/*! 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 58 Two charges are placed on the \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Two charges are placed on the \(x\) -axis: one, of \(2.50 \mu \mathrm{C}\), at the origin and the other, of \(-3.50 \mu \mathrm{C},\) at \(x=0.600 \mathrm{~m}\) (Fig. \(\mathrm{P} 21.58\) ). Find the position on the \(x\) -axis where the net force on a small charge \(+q\) would be zero.

Short Answer

Expert verified
The two positions where the net force on a small charge \(+q\) would be zero are at \(x = -0.166 m\) and \(x = 0.360 m\).

Step by step solution

01

Identify the Problem

Identify the two charges that are placed on the x-axis: a positive 2.50 µC at the origin and a negative 3.50 µC at \(x=0.600 m\). We want to find a location where we could place a third charge such that it experiences no net force.
02

Establish Variables

In the problem, two possible positions arise for +q to experience zero net force - one to the left (negative x) of the positive charge, and one to the right (positive x) in between positive and negative charges. First, let's assume the positive force location x.
03

Apply Coulomb's Law

Apply Coulomb's Law to get the electric forces. Coulomb's law is given as \(F = k\frac{q1*q2}{r^2}\), where k is Coulomb’s constant, \(q1\) and \(q2\) are the charges, and r is the distance between them.
04

Establish Equations

Establish the equations using Coulomb's law and conditions of the problem. Due to the +q experiencing zero net force, the forces from both 2.50 µC and -3.50 µC must balance each other. For positive x, the force due to 2.50 µC charge is \(F_1 = k \frac{2.50*q} {x^2}\) and due to -3.50 µC charge is \(F_2 = k \frac{3.50*q} {(0.600-x)^2}\). Set \(F_1 = F_2\).
05

Solve for x and Repeat

Now solve for x. If we use \( k = 8.99 × 10^9 Nm^2/C^2 \) and solve the \( F_1=F_2 \) equation, for positive x we should get x = 0.360 m. Repeat steps 3-5 for negative x.
06

Use Negative x Value

For negative x, the force due to 2.50 µC charge is \( F_1 = k \frac{2.50*q} {x^2} \), set this equal to the force due to -3.50 µC, \( F_2 = k \frac{3.50*q} {(0.600+x)^2} \), resulting in \( F_1 = F_2 \).
07

Solve for Negative x

Solving the equation \( F_1=F_2 \), for negative x we should get approximately x = -0.166 m.
08

Summary

The charges +q placed at both calculated locations (x = -0.166 m and x = 0.360 m) would then experience zero net force due to the other charges.

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Key Concepts

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

Electric Force
The electric force is a fundamental interaction between charged particles. According to Coulomb's Law, the magnitude of the force \( F \) between two point charges \( q1 \) and \( q2 \) is directly proportional to the product of the charges and inversely proportional to the square of the distance \( r \) between them:

\[ F = k\frac{q1 * q2}{r^2} \]
where \( k \) represents Coulomb's constant. The direction of the force depends on the nature of the charges: opposite charges attract, and like charges repel each other. In the exercise provided, this law is used to calculate the positions on the \( x \) -axis where a third charge \( +q \) would experience no net electric force. It's essential to identify that the electric force can change depending on the position of the charge in the electric field created by other charges. In creating a clear solution for students, the explanation of electric force should emphasize the vector nature of force and the significance of the sign and magnitude of charges involved.
Charge Distribution
Charge distribution refers to how electric charge is allocated or spread out in space. In the context of the exercise, we have a discrete charge distribution with two point charges fixed along the \( x \) -axis. Understanding the arrangement and values of these charges is crucial because it determines the electric field configuration and potential points of equilibrium. The charge distribution can be uniform or non-uniform, and in our case, the charges are unevenly spaced and opposite in sign, creating a unique set of electric fields.

In analyzing situations like this, students should visualize the charges and their respective electric fields, noting areas where the fields may overlap. It's also vital to recognize that the charge's influence extends into the space around it, and thus, the presence of another charge within this space will experience a force. When explaining the concept, use diagrams to illustrate the distribution and resultant electric field lines. This visual aid helps to internalize the abstract concept of charge distribution and its significant role in determining electric force.
Equilibrium Point
The equilibrium point in an electric field is the specific location at which a charged object experiences no net electric force. In other words, it's the position where the attractive and repulsive forces on a charge cancel out each other. Identifying the equilibrium point involves analyzing the forces acting on a charge due to the surrounding charge distribution.

For instance, in the exercise we're considering, the equilibrium point is found by setting the force exerted by one charge on \( +q \) equal to the force exerted by another charge on \( +q \) and solving for the position \( x \). Finding such points is essential for understanding electric field interactions and can also be a principle used in designing sensors and switches that respond to electrical charge distribution.
To convey the idea effectively to students, it is important to explain that while the charge \( +q \) is at the equilibrium point, it's in a stable or unstable equilibrium based on the surrounding charges and could be displaced with a slight disturbance, depending on the nature of the equilibrium. To enhance understanding, provide practice problems where students can calculate equilibrium points under various charge configurations and conditions.

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Most popular questions from this chapter

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{~m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{~m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e,\) enter the axon. How many coulombs of charge enter a \(1.5 \mathrm{~cm}\) length of the axon during this process?

An American penny is \(97.5 \%\) zinc and \(2.5 \%\) copper and has a mass of \(2.5 \mathrm{~g}\). (a) Use the approximation that a penny is pure zinc, which has an atomic mass of \(65.38 \mathrm{~g} / \mathrm{mol},\) to estimate the number of electrons in a penny. (Each zinc atom has 30 electrons.) (b) Estimate the net charge on all of the electrons in one penny. (c) The net positive charge on all of the protons in a penny has the same magnitude as the charge on the electrons. Estimate the force on either of two objects with this net magnitude of charge if the objects are separated by \(2 \mathrm{~cm}\). (d) Estimate the number of leaves on an oak tree that is 60 feet tall. (e) Imagine a forest filled with such trees, arranged in a square lattice, each \(10 \mathrm{~m}\) distant from its neighbors. Estimate how large such a forest would need to be to include as many leaves as there are electrons in one penny. (f) How does that area compare to the surface area of the earth?

An insulating rigid rod of length \(2 a\) and negligible mass is attached at its center to a pivot at the origin and is free to rotate in the \(x y\) -plane. A small ball with mass \(M\) and charge \(Q\) is attached to one end of the rod. A second small ball with mass \(M\) and no charge is attached to the other end. A constant electric field \(\vec{E}=-E \hat{\imath}\) is present in the region \(y>0\) while the region \(y<0\) has a vanishing electric field. Define \(\vec{r}\) as the vector that points from the center of the rod to the charged end of the rod, and \(\theta\) as the angle between \(\vec{r}\) and the positive \(x\) -axis. The rod is oriented so that \(\theta=0\) and is given an infinitesimal nudge in the direction of increasing \(\theta\). (a) Write an expression for the vector \(\vec{r}\). (b) Determine the torque \(\vec{\tau}\) about the center of the rod when \(0 \leq \theta \leq \pi\). (c) Determine the torque on the rod about its center when \(\pi \leq \theta \leq 2 \pi\). (d) What is the moment of inertia \(I\) of the system about the \(z\) -axis? (e) The potential energy \(U(\theta)\) is determined by \(\tau=-d U / d \theta .\) Use this equation to write an expression for \(U(\theta)\) over the range \(0 \leq \theta \leq 4 \pi\) using the convention that \(U(0)=0 .\) Make sure that \(U(\theta)\) is continuous. (f) The angular velocity of the rod is \(\omega=\omega(\theta) .\) Using \(\tau=I d^{2} \theta / d t^{2}\) show that the energy \(\frac{1}{2} I \omega^{2}+U(\theta)\) is conserved. (g) Using energy conservation, determine an expression for the angular velocity at the \(n\) th time the positive charge crosses the negative \(y\) -axis.

A charge \(+Q\) is located at the origin, and a charge \(+4 Q\) is at distance \(d\) away on the \(x\) -axis. Where should a third charge, \(q\), be placed, and what should be its sign and magnitude, so that all three charges will be in equilibrium?

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