/*! 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 \(\mathrm{mmh}\) A small object ... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{mmh}\) A small object has a mass of \(3.0 \times 10^{-3} \mathrm{kg}\) and a charge of \(-34 \mu \mathrm{C}\). It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.

Short Answer

Expert verified
The electric field magnitude is \( 2.206 \times 10^{5} \) N/C and is in the \(-x\) direction.

Step by step solution

01

Understand the Problem

We need to determine the magnitude and direction of the electric field that causes a charged object to accelerate. The object has a given mass, charge, and acceleration.
02

Write Down the Given Values

Mass of the object, \( m = 3.0 \times 10^{-3} \) kg.Charge of the object, \( q = -34 \times 10^{-6} \) C.Acceleration of the object, \( a = 2.5 \times 10^{3} \) m/s².
03

Use Newton’s Second Law

Newton's second law states that \( F = ma \). Since force \( F \) is also given by \( F = qE \) in an electric field, we can equate the two expressions: \[ ma = qE \]
04

Solve for the Electric Field Intensity \( E \)

Rearrange the equation from Step 3 to solve for \( E \):\[ E = \frac{ma}{q} \]Substitute the given values: \[ E = \frac{(3.0 \times 10^{-3} \text{ kg})(2.5 \times 10^{3} \text{ m/s}^2)}{-34 \times 10^{-6} \text{ C}} \]
05

Calculate the Magnitude of \( E \)

Calculate the expression obtained in Step 4:\[ E = \frac{7.5}{-34 \times 10^{-6}} \]\[ E \approx -2.206 \times 10^{5} \text{ N/C} \]The magnitude is \( 2.206 \times 10^{5} \text{ N/C} \).
06

Determine the Direction of \( E \)

Since the acceleration is in the \(+x\) direction but the charge is negative, the electric field \( E \) is in the opposite direction of the acceleration vector, meaning it is in the \(-x\) direction.

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is one of the foundational principles of physics. It establishes the relationship between the motion of an object and the forces acting on it. According to this law, the force (\( F \)) acting on an object is equal to the mass (\( m \)) of that object multiplied by its acceleration (\( a \)). In mathematical terms, this is expressed as:\[ F = ma \]This equation tells us that a stronger force will result in a greater acceleration if the mass remains constant. Additionally, for the same mass, an increase in acceleration demands a greater force. Newton's Second Law connects to the concept of an electric field because when a charged object is placed in an electric field, the electric force acting upon the object causes it to accelerate. By using this principle, we can relate the electric force (\( F = qE \)) to Newton's formula (\( F = ma \)) and solve for the electric field (\( E \)). It serves as a bridge to understanding the dynamics of charged objects in electric fields.
Acceleration
Acceleration is a physical concept that describes how the velocity of an object changes over time. When an object speeds up, slows down, or changes direction, it is accelerating. In this particular exercise, the object experiences an acceleration given as (\( 2.5 \times 10^3 \text{ m/s}^2 \)). This acceleration is crucial as it helps us determine the force acting on the mass of the object, and thus the strength of the electric field.- Acceleration is a vector quantity, meaning it has both a magnitude (how much) and a direction (which way).- The unit of acceleration is meters per second squared (\( \text{m/s}^2 \)).Knowing the acceleration of the object, we can use Newton’s Second Law to find the force causing this acceleration. This force, in an electric context, is due to the electric field acting on the charge.
Charge
Charge is a fundamental property of matter that causes it to experience a force in an electric field. Electric charge comes in two types: positive and negative. In this exercise, the object has a charge of \(-34 \mu C \), which is equal to \(-34 \times 10^{-6} \text{ C}\). This negative charge plays a critical role in determining the direction of the electric force and the electric field itself.- Charges exert forces on each other: like charges repel, unlike charges attract.- The magnitude of the charge is the numerical value representing how much charge an object has.In an electric field, a charged object experiences a force (\( F = qE \)), which results in acceleration (\( F = ma \)). The sign of the charge affects the direction of the force. In our scenario, since the charge is negative, the electric field direction will be opposite to the acceleration direction.
Magnitude and Direction
The concepts of magnitude and direction are essential when discussing vectors in physics. The magnitude refers to the size or length of the vector, while the direction indicates where the vector points. For an electric field, the magnitude describes the strength of the field, and direction shows the path along which a positive charge would naturally move.In this problem:- The calculated magnitude of the electric field is \( 2.206 \times 10^5 \text{ N/C} \). This quantifies how strong the electric field is in influencing the charged object.- The direction is determined by the acceleration and charge. Given that the acceleration is in the \(+x\) direction, and the charge is negative, the electric field's direction is opposite to the acceleration, in the \(-x\) direction.Understanding magnitude and direction helps us not only calculate the force that causes the motion but also predict how the object will behave within the field.

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Most popular questions from this chapter

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol}\), and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?

Four point charges have the same magnitude of \(2.4 \times 10^{-12} \mathrm{C}\) and are fixed to the corners of a square that is \(4.0 \mathrm{cm}\) on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

ssm Two point charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{cm},\) and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{cm} .\) Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{cm},\) and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at the origin.

ssm A tiny ball (mass \(=0.012 \mathrm{kg}\) ) carries a charge of \(-18 \mu \mathrm{C}\). What electric field (magnitude and direction) is needed to cause the ball to float above the ground?

Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. However, the electric field between the plates of the first capacitor is \(2.2 \times 10^{5} \mathrm{N} / \mathrm{C},\) whereas the field within the second capacitor is \(3.8 \times 10^{5} \mathrm{N} / \mathrm{C} .\) Determine the ratio \(r_{2} / r_{1}\) of the plate radius for the second capacitor to the plate radius for the first capacitor.

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