/*! 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 21 A positive charge \(q\) is fixed... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(x>\) a? Explain why this result is obtained.

Short Answer

Expert verified
Diagram has charges at \(x=0\) and \(x=a\); potential \(V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|}\); \(V=0\) at \(x=a/3\); potential graph shows symmetry; as \(x>a\), \(V\) approaches zero.

Step by step solution

01

Diagram the Charge Positions

In a coordinate plane, place the positive charge \( q \) at \((0, 0)\). Place a negative charge \(-2q\) at \((a, 0)\). Indicate these points on the x-axis showing the charges at \(x = 0\) and \(x = a\).
02

Derive Potential Expression on x-axis

The electric potential due to a point charge \(q\) at a distance \(r\) is given by \( V = \frac{kq}{r} \), where \(k\) is Coulomb's constant. At a point \(x\) on the x-axis, the potential \(V\) due to both charges is the sum: \[ V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|}. \]
03

Determine Positions Where \(V = 0\)

To find the points on the x-axis where the potential is zero, set \(V(x) = 0\): \[ \frac{kq}{|x|} = \frac{2kq}{|x-a|}. \] Solving this gives \( |x| = 2|x-a| \), resulting in values: \( x = a/3 \) and \( x = a \). Since \( x = a \) is one of the charge points, the solution of interest is \( x = a/3 \).
04

Graph Potential Function on x-axis

Create a graph of \( V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|} \) from \(x = -2a\) to \(x = 2a\). Note the behavior of the graph showing values tend to zero when \( x \) is large and infinite potential changes near the charge points.
05

Evaluate Potential for \(x > a\)

For \(x > a\), the distances to the charges are large compared to \(a\). Thus, \( V(x) \approx \frac{-qk}{x} \) which becomes small as \(x\) increases. This indicates potential approaches zero because the effects of each charge decrease at large distances.

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.

Coulomb's Law: Understanding the Force Between Charges
Coulomb's Law is a cornerstone principle in the study of electromagnetism. This law describes the force between two point charges. According to Coulomb's Law, the electric force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it’s expressed as:
  • \( F = \frac{k |q_1 q_2|}{r^2} \)
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.

Coulomb's Law helps us understand that:
  • Like charges repel each other.
  • Opposite charges attract each other.
This principle explains the interaction between the positive and negative charges in the exercise, helping us to predict how they influence the electric potential in their vicinity.
Electric Field: The Force Field Around Charges
An electric field is essentially a region around a charged particle where other charges would experience a force. It is a vector field, meaning it has both magnitude and direction. We use the symbol \( E \) to denote electric field, and the formula for calculating it due to a point charge is:
  • \( E = \frac{kq}{r^2} \)
where \( q \) is the charge creating the field and \( r \) is the distance from the charge.

The electric field at any point in space is defined as the force experienced by a positive test charge placed at that point divided by the magnitude of the charge. Importantly, the direction of \( E \) is radial relative to the charge:
  • Outward for positive charges.
  • Inward for negative charges.
This helps visualize how the fixed charges at \((0,0)\) and \((a,0)\) in the exercise influence space around them, with resulting fields capable of moving other charges found within these fields.
Point Charges: The Basic Building Blocks of Electric Fields
Point charges are hypothetical charges that occupy a single point in space with no actual size or structure. They act as simplifications to model how charges interact, as seen in the exercise where a point charge represents each charge located in space. The simplicity of point charges allows us to easily calculate the electric field, potential, and forces without worrying about the shape or volume of the charge.

In the context of the exercise, point charges simplify the task of calculating the electric potential on the x-axis. Given their positions at \((0,0)\) and \((a,0)\), these charges produce distinct effects in terms of the field and potential, directly governed by their magnitudes and positions:
  • A positive point charge produces a field pointing outwards, reducing potential.
  • A negative point charge creates a field pointing inwards, increasing potential on nearby points relative to the positive charge.
Point charges facilitate immense ease in theoretical studies and calculations in eletromagnetic theory through precise formulas.

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

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+3.00-\mu C\) charged ball of mass 1.50 \(\mathrm{g}\) from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times\) \(10^{-11} \mathrm{m} .\) Give your numerical result in joules and in electron volts.

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{V} .\) The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{cm} .\) What is the potential difference between the center of the sphere and the surface of the sphere?

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y-\) , and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

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.