/*! 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 77 A particle of charge \(-q_{1}\) ... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 77

A particle of charge \(-q_{1}\) is at the origin of an \(x\) axis (a) At what location on the axis should a particle of charge \(-4 q_{1}\) be placed so that the net electric field is zero at \(x=2.0 \mathrm{~mm}\) on the axis? (b) If, instead, a particle of charge \(+4 q_{1}\) is placed at that location, what is the direction (relative to the positive direction of the \(x\) axis) of the net electric field at \(x=2.0 \mathrm{~mm} ?\)

Short Answer

Expert verified
(a) Place \\( -4q_1 \\\) at \\( x = -2 \\\) mm. (b) The direction is positive \\( x \\\).

Step by step solution

01

Identify the Charges and Positions

We have two charges: 1. Charge \( -q_1 \) is located at the origin \( x = 0 \).2. We need to find the position \( x \) for charge \( -4q_1 \) such that the net electric field at \( x = 2 \) mm is zero.
02

Set Up the Equation for Electric Field

The electric field due to a point charge is given by \( E = \frac{kq}{r^2} \), where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.We want the net electric field at \( x = 2 \) mm to be zero. Thus, the electric field contributions from both charges must cancel each other out: \( \frac{k(-q_1)}{(0.002)^2} = \frac{k(-4q_1)}{(0.002 - x)^2} \).
03

Simplify the Equation

After setting the expressions for the electric fields equal, we can eliminate common terms: \( \frac{-q_1}{(0.002)^2} = \frac{-4q_1}{(0.002 - x)^2} \).We can cancel \( -kq_1 \) from both sides of the equation, leading to:\( \frac{1}{(0.002)^2} = \frac{4}{(0.002 - x)^2} \).
04

Solve for x

Cross-multiply to find:\( (0.002 - x)^2 = 4 \times (0.002)^2 \).Taking the square root of both sides gives:\( 0.002 - x = 0.004 \).Solving for \( x \), we get:\( x = 0.002 - 0.004 = -0.002 \) meters or \( -2 \) mm.
05

Determine the Direction of the Electric Field with +4q1

If the charge at \( x = -2 \) mm is \( +4q_1 \), then both charges at \( x = 0 \) and \( x = -2 \) mm contribute electric fields in the same direction at \( x = 2 \) mm.The charge \( -q_1 \) at the origin will still produce an electric field pointing towards it (positive \( x \) direction), and the charge \( +4q_1 \) on the left will also push the electric field in the positive \( x \) direction due to repulsion.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is fundamental in understanding how electric charges interact with each other. It describes the force between two point charges. According to this law, the electric force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[ F = k \frac{|q_1 q_2|}{r^2} \]Here, \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant which is approximately \( 8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2 \).
  • This law helps predict the behavior of charges at different distances.
  • It applies to point charges, which are idealized charges that occupy no volume.
As per the given problem, using Coulomb's law helps in determining where another charge should be placed to achieve equilibrium in the electric field setup.
Electric Charge
Electric charge is a property of matter that causes it to experience a force when placed in an electric field. Charges come in two types: positive and negative. Like charges repel each other, while opposite charges attract each other.
  • Charges are measured in coulombs (C).
  • Common subatomic particles with charge are electrons (negative) and protons (positive).
In the provided exercise, we deal with two charges:
  • A charge of \(-q_1\) at the origin.
  • The determination of a second charge, \(-4q_1\), and its hypothetical counterpart, \(+4q_1\), to assess the electric field behavior.
Understanding electric charge is crucial in determining how these charges will influence the electric field and how they interact with each other.
Electric Field Direction
The direction of an electric field represents the path that a positive test charge would follow if placed in the field. It is often depicted using field lines, which start on positive charges and end on negative charges.
In this exercise, we're asked to determine the net electric field direction at a specific point (\(x = 2\, \text{mm}\)).
  • If a charge is negative, it pulls field lines towards itself.
  • If a charge is positive, it pushes field lines away from itself.
In part (b) of the exercise, when a \(+4q_1\) charge is positioned, both the negative charge at the origin and the positive charge \(+4q_1\) contribute fields that point in the positive \(x\) direction at \(x = 2\, \text{mm}\). The electric field lines emanate from the positive charge, reinforcing the field due to the original negative charge.
Point Charges
Point charges are an idealization used in physics to simplify the study of electric fields and forces. They are considered to be a charge with infinitesimal size, allowing us to concentrate on the electric field and force calculations without worrying about size or shape.
  • Point charges are assumed to have no dimensions.
  • Their effect is calculated solely based on charge magnitude and distance from another charge.
In the exercise, both \(-q_1\) and the additional charge \((-4q_1\) or \(+4q_1\)) are considered point charges. This simplifies the task of finding the position where the electric field from these charges cancels or aligns.
  • By treating them as point charges, calculations are more straightforward.
  • It allows us to easily use Coulomb's Law to find the forces and, subsequently, the electric fields involved.
Understanding this concept is essential for solving the exercise and determining positions where the electric field conditions are met.

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Most popular questions from this chapter

At some instant the velocity components of an electron moving between two charged parallel plates are \(v_{x}=1.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\) and \(v_{y}=3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Suppose the electric field between the plates is uniform and given by \(\vec{E}=(120 \mathrm{~N} / \mathrm{C}) \hat{j} .\) In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its \(x\) coordinate has changed by \(2.0 \mathrm{~cm}\) ?

A proton and an electron form two corners of an equilateral triangle of side length \(2.0 \times 10^{-6} \mathrm{~m}\). What is the magnitude of the net electric field these two particles produce at the third corner?

The nucleus of a plutonium- 239 atom contains 94 protons. Assume that the nucleus is a sphere with radius \(6.64 \mathrm{fm}\) and with the charge of the protons uniformly spread through the sphere. At the surface of the nucleus, what are the (a) magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons?

A clock face has negative point charges \(-q,-2 q,-3 q, \ldots\), \(-12 q\) fixed at the positions of the corresponding numerals. The clock hands do not perturb the net field due to the point charges. At what time does the hour hand point in the same direction as the electric field vector at the center of the dial? (Hint: Use symmetry.)

Assume that a honeybee is a sphere of diameter \(1.000\) \(\mathrm{cm}\) with a charge of \(+45.0 \mathrm{pC}\) uniformly spread over its surface. Assume also that a spherical pollen grain of diameter \(40.0 \mu \mathrm{m}\) is electrically held on the surface of the bee because the bee's charge induces a charge of \(-1.00 \mathrm{pC}\) on the near side of the grain and a charge of \(+1.00 \mathrm{pC}\) on the far side. (a) What is the magnitude of the net electrostatic force on the grain due to the bee? Next, assume that the bee brings the grain to a distance of \(1.000 \mathrm{~mm}\) from the tip of a flower's stigma and that the tip is a particle of charge \(-45.0 \mathrm{pC}\) (b) What is the magnitude of the net electrostatic force on the grain due to the stigma? (c) Does the grain remain on the bee or does it move to the stigma?
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