Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 76

A point charge \(q_{1}=100 . \mathrm{nC}\) is at the origin of an \(x y\) -coordinate system, a point charge \(q_{2}=-80.0 \mathrm{nC}\) is on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a point charge \(q_{3}=-60.0 \mathrm{nC}\) is on the \(y\) -axis at \(y=-2.00 \mathrm{~m} .\) Determine the net force (magnitude and direction) on \(q_{1}\).

Short Answer

Expert verified
Answer: The magnitude of the net force on \(q_1\) is \(22.5\:nN\), and the direction is \(36.9^\circ\) counterclockwise from the negative x-axis.

Step by step solution

01

Understanding the problem and setting up the variables

Let's first identify the given values and variables to set up the problem. The charges are: \(q_{1}=100\:nC\) (at the origin), \(q_{2}=-80.0\:nC\) (at \((2.00\:m,0)\)), and \(q_{3}=-60.0\:nC\) (at \((0,-2.00\:m)\)). We need to find the net force on \(q_1\), which is the sum of the individual forces acting on \(q_1\) due to \(q_2\) and \(q_3\).
02

Calculate the force between \(q_1\) and \(q_2\)

To calculate the force between \(q_1\) and \(q_2\), we can use Coulomb's Law which states: \(F=\frac{kq_1q_2}{r^2}\). In this case, the distance \(r_{12}\) between \(q_1\) (origin) and \(q_2\) (2.00 meters along the x-axis) is 2.00 meters. So, the force between \(q_1\) and \(q_2\) is given by: \(F_{12}=\frac{(9*10^9)(100*10^{-9})((-80)*10^{-9})}{(2)^2}=-18*10^{-9}N\) This force is acting in the negative x-direction.
03

Calculate the force between \(q_1\) and \(q_3\)

Now, we will calculate the force between \(q_1\) and \(q_3\) using Coulomb's Law, with the distance \(r_{13}\) being \(2.00\) meters. The force between \(q_1\) and \(q_3\) is given by: \(F_{13}=\frac{(9*10^9)(100*10^{-9})((-60)*10^{-9})}{(2)^2}=-13.5*10^{-9}N\) This force is acting in the positive y-direction.
04

Finding the net force on \(q_1\)

To find the net force on \(q_1\), we will sum the individual forces acting on \(q_1\) due to \(q_2\) and \(q_3\). The net force on \(q_1\) can be represented as a 2D vector: \(\vec{F}_{net} = \vec{F}_{12}+\vec{F}_{13}\) Given the direction of \(F_{12}\) (negative x-direction) and \(F_{13}\) (positive y-direction), we can rewrite \(\vec{F}_{net}\) as: \(\vec{F}_{net} = (-18*10^{-9}\hat{i}) + (-13.5*10^{-9}\hat{j})\)
05

Compute the magnitude and direction of the net force

We can compute the magnitude of the net force using the Pythagorean theorem: \(F_{net} = \sqrt{(-18*10^{-9})^2+(-13.5*10^{-9})^2}= 22.5*10^{-9}N\) Next, we need to find the angle between the net force and the negative x-axis. We can use the arctangent function for this purpose: \(\theta = \arctan\left(\frac{-13.5*10^{-9}}{-18*10^{-9}}\right)=36.9^\circ\) So, the net force magnitude on \(q_1\) is \(22.5\:nN\), and the direction is \(36.9^\circ\) counterclockwise from the negative x-axis.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Identical point charges \(Q\) are placed at each of the four corners of a rectangle measuring \(2.0 \mathrm{~m}\) by \(3.0 \mathrm{~m} .\) If \(Q=32 \mu C,\) what is the magnitude of the electrostatic force on any one of the charges?

Three 5.00-g Styrofoam balls of radius \(2.00 \mathrm{~cm}\) are coated with carbon black to make them conducting and then are tied to 1.00 -m-long threads and suspended freely from a common point. Each ball is given the same charge, q. At equilibrium, the balls form an equilateral triangle with sides of length \(25.0 \mathrm{~cm}\) in the horizontal plane. Determine \(q\)

A 10.0 -g mass is suspended \(5.00 \mathrm{~cm}\) above a nonconducting flat plate, directly above an embedded charge of \(q\) (in coulombs). If the mass has the same charge, \(q\), how much must \(q\) be so that the mass levitates (just floats, neither rising nor falling)? If the charge \(q\) is produced by adding electrons to the mass, by how much will the mass be changed?

A silicon sample is doped with phosphorus at 1 part per \(1.00 \cdot 10^{6} .\) Phosphorus acts as an electron donor, providing one free electron per atom. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3},\) and its atomic mass is \(28.09 \mathrm{~g} / \mathrm{mol}\) a) Calculate the number of free (conduction) electrons per unit volume of the doped silicon. b) Compare the result from part (a) with the number of conduction electrons per unit volume of copper wire (assume each copper atom produces one free (conduction) electron). The density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3},\) and its atomic mass is \(63.54 \mathrm{~g} / \mathrm{mol}\)

Two charged objects experience a mutual repulsive force of \(0.10 \mathrm{~N}\). If the charge of one of the objects is reduced by half and the distance separating the objects is doubled, what is the new force?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Energy Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Nuclear Physics

Read Explanation

Scientific Method Physics

Read Explanation

Physical Quantities And Units

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.

Sign-up for free