/*! 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 44 A spring with an unstrained leng... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring with an unstrained length of \(0.074 \mathrm{m}\) and a spring constant of \(2.4 \mathrm{N} / \mathrm{m}\) hangs vertically downward from the ceiling. A uniform electric field directed vertically upward fills the region containing the spring. A sphere with a mass of \(5.1 \times 10^{-3} \mathrm{kg}\) and a net charge of \(+6.6 \mu \mathrm{C}\) is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is \(0.059 \mathrm{m} .\) What is the magnitude of the external electric field?

Short Answer

Expert verified
The magnitude of the external electric field is approximately 2118.33 N/C.

Step by step solution

01

Understand the forces acting on the sphere

At equilibrium, the forces acting on the sphere include the gravitational force pulling it downward, the electric force pushing it upward, and the spring force (Hooke's Law) opposing the spring's compression or elongation. These forces must balance each other out.
02

Calculate gravitational force

The gravitational force on the sphere can be calculated using the formula: \[ F_g = m \cdot g \]where \( m = 5.1 \times 10^{-3} \text{ kg} \) is the mass of the sphere and \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. Substitute the given values:\[ F_g = 5.1 \times 10^{-3} \times 9.81 = 0.049981 \text{ N} \]
03

Calculate spring force using Hooke's Law

Using Hooke's Law, the spring force is given by \[ F_s = k \cdot \Delta x \]where \( k = 2.4 \text{ N/m} \) is the spring constant and \( \Delta x = L_0 - L \) is the change in spring length, with \( L_0 = 0.074 \text{ m} \) being the unstrained length, and \( L = 0.059 \text{ m} \) being the equilibrium length.Calculate the change in length:\[ \Delta x = 0.074 - 0.059 = 0.015 \text{ m} \]Now, substitute into Hooke's law:\[ F_s = 2.4 \times 0.015 = 0.036 \text{ N} \]
04

Calculate the electric force and set equation for equilibrium

At equilibrium, the sum of the forces must be zero. Hence, \[ F_g = F_s + F_e \]where \( F_e \) (electric force) is given by\[ F_e = q \cdot E \]for charge \( q = 6.6 \times 10^{-6} \text{ C} \) and unknown electric field magnitude \( E \). Rearrange the equation for equilibrium:\[ F_e = F_g - F_s \]Substitute the known forces:\[ F_e = 0.049981 - 0.036 = 0.013981 \text{ N} \]
05

Solve for the electric field

Using the expression from Step 4 for \( F_e \), solve for \( E \):\[ E = \frac{F_e}{q} = \frac{0.013981}{6.6 \times 10^{-6}} \]Calculate:\[ E \approx 2118.33 \text{ N/C} \]
06

Verify the calculation and result

Ensure that all calculations are correct and consistent with logical expectations. The result verifies the conceptual understanding of equilibrium and the balance of forces, confirming the result is coherent.

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Key Concepts

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

Hooke's Law
Hooke's Law is an essential principle in physics that explains how springs behave when force is applied to them. It signifies the relationship between the force exerted by a spring and the distance it is stretched or compressed. Formally, Hooke's Law is described by the equation, \( F_s = k \cdot \Delta x \), where \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( \Delta x \) is the change in the spring's length from its natural length. The spring constant \( k \) is crucial as it indicates how stiff or flexible the spring is. A large \( k \) means the spring is stiffer, requiring more force to stretch or compress. In the given exercise, the spring's constant is \( 2.4 \text{ N/m} \), showing the spring is relatively flexible since it alters its length under the sphere's weight and the external force.
Gravitational Force
The gravitational force is a fundamental force that attracts two bodies with mass. It's the reason objects fall towards the Earth when dropped. This force can be calculated with the formula \( F_g = m \cdot g \), where \( F_g \) is the gravitational force, \( m \) is the object's mass, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) on Earth's surface. In the problem, the sphere has a mass of \( 5.1 \times 10^{-3} \, \text{kg} \), producing a downward force. Calculating \( F_g \) provides \( 0.049981 \text{ N} \), reflecting the gravitational pull on the sphere. This force is essential as it influences the equilibrium position of the spring-sphere system against the upward electric force.
Electric Force
Electric force is the force between charged objects. It can either attract or repel depending on the nature of the charges. The electric force acting on a charge in an electric field is given by \( F_e = q \cdot E \), where \( F_e \) is the electric force, \( q \) is the charge on the object, and \( E \) is the electric field strength. In our scenario, the sphere has a charge of \( +6.6 \mu \text{C} \) and interacts with an external electric field that counteracts the gravitational and spring forces. By rearranging the equilibrium equation to solve for the electric force, we obtain \( F_e = 0.013981 \text{ N} \). Using this, we can find \( E \), the desired magnitude of the electric field, equaling approximately \( 2118.33 \text{ N/C} \). Electric forces like this are integral in balancing other forces acting on the system, achieving equilibrium.
Equilibrium
Equilibrium refers to the state of a system where all forces acting upon it balance out, resulting in no net force. This means the object remains stationary or moves at a constant speed. In the exercise, we observe a spring hanging with a sphere attached to its end, influenced by gravitational and electric forces. At equilibrium, the downward gravitational force equals the sum of the upward spring force and electric force. Mathematically, \( F_g = F_s + F_e \). This relationship allows us to understand where and why the sphere stops moving, showing that balance occurs when the net force is zero. Recognizing equilibrium is crucial in solving complex physics problems, as it guides you through understanding how forces interact and balance in various physical systems.

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Most popular questions from this chapter

ssm Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{cm}\) and (b) \(x=+6.0 \mathrm{cm}\)

Four identical metallic objects carry the following charges: +1.6 \(+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into contact, so that each touches the others. Then they are separated. (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

The total electric field \(\overrightarrow{\mathbf{E}}\) consists of the vector sum of two parts. One part has a magnitude of \(E_{1}=1200 \mathrm{N} / \mathrm{C}\) and points at an angle \(\theta_{1}=35^{\circ}\) above the \(+x\) axis. The other part has a magnitude of \(E_{2}=1700 \mathrm{N} / \mathrm{C}\) and points at an angle \(\theta_{2}=55^{\circ}\) above the \(+x\) axis. Find the magnitude and direction of the total field. Specify the directional angle relative to the \(+x\) axis.

A spherical surface completely surrounds a collection of charges. } . Find the electric flux through the surface if the collection consists of (a) a single \(+3.5 \times 10^{-6} \mathrm{C}\) charge, (b) a single \(-2.3 \times 10^{-6} \mathrm{C}\) charge, and (c) both of the charges in (a) and (b).

A charge of \(-3.00 \mu C\) is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of \(0.100 \mathrm{m} .\) The charges on the circle are \(-4.00 \mu \mathrm{C}\) at the position due north and \(+5.00 \mu \mathrm{C}\) at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east.

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