/*! 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 12 A positive charge \(q\) is fixed... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 12

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 \(V=0 ?\) (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
(a) Charges are shown at (0, 0) and (a, 0). (c) \(V=0\) at \(x = -a\) and \(x = \frac{2}{3} a\). (e) \(V \approx -\frac{kq}{x}\) when \(x \gg a\).

Step by step solution

01

Diagram of Charge Positions

Draw a horizontal line to represent the x-axis. Place a positive charge \(q\) at the origin \((0, 0)\) and a negative charge \(-2q\) at the point \((a, 0)\) on the x-axis.
02

Potential at Points on the X-axis

The electric potential \(V\) due to a point charge \(Q\) is given by \(V = \frac{kQ}{r}\), where \(r\) is the distance from the charge, and \(k\) is Coulomb's constant. For any point \((x, 0)\) on the x-axis, the potential due to charge \(q\) is \(V_1 = \frac{kq}{|x|}\), and the potential due to charge \(-2q\) is \(V_2 = -\frac{2kq}{|x-a|}\). The total potential is \(V = V_1 + V_2 = \frac{kq}{|x|} - \frac{2kq}{|x-a|}\).
03

Find Positions Where V = 0

Set the expression for \(V\) to zero: \(\frac{kq}{|x|} - \frac{2kq}{|x-a|} = 0\). Solving for \(x\) gives \(\frac{1}{|x|} = \frac{2}{|x-a|}\). Solving the equation \(|x-a| = 2|x|\), we get potential null points at \(x = -a\) and at some point between 0 and \(a\) which require solving \((x-a)^2 = 4x^2\), which on solving yields \(x = \frac{2}{3}a\).
04

Graph V as a Function of x

Sketch the expression \(V = \frac{kq}{|x|} - \frac{2kq}{|x-a|}\) over the domain \(x = [-2a, 2a]\). Note that \(V\) approaches negative infinity as \(x\) approaches 0 from the right and positive from the left, and positive infinity as \(x\) approaches \(a\) from the left, and rapidly decreases to negative after \(x = a\).
05

Analyze V When x 鈮� a

When \(x \gg a\), the terms \(x-a \approx x\), resulting in \(V \approx \frac{kq}{x} - \frac{2kq}{x} = -\frac{kq}{x}\). This implies that at large distances, the charges behave as a net point charge equivalent to \(-q\) at the origin, because these distances can no longer resolve the individual contribution of each charge due to their relative proximity.

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 cornerstone principle in electrostatics. It describes the force between two charged objects. The law states that the electric force (F) between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
  • \[ F = k \frac{|q_1 q_2|}{r^2} \]
where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them. This law is crucial for calculating interactions in systems like the one described in the exercise, where we consider properties derived from charge interactions, such as electric potential and fields.
Electric Charge Distribution
In any system involving charges, understanding the distribution of charge is essential to analyze electric fields and potentials. In our exercise, we have two charges:
  • A positive charge \(q\) at the origin \((0, 0)\)
  • A negative charge \(-2q\) at \((a, 0)\)
This specific arrangement defines how potentials and fields behave around these charges. The total electric potential at any point is influenced by the relative position and magnitude of these charges, determining how potential varies along the x-axis. By understanding charge distribution, we can predict electric field lines, potential contours, and neutrality points on the axis.
Electric Field Potential
Electric potential is a measure of the work done in moving a unit positive charge from a reference point to a specific point in the field without any acceleration. For point charges, the potential (V) is calculated using the formula:
  • \[ V = \frac{kQ}{r} \]
where \(Q\) is the charge and \(r\) is the distance from the charge. In the exercise, for a point \((x, 0)\) on the x-axis, the total potential is the sum of potentials from both charges:
  • \[ V = \frac{kq}{|x|} - \frac{2kq}{|x-a|} \]
This expression shows how potentials from different charges combine, sometimes canceling each other out, which is key when finding points where the potential \(V = 0\).
Electric Dipole Moment
An electric dipole consists of two equal and opposite charges separated by a distance. It is characterized by its dipole moment \(\vec{p}\), a vector pointing from the negative to the positive charge. The dipole moment is given by:
  • \[ \vec{p} = q \times \vec{d} \]
where \(q\) is the magnitude of one of the charges, and \(\vec{d}\) is the displacement vector between the charges. In a simplified context like when \(x \gg a\), the charges in the exercise act like a dipole averaged to a single point charge potential,
  • potential \[-\frac{kq}{x}\] because at large distances, we observe the cumulative effect rather than individual contributions. Studying dipoles helps us understand diverse phenomena in electric fields, ranging from microscopic interactions in molecules to macroscopic field analysis.

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

An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

(a) Five large identical metal plates of area \(A\) are arranged horizontally, parallel to each other, with gap \(d\) in between adjacent plates. Counted from the top, they are given charge \(+q,+2 q\), \(-3 q,-q,+3 q\) respectively. Find the potential difference between the topmost and bottommost plates. (b) What charge will flow through a conducting wire if the topmost and bottommost plates are shorted?

A conducting sphere of radius \(R\) has two spherical cavities of radius \(a\) and \(b\). The cavities have charges \(q_{a}\) and \(q_{b}\) respectively at their centres. Find : distance \(r\) (a) The electric field and electric potential at a (i) \(r\) (distance from \(O\), the centre of sphere \(>R\) ) (ii) \(r\) (distance from \(B\), the centre of cavity \(b)

A very long wire carries a uniform linear charge density \(\lambda\). Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed \(2.50 \mathrm{~cm}\) from the wire and the other probe is \(1.00 \mathrm{~cm}\) farther from the wire, the meter reads \(575 \mathrm{~V}\). (a) What is \(\lambda\) ? (b) If you now place one probe at \(3.50 \mathrm{~cm}\) from the wire and the other probe \(1.00 \mathrm{~cm}\) farther away, will the voltmeter read \(575 \mathrm{~V} ?\) If not, will it read more or less than \(575 \mathrm{~V}\) ? Why? (c) If you place both probes \(3.50 \mathrm{~cm}\) from the wire but \(17.0 \mathrm{~cm}\) from each other, what will the voltmeter read? Given: \(\ln \left(\frac{7}{5}\right)=0.337\)

" (a) How much excess charge must be placed on a copper sphere \(25.0 \mathrm{~cm}\) in diameter so that the potential of its center, relative to infinity, is \(1.50 \mathrm{kV} ?\) (b) What is the potential of the sphere's surface relative to infinity?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Fluids

Read Explanation

Linear Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation

Astrophysics

Read Explanation

Physical Quantities And Units

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.