate plane. The two charges lie on the x-axis.
02

Derive the potential expression

The electric potential at a point \( x \) on the x-axis due to a charge \( q \) at position \( x_0 \) is given by \( V = \frac{k q}{|x - x_0|} \). Therefore, for the charge \( q \) at \( x=0 \), the potential is \( V_1 = \frac{k q}{|x|} \), and for the charge \(-2q\) at \( x=a \), the potential is \( V_2 = \frac{-2k q}{|x-a|} \). The total potential \( V \) at any point \( x \) is then \( V = V_1 + V_2 = \frac{k q}{|x|} + \frac{-2k q}{|x-a|} \).
03

Find where potential V = 0

Set the total potential expression to zero: \( \frac{k q}{|x|} + \frac{-2k q}{|x-a|} = 0 \). Simplify to find: \( \frac{1}{|x|} = \frac{2}{|x-a|} \). Solving gives two cases: \( x = -a \) and \( x = \frac{3a}{2} \). These are the positions on the x-axis where the potential is zero.
04

Graph V as a function of x

To plot the graph of potential \( V \) as a function of \( x \) from \( x=-2a \) to \( x=2a \), calculate \( V \) at various points in this range using the expression derived in Step 2. The graph will show the potential approaching zero far from the charges and having changes in sign due to the different contributions of each charge as they are approached.
05

Analyze V for x much greater than a

When \( x \gg a \), both \( x \) and \( x-a \) are approximately equal to \( x \). The potential simplifies to \( V \approx \frac{k q}{x} - \frac{2k q}{x} = \frac{-k q}{x} \). This occurs because the negative charge dominates; far from the origin, it behaves like a single charge. Thus, \( V \) approximates the potential of a single effective charge \(-q\) positioned far along \( x \).

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.

Coulomb's Law
Coulomb's Law is a fundamental principle in physics that describes the force between two point charges. The law states that the electric force \( F \) between two charges is directly proportional to the product of the charges' magnitudes and inversely proportional to the square of the distance between their centers. The mathematical expression for Coulomb's Law is given by:
  • \( F = k \frac{|q_1 q_2|}{r^2} \) where \( F \) is the force, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.
The direction of the force is along the line joining the charges:
  • If both charges have the same sign, the force is repulsive, pushing them apart.
  • If the charges have opposite signs, the force is attractive, pulling them together.
Understanding this concept helps us predict how charges interact at a distance and is foundational for many topics in electromagnetism.
Point Charges
Point charges are idealized models of charges used in physics to simplify problems involving electric fields and forces. They are considered to have no physical dimensions; instead, all of the charge is assumed to be concentrated at a single point in space. This abstraction is useful because:
  • It allows for easy application of Coulomb's Law, as seen in our original problem where the charges were fixed at certain locations along the x-axis.
  • Point charges create radial electric fields, which are symmetrical and point directly towards or away from the charge, depending on its sign.
Despite being theoretical, point charges offer a great way to approach complex scenarios, such as calculating the electric field or electric potential of distributed charges by considering them as a series of point charges.
Electric Field
The electric field is a concept that describes the influence a charge exerts on other charges in its vicinity. It is a vector field, meaning it has both magnitude and direction at every point in space. The strength of the electric field \( E \) produced by a point charge \( q \) at a distance \( r \) from the charge is given by:
  • \( E = k \frac{|q|}{r^2} \)
  • The direction of the electric field is radially outward for positive charges and radially inward for negative charges.
The electric field represents the force per unit charge that a positive test charge would experience if it were placed in the field. This concept is especially useful in determining how electric charges influence each other without directly considering their interactions, as electric field lines can be sketched to visualize the field's behavior, especially when dealing with multiple charges, like in our exercise.Understanding electric fields helps in visualizing the complex interactions in electrostatics and simplifying the calculation of forces and potentials.

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

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 \(\mathrm{cm}\) . (a) If the nsurface charge density for each plate has magnitude 47.0 \(\mathrm{nC} / \mathrm{m}^{2}\) , what is the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

A potential difference of 480 \(\mathrm{V}\) is established between large, parallel, metal plates. Let the potential of one plate be 480 \(\mathrm{V}\) and the other be 0 \(\mathrm{V}\) . The plates are separated by \(d=1.70 \mathrm{cm} .\) (a) Sketch the equipotential surfaces that correspond to \(0,120,\) \(240,360,\) and 480 \(\mathrm{V}\) . (b) In your sketch, show the electric field lines. Does your sketch confirm that the field lines and equipotential surfaces are mutually perpendicular?

The electric potential \(V\) in a region of space is given by $$ v(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C}\) test charge moves from the point \((x, y, z)=\) 1.50-\muC test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{J}\) . Determine \(A\) . (c) Determine the electric field at the point \((0,0,0.250\) m). (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{V}\) and \(y=2.00 \mathrm{m} ?\)

Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 \(\mathrm{V}\) higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode.

Alpha particles \(\left(\text { mass }=6.7 \times 10^{-27} \mathrm{kg}, \text { charge }=+2 e\right)\) are shot directly at a gold foil target. We can model the gold nucleus as a uniform sphere of charge and assume that the gold does not move. (a) If the radius of the gold nucleus is \(5.6 \times\) \(10^{-15} \mathrm{m}\) , what minimum speed do the alpha particles need when they are far away to reach the surface of the gold nucleus? (lanore relativistic effects.) (b) Give good physical reasons why we can ignore the effects of the orbital electrons when the alpha particle is (i) outside the electron orbits and (ii) inside the electron orbits.
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Quantum Physics

Read Explanation

Medical Physics

Read Explanation

Fields in Physics

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.