/*! 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 17 Two point charges are located on... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 17

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m},\) and charge \(q_{2}=+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)

Short Answer

Expert verified
The total force exerted on the third charge \(q_3\) is 1.2375 N directed upwards along the y-axis.

Step by step solution

01

Arrange the charges

Let's arrange the three charges along the y-axis, where \(y_{1}\) = -0.600 m, \(y_2\) = 0 and \(y_3\) = -0.400 m, representing the positions of \(q_1\), \(q_2\) and \(q_3\) respectively.
02

Calculate the distances

The distance \(r_1\) between \(q_1\) and \(q_3\) is \(r_1 = |y_3 - y_1| = |-0.4 + 0.6| = 0.2 m\). The distance \(r_2\) between \(q_2\) and \(q_3\) is \(r_2 = |y_3 - y_2| = |-0.4 - 0| = 0.4 m\).
03

Apply Coulomb鈥檚 law to calculate the forces

The force \(F_1\) exerted on \(q_3\) by \(q_1\) is given by Coulomb's law: \(F_1 = k*\frac{|q_1*q_3|}{r_1^2}\), where \(k = 9.0*10^9 N*m^2/C^2\) is the electrostatic constant. Substituting the values, we get \(F_1 = 9.0*10^9*\frac{|-1.5*10^{-9}*5*10^{-9}|}{0.2^2} = 0.3375 N\). The force is directed from \(q_1\) to \(q_3\), or upwards along the y-axis since \(q_1\) and \(q_3\) are of opposite signs. Similarly, the force \(F_2\) exerted on \(q_3\) by \(q_2\) is \(F_2 = k*\frac{|q_2*q_3|}{r_2^2} = 9.0*10^9*\frac{|3.2*10^{-9}*5*10^{-9}|}{0.4^2} = 0.9 N\). The force is directed from \(q_3\) to \(q_2\), or upwards along the y-axis since \(q_2\) and \(q_3\) are of the same sign.
04

Calculate the total force

The total force F exerted on \(q_3\) is the vector sum of \(F_1\) and \(F_2\), both of which are along the y-axis and in the same direction (upwards). So, \(F = F_1 + F_2 = 0.3375 N + 0.9 N = 1.2375 N\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Electric Charge
Electric charge is a fundamental property of particles that determines how they interact with other charged objects. It comes in two types: positive and negative. Charges with the same sign repel each other, while charges with opposite signs attract.

Charges are measured in a unit called the Coulomb (C), named after Charles-Augustin de Coulomb who formulated Coulomb's law. The smallest charge is carried by an electron or a proton, where the electron carries a negative charge, and the proton carries a positive charge, both having the same magnitude of roughly \(1.6 \times 10^{-19}\) Coulombs.

Understanding electric charge is crucial when studying electric forces and fields, as the presence of charge leads to various electrical phenomena. The textbook problem about determining the total force experienced by a third charge due to two other charges is heavily reliant on understanding the nature of these charges and how they exert forces over a distance.
Electric Force
Electric force is the push or pull exerted by charged particles or objects. It's a non-contact force, meaning that the objects do not have to touch for the force to operate. This concept is described quantitatively by Coulomb's law, which gives the magnitude and direction of the electric force between two point charges.

The law states that the force (\(F\)) between two point charges is proportional to the product of the charges (\(q_1\) and \(q_2\)), and inversely proportional to the square of the distance (\(r\)) between them: \[ F = k\frac{q_1 q_2}{r^2} \], where \(k\) is the Coulomb's constant, approximately equal to \(9.0 \times 10^9\) Nm虏/C虏.

The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign. In the textbook exercise, the use of Coulomb's law helps us calculate the force exerted by each charge on the third charge, eventually leading to an understanding of the resulting net force.
Superposition Principle
The superposition principle is a fundamental concept in physics that applies to a wide range of phenomena, including electric forces. It states that when two or more forces are acting at a point, the resulting force, also known as the net force, is the vector sum of all individual forces.

In the context of our textbook problem, we apply the superposition principle to determine the total force exerted on a charge by multiple other charges. When calculating the net force on the third charge due to charges \(q_1\) and \(q_2\), we add the individual forces vectorially, considering both their magnitudes and directions.

Since the charges in this case are arranged along a line (the y-axis), the forces they exert on \(q_3\) are also along this axis, making the problem a one-dimensional vector addition. When dealing with more complex configurations, we might need to resolve forces into components and apply the principle in two or three dimensions. The superposition principle simplifies the analysis of interactions between multiple charges, allowing us to calculate the total effect on a single charge.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 -3.00 nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{~m}\). A second point charge, \(Q,\) is on the \(x\) -axis at \(-0.600 \mathrm{~m}\). What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) \(45.0 \mathrm{~N} / \mathrm{C}\) in the \(+x\) -direction, (b) \(45.0 \mathrm{~N} / \mathrm{C}\) in the \(-x\) -direction?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0.800 \mathrm{~m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

(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 \(A ?\) (b) 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 \(B ?\)

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \mathrm{~N} / \mathrm{C} ;\) (b) \(16 \mathrm{~N} / \mathrm{C} ;\) (c) \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\).
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Wave Optics

Read Explanation

Fluids

Read Explanation

Electric Charge Field and Potential

Read Explanation

Nuclear Physics

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.