/*! 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 40 \( \mathrm{~A}-4.0-\mu \mathrm{C... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\( \mathrm{~A}-4.0-\mu \mathrm{C}\) charge lies \(20.0 \mathrm{~cm}\) to the right of a \(2.0-\mu \mathrm{C}\) charge on the \(x\) -axis. What is the force on the \(2.0-\mu C\) charge?

Short Answer

Expert verified
Question: Determine the force exerted on the 2.0-μC charge, given two charges of -4.0-μC and 2.0-μC are separated by a distance of 20.0 cm. Answer: The force exerted on the 2.0-μC charge is -1.798 x 10^3 N. The negative sign indicates that the force is attractive, meaning the charges are opposite in nature and are pulling each other together.

Step by step solution

01

Convert distances to meters

Given the distance is \(20.0\) cm, we will convert it to meters by dividing by 100: \(20.0 \mathrm{~cm} = \frac{20.0}{100} \mathrm{~m} = 0.200 \mathrm{~m}\)
02

Identify charges and distance

We have two charges, \(q_1 = -4.0-\mu C\) and \(q_2 = 2.0-\mu C\), and the distance between them is \(r = 0.200 \mathrm{~m}\).
03

Apply Coulomb's Law formula

We will use the following formula: \(F = k\frac{q_1q_2}{r^2}\). Plugging in the charges, distance, and constant values: \(F = (8.99 \times 10^9 \ Nm^2C^{-2})\frac{(-4.0 \times 10^{-6} \ C)(2.0 \times 10^{-6} \ C)}{(0.200 \ m)^2}\)
04

Calculate the force

Now, we will perform the calculations to find the force: $F = (8.99 \times 10^9 \ Nm^2C^{-2})(-8.0 \times 10^{-12} \ C^2) \div (0.0400 \ m^2) \\ F = (-71.92 \times 10^{-3} \ Nm^2) \div (0.0400 \ m^2) \\ F = -1.798 \times 10^3 \ N$ The force on the \(2.0-\mu C\) charge is \(-1.798 \times 10^3 \ N\). The negative sign indicates that the force is attractive, meaning the charges are opposite in nature and are pulling each other together.

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 91Ó°ÊÓ!

Key Concepts

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

Electric Charge
Understanding electric charges is fundamental to studying electric forces and how they interact. An electric charge is a property of matter that allows it to experience a force when near other electrically charged matter. There are two types of electric charges: positive (\texttt{+}) and negative (\texttt{-}). Like charges repel each other while opposite charges attract. The unit of electric charge is the coulomb (C). In our problem, we dealt with microcoulombs (\texttt{\(\mu\)C}), which are one millionth of a coulomb (\(10^{-6}\) C).

It is crucial to know that electric charges can be measured by static (stationary) or dynamic (flowing, as in current) means, and they can exist on objects as excess or deficit of electrons. The balance or imbalance of these charges leads to the presence of an electric force.
Electric Force
The electric force is the attraction or repulsion between electric charges. It is described quantitatively by Coulomb's Law. This law states that the magnitude of the electric force (\texttt{F}) between two point charges is directly proportional to the product of the magnitudes of charges (\texttt{\(q_1\)} and \texttt{\(q_2\)}) and inversely proportional to the square of the distance (\texttt{\(r\)}) between them. The constant of proportionality is denoted by \texttt{k}, known as Coulomb's constant (\(8.99 \times 10^9 \texttt{ Nm}^2\texttt{C}^{-2}\)).

Mathematically, Coulomb's Law is expressed as \texttt{\(F = k \frac{q_1q_2}{r^2}\)}. In the problem presented, we used this formula to calculate the electric force between two charges. It's important to note that the direction of the force is not given by this equation and depends on the nature of the charges: if the product of charges (sign-wise) is positive, the forces are repulsive; if negative, they are attractive, as was the case in our problem.
Physics Problem Solving
Solving physics problems requires a systematic approach that often involves several key steps: understanding the problem, visualizing it, identifying the relevant concepts, and applying mathematical equations.

In our example, we followed these steps to solve for the electric force between two charges. First, we converted units to ensure consistency. Then, the problem was assessed, identifying the two charges and the distance between them. Next, we applied the appropriate physics principle, Coulomb's Law. Finally, we executed the calculations to arrive at a quantifiable force.

When solving physics problems, it's vital to keep track of units, signs (which indicate directions), and constants. Moreover, one should check the final answer to see if it makes physical sense. For instance, in our problem, the negative force indicated an attractive interaction between the charges, consistent with their opposite natures. This logical check can be an invaluable tool for students to confirm their understanding of the concepts at play.

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

A small ball with a mass of \(30.0 \mathrm{~g}\) and a charge of \(-0.200 \mu \mathrm{C}\) is suspended from the ceiling by a string. The ball hangs at a distance of \(5.00 \mathrm{~cm}\) above an insulating floor. If a second small ball with a mass of \(50.0 \mathrm{~g}\) and a charge of \(0.400 \mu \mathrm{C}\) is rolled directly beneath the first ball, will the second ball leave the floor? What is the tension in the string when the second ball is directly beneath the first ball?

Another unit of charge is the electrostatic unit (esu). It is defined as follows: Two point charges, each of 1 esu and separated by \(1 \mathrm{~cm}\), exert a force of exactly 1 dyne on each other: 1 dyne \(=1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}=1 \cdot 10^{-5} \mathrm{~N}\). a) Determine the relationship between the esu and the coulomb. b) Determine the relationship between the esu and the elementarv charoe

A metal plate is connected by a conductor to a ground through a switch. The switch is initially closed. A charge \(+Q\) is brought close to the plate without touching it, and then the switch is opened. After the switch is opened, the charge \(+Q\) is removed. What is the charge on the plate then? a) The plate is uncharged. b) The plate is positively charged. c) The plate is negatively charged. d) The plate could be either positively or negatively charged, depending on the charge it had before \(+Q\) was brought near.

In gaseous sodium chloride, chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.24 \mathrm{nm}\). Suppose a free electron is located \(0.48 \mathrm{nm}\) above the midpoint of the sodium chloride molecule. What are the magnitude and the direction of the electrostatic force the molecule exerts on it?

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?

See all solutions

Recommended explanations on Physics Textbooks

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.