/*! 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 95 CALC Positive charge \(+Q\) is d... [FREE SOLUTION] | 91Ó°ÊÓ

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CALC Positive charge \(+Q\) is distributed uniformly along the \(+x\) -axis from \(x=0\) to \(x=a .\) Negative charge \(-Q\) is distributed uniformly along the \(-x\) -axis from \(x=0\) to \(x=-a\) . (a) A positive point charge \(q\) lies on the positive \(y\) -axis, a distance \(y\) from the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on \(q .\) Show that this force is proportional to \(y^{-3}\) for \(y>>\) a. (b) Suppose instead that the positive point charge \(q\) lies on the positive \(x\) -axis, a distance \(x>a\) from the origin. Find the force (magnitude and direction) that the charge distribution exerts on \(q .\) Show that this force is proportional to \(x^{-3}\) for \(x>>a\) .

Short Answer

Expert verified
(a) Force is \( \mathbf{F} = \frac{2kQq}{y^3} \; \hat{j} \) for \( y >> a \). (b) Force is \( \propto x^{-3} \) for \( x >> a \).

Step by step solution

01

Understand the Distribution of Charges

We have a line of positive charge +Q from 0 to a along the x-axis, and a line of negative charge -Q from 0 to -a along the x-axis. A point charge q is located at a point along the y-axis, a distance y from the origin.
02

Express Charge Density and Forces

The linear charge density \( \lambda = \frac{Q}{a} \) for both the positive and negative charge distributions, as each segment has length \(a\). The differential force on an element of charge \(dq\) on the x-axis is \( d\mathbf{F} = \frac{kq \cdot dq}{r^2} \), where \(r\) is the distance between \(dq\) and point charge \(q\).
03

Determine the Differential Charge Element

For a small element of charge located at a position \(x\) along the line, the charge element is \( dq = \lambda \, dx = \frac{Q}{a} \, dx \). The distance from this element to the charge \(q\) is \( \sqrt{x^2 + y^2} \).
04

Calculate Force Contributions from the Line Charges

The elements from the positive and negative line charges exert forces in opposite x directions but similar y directions. Each differential force is given by \( d\mathbf{F} = \frac{kq(dq)}{x^2 + y^2}(\frac{-x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}) \) for the positive x-axis segment, and similarly for the negative x-axis segment.
05

Integrate to Find the Total Force

Integrate the \( x \) components of the force over the interval \( [0, a] \) for both the positive and negative charge distributions, and the \( y \) component from the respective charges. For large \(y \, (y >> a)\), these simplify to show the force being inversely proportional to \( y^3 \).
06

Analyze the Proportionality

For part (a), the force simplifies to \( \mathbf{F} \approx \frac{2kQq}{y^3} \; \hat{j} \) showing the inverse cube relationship. For part (b), use a similar method for a point on the positive x-axis with \(x > a\) and derive that the correlations are \( \mathbf{F} \propto x^{-3} \) for large \(x\).
07

Solve Part (b): Forces Along x-axis

In this case, focus on calculating the net force exerted by both lines on the charge \(q\) located on the positive x-axis, with integration over the linear distributions leading again to force proportional to \( x^{-3} \) when \(x >> a\).

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Key Concepts

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

electric forces
Electric forces are the interactions that occur between charged particles. These forces can either attract or repel particles depending on their charges. Like charges repel each other, while opposite charges attract. It's fascinating how these forces dictate the structure and behavior of matter at a molecular level.

In the context of our problem, the force exerted on a point charge by a line of charge depends on both the magnitude of the charges involved and their spatial arrangement. The positive and negative line charges generate electric fields that interact with the point charge, resulting in a net force.

The classic Coulomb's law, which describes these forces, states that the force 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 can be expressed as:
  • \( F = \frac{k \, |q_1 \, q_2|}{r^2} \)
where \( k \) is the electrostatic constant. In our scenario, differential charge elements along the line produce forces that must be summed (or integrated) to find the total force on the point charge. Each of these forces has both magnitude and direction, which is carefully considered when towing up the total force.
line charge distribution
A line charge distribution refers to how charge is spread out along a line, rather than being concentrated at a single point. In the problem, charge is uniformly distributed along segments of the x-axis. Positive charge is spread from zero to some point 'a', and negative charge from zero to negative 'a'.

The linear charge density \( \lambda \) is a crucial concept here. It describes how much charge exists per unit length of the line. Mathematically, this is expressed as:
  • \( \lambda = \frac{Q}{a} \)
where \( Q \) is the total charge and \( a \) is the length over which the charge is distributed. Understanding this concept allows us to calculate differential charges \( dq \) at any small segment of the line for our calculations. This differential charge \( dq \) is essential when calculating differential forces that contribute to the overall force in electrostatics.In the problem solution, we consider differential charge elements to accurately compute electrostatic forces. This approach is essential because charges along a line don't all interact with the point charge in the same way due to varying distances and angles. By calculating and then integrating these differential effects, we can determine the net effect accurately.
inverse cube law
The inverse cube law describes a specific situation where the relationship between two quantities involves one quantity being inversely proportional to the cube of another. In electrostatics, this often pertains to scenarios involving dipoles or particular configurations where the field or force falls off more rapidly than the inverse square law.In this exercise, we demonstrate how the force exerted by the line charge distribution on a point charge scales with distance for large distances. When the charge is far from the line (either on the y-axis or far on the x-axis), the situation simplifies, and this naturally leads to an inverse cube dependency.

For example, when the point charge is along the y-axis far from the origin, we see the force magnitude simplify to:
  • \( \mathbf{F} \approx \frac{2kQq}{y^3} \)
This indicates that the force lessens sharply as the charge moves further away, decreasing as the cube of the distance \( y \). Similarly, a charge further along the x-axis sees a force of:
  • \( \mathbf{F} \propto x^{-3} \)
Such laws are important in understanding how electric fields and forces behave over large distances and why certain simplifying assumptions are valid in electrostatic calculations in these scenarios.

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Most popular questions from this chapter

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

CP A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{m} / \mathrm{s} .\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 \(\mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

BIO Electrophoresis. Electrophoresis is a process used by biologists to separate different biological molecules (such as proteins) from each other according to their ratio of charge to size. The materials to be separated are in a viscous solution that produces a drag force \(F_{\mathrm{D}}\) proportional to the size and speed of the molecule. We can express this relation- ship as \(F_{\mathrm{D}}=K R v,\) where \(R\) is the radius of the molecule (modeled as being spherical), \(v\) is its its speed, and \(K\) is a constant that depends on the viscosity of the solution. The solution is placed in an external electric field \(E\) so that the electric force on a particle of charge \(q\) is \(F=q E\) . (a) Show that when the electric field is adjusted so that the two forces (electric and viscous drag) just balance, the ratio of \(q\) to \(R\) is \(K v / E\) . (b) Show that if we leave the electric field on for a time \(T,\) the distance \(x\) that the molecule moves during that time is \(x=(E T / k)(q / R)\) . (c) Suppose you have a sample containing three different biological molecules for which the molecular ratio \(q / R\) for material 2 is twice that of material 1 and the ratio for material 3 is three times that of material 1. Show that the distances migrated by these molecules after the same amount of time are \(x_{2}=2 x_{1}\) and \(x_{3}=3 x_{1}\) . In other words, material 2 travels twice as far as material \(1,\) and material 3 travels three times as far as material \(1 .\) Therefore, we have separated these molecules according to their ratio of charge to size. In practice, this process can be carried out in a special gel or paper, along which the biological molecules migrate. (Fig. P21.94). The process can be rather slow, requiring several hours for separations of just a centimeter or so.

CP Operation of an Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of \(1.4 \times 10^{-8}\) g each, leave the nozzle and travel toward the paper at \(20 \mathrm{m} / \mathrm{s},\) passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 \(\mathrm{cm}\) long where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) . If a drop is to be deflected 0.30 \(\mathrm{mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

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