/*! 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 43 Two positive point charges \(q\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Two positive point charges \(q\) are placed on the \(x\) -axis, one at \(x=a\) and one at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0\) . (b) Derive an expression for the electric field at points on the \(x\) -axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x\) , for values of \(x\) between \(-4 a\) and \(+4 a\) .

Short Answer

Expert verified
(a) Electric field at x=0 is zero. (b) \( E_x = \frac{4kqax}{(x^2-a^2)^2} \) for x-axis field; plot shows symmetry around x=0.

Step by step solution

01

Set up the Problem

We have two point charges, each with charge \( q \), placed symmetrically at positions \( x = a \) and \( x = -a \) on the x-axis. We want to find the electric field at various points on the x-axis due to these charges.
02

Electric field at x=0

The electric field due to a point charge \( q \) at a distance \( r \) is given by \( E = \frac{kq}{r^2} \) where \( k \) is the electrostatic constant. At \( x = 0 \), the distance to both charges is \( a \). The electric field due to the charge at \( x = a \) is directed towards \( x = a \) and has magnitude \( \frac{kq}{a^2} \). The field due to charge at \( x = -a \) is directed towards \( x = -a \) and also has magnitude \( \frac{kq}{a^2} \). Since these fields are equal and opposite, they cancel each other out at \( x = 0 \). Thus, the net electric field at \( x = 0 \) is zero.
03

Expression for electric field along the x-axis

Consider a point at position \( x \) on the x-axis. The electric field due to charge at \( x = a \) has magnitude \( E_1 = \frac{kq}{(x-a)^2} \) and points towards \( -x \) (if \( x > a \), otherwise towards \( x \)). The field due to charge at \( x = -a \) is \( E_2 = \frac{kq}{(x+a)^2} \) and points in the opposite direction of \( E_1 \) (towards \( +x \) if \( x < -a \), otherwise towards \( x \)). The total electric field is thus the vector sum: \[ E_x = \frac{kq}{(x-a)^2} - \frac{kq}{(x+a)^2} \].
04

Simplify the Expression

The expression for the total electric field at position \( x \) is: \[ E_x = kq \left( \frac{1}{(x-a)^2} - \frac{1}{(x+a)^2} \right) \]. This can be simplified further to \[ E_x = \frac{4kqax}{(x^2-a^2)^2} \]. This expression gives the x-component of the electric field along the x-axis.
05

Graph the Electric Field Expression

Graph the expression \( E_x = \frac{4kqax}{(x^2-a^2)^2} \) for \( x \) ranging from \(-4a\) to \(+4a\). The graph shows the symmetric nature of the electric field about \( x = 0 \), with peaks near \( x = \pm a \) and zero crossings at the origin. These characteristics are expected given the presence of the two charges at \( x = \pm a \).

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Key Concepts

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

Point Charges
Point charges are a fundamental concept in electrostatics, representing charged particles with negligible size compared to the distances involved. In our exercise, each point charge has a charge of \( q \), located symmetrically on the x-axis at positions \( x = a \) and \( x = -a \). These charges create electric fields around them, influencing the space and other charges within it. Understanding the behavior of point charges is crucial to analyzing the electric fields they produce. It helps to predict how forces act on other charged particles within the vicinity. Keep in mind that the main characteristic of point charges is that they act as idealized entities where the entire charge is considered to be concentrated at one point.
Electrostatic Force
The electrostatic force is the force of attraction or repulsion between electrically charged objects. For point charges, this force is described by Coulomb's law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance \( r \) between them. Mathematically, it is represented as: \[ F = k \frac{q_1 q_2}{r^2} \] where \( k \) is Coulomb's constant. In our exercise, this principle manifests in the interactions between the two point charges at \( x = a \) and \( x = -a \). The electric field resulting from each charge at specific points along the x-axis showcases how these forces distribute.
Symmetry in Electric Fields
Symmetry plays a vital role in determining the patterns of electric fields. In this scenario, with equal charges symmetrically placed at \( x = a \) and \( x = -a \), any point along the x-axis experiences their influence symmetrically. For example, at \( x = 0 \), the electric fields generated by each charge are equal in magnitude but opposite in direction, leading to a net electric field of zero. This symmetry simplifies calculations and enhances our understanding of how electric fields behave around symmetrical setups. It also provides useful insights when predicting the effects at points that lie within symmetrical configurations. Graphically, this symmetry is evident, as the pattern of electric fields mirrors about the y-axis.
Graphing Electric Fields
Graphing electric fields provides a visual representation of how electric field strengths vary across different regions. In our example, the expression for the electric field \( E_x = \frac{4kqax}{(x^2-a^2)^2} \) along the x-axis allows us to explore how the field changes between \( x = -4a \) and \( x = +4a \). The graph illustrates peaks near the point charges \( x = \pm a \), where the effect of each charge is most intense. It also shows symmetrical characteristics, with zero crossings at the origin \( x = 0 \). Plotting such graphs aids in visualizing electric field interactions, helping to predict potential impacts and behaviors of charged particles in these regions. Graphing tools can be particularly effective for illustrating complicated field distributions in educational settings.

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Most popular questions from this chapter

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A ?(\mathrm{b}) \mathrm{A}\) proton is moving in the uniform electric field of part (a). At point \(A .\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s}\) , east. What is the speed of the proton at point \(B ?\)

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.

Two very small \(8.55-\) g spheres, 15.0 \(\mathrm{cm}\) apart from center to center, are charged by adding equal numbers of electrons to each of them. Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0 \(\mathrm{g}\) when released? Which way will they accelerate?

An average human weighs about 650 \(\mathrm{N}\) . If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650 - \(\mathrm{N}\) weight?

A positive point charge \(q\) is placed on the \(+y\) -axis at \(y=a\) and a negative point clarge \(-q\) is placed on the \(-y\) -axis at \(y=-a\) . A negative point charge \(-Q\) is located at some point on the \(+x\) -axis, (a) In a free- body diagram, show the forces that act on the charge \(-Q\) . (b) Find the \(x\) - and \(y\) -components of the net force that the two charges \(q\) and \(-q\) exert on \(-Q .\) (Your answer should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge. \()\) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0)\) ? (d) Graph the \(y\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

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