/*! 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 79 Positive charge \(Q\) is distrib... [FREE SOLUTION] | 91Ó°ÊÓ

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Positive charge \(Q\) is distributed uniformly along the \(x\) axis from \(x=0\) to \(x=a\). A positive point charge \(q\) is located on the positive \(x\) -axis at \(x=a+r,\) a distance \(r\) to the right of the end of \(Q\) (Fig. P21.79). (a) Calculate the \(x\) - and \(y\) -components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\) -axis where \(x>a\). (b) Calculate the force (magnitude and direction) that the charge distribution \(Q\) exerts on \(q\). (c) Show that if \(r \gg a\), the magnitude of the force in part (b) is approximately \(Q q / 4 \pi \epsilon_{0} r^{2}\). Explain why this result is obtained.

Short Answer

Expert verified
The x-component of the electric field produced by the uniformly distributed charge Q at points where x>a is \(kQa /(r^2+\a^2 )^{3/2} \), whereas the y-component is 0. The force that Q exerts on q is \( q k Q a / (a^2 + r^2)^{3/2}\). If r \gg a, then the force will approximately be \(\frac{Q q }{4 \pi \epsilon_{0} r^{2}}\), which is the formula for a force between two point charges separated by a distance r.

Step by step solution

01

Calculate the Electric Field Due to a Small Charge Element

Imagine a small charge element \ud Q at position x on the x-axis. The electric field \ud E at a position x=a+r due to \ud Q is given by Coulomb's law: \ud E = k \ud Q/ R², where k is Coulomb's constant, R is the distance between \ud Q and the point at x=a+r, and \ud E points in the direction from \ud Q to this point. The components of \ud E are \ud E_x = \ud E cos(\θ) and \ud E_y = \ud E sin(\θ), where \(\theta\) is the angle between the x-axis and the line joining \ud Q and the point at x=a+r.
02

Integrate to Find the Total Electric Field

Due to the symmetry of the situation, the vertical (y) components of the electric field \ud E due to each charge element \ud Q will cancel out when summed over the entire charge distribution. Therefore, the total electric field E at x=a+r due to the charge distribution Q will only have an x-component. This is found by integrating the x-components of the electric field \ud E for all charge elements from x=0 to x=a, which gives E=\int_0^a \ud E_x. This integral can be evaluated using calculus.
03

Calculate the Force on the Point Charge

The electric force F on the point charge q due to the charge distribution Q is given by F=qE. By substituting E obtained from Step 2 into this equation, we find the force exerted on the point charge.
04

Derive the Approximate Force for Large r

Consider the condition where r \gg a. In this case, the denominator of the equation obtained in Step 2 can be approximated by r, leading to a simplification of the electric field E for x>a. Substituting this in the force equation obtained in Step 3, we find that the force function simplifies drastically. Compare this result with the standard formula for force between two point charges, which provides an explanation for the result.

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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 a fundamental principle in physics that describes the electric force between two charged objects. It states that the electric force (\( F \)) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (\( R \)) between them. Coulomb's law can be expressed mathematically as:\[F = k \frac{Q q}{R^2}\]where:
  • \( k \) is Coulomb's constant (approximately \( 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \)).
  • \( Q \) and \( q \) are the magnitudes of the charges.
  • \( R \) is the distance between the charges.
Coulomb's law is essential because it helps us calculate the strength and direction of the force exerted by a charged object on another. The force acts along the line joining the two charges and it can be either attractive or repulsive depending on the sign of the charges. Positive charges repel each other, while opposite charges attract each other.
In the context of the original problem, Coulomb's law is used to find the electric field generated by a small element of charge along a distribution and the resulting force on a nearby point charge.
Point Charge
A point charge is an idealized model of a charged object where the charge is concentrated at a single point in space. This concept simplifies solving problems in electrostatics because it allows us to treat the charge as having no volume or spatial extent.
In practical terms, any small charged object can be approximated as a point charge if it's small enough compared to the distances involved in the problem. For instance, electrons or protons can often be treated as point charges in calculations due to their minuscule size relative to distances in typical physical scenarios.
In the context of the exercise, the point charge \( q \) is located at a specific position on the x-axis, and applying the concept of a point charge allows us to use Coulomb's law to calculate the electrical interactions due to the distributed charge \( Q \). These calculations involve determining how the electric field and thus the electrostatic force affect this point charge.
Electric Force
The electric force is the influence that charged objects exert on each other, due to their charge. This force can cause objects to move if they are free or, if not, to experience stress if they are constrained.
Electric forces are one of the fundamental forces in nature. They are explained and calculated using Coulomb's law and expressed mathematically as:\[F = qE\]where:
  • \( F \) is the force experienced by a charge.
  • \( q \) is the charge experiencing the force.
  • \( E \) is the electric field intensity created by other charges.
In the exercise context, the electric force is calculated by analyzing how distributed charge \( Q \) interacts with a point charge \( q \), accounting for the geometric arrangement of the charges and the distance separating them. When the point charge is in the field generated by the continuous charge distribution, it experiences a force that can be determined through integration over the electric field components. Additionally, when the distance \( r \) is much larger than the extension \( a \), the situation simplifies, resembling the standard interaction between distant point charges.

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Most popular questions from this chapter

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

A charge of \(-3.00 \mathrm{nC}\) is placed at the origin of an \(x y-\)coordinate system, and a charge of \(2.00 \mathrm{nC}\) is placed on the \(y\) -axis at \(y=4.00 \mathrm{~cm} .\) (a) If a third charge, of \(5.00 \mathrm{nC}\), is now placed at the point \(x=3.00 \mathrm{~cm}, y=4.00 \mathrm{~cm},\) find the \(x-\) and \(y-\) components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

If we rub a balloon on our hair, the balloon sticks to a wall or ceiling. This is because the rubbing transfers electrons from our hair to the balloon, giving it a net negative charge. When the balloon is placed near the ceiling, the extra electrons in it repel nearby electrons in the ceiling, creating a separation of charge on the ceiling, with positive charge closer to the balloon. Model the interaction as two point-like charges of equal magnitude and opposite signs, separated by a distance of \(500 \mu \mathrm{m}\). Neglect the more distant negative charges on the ceiling. (a) A typical balloon has a mass of \(4 \mathrm{~g}\). Estimate the minimum magnitude of charge the balloon requires to stay attached to the ceiling. (b) since a balloon sticks handily to the ceiling after being rubbed, assume that it has attained 10 times the estimated minimum charge. Estimate the number of electrons that were transferred to the balloon by the process of rubbing.

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

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