/*! 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 28 Two identically charged particle... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two identically charged particles are fastened to the two ends of a spring of spring constant \(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(10 \mathrm{~cm}\). The system rests on a smooth horizontal table. If the charge on each particle is \(2 \cdot 0 \times 10^{-8} \mathrm{C}\), find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Short Answer

Expert verified
The extension in the spring is approximately 0.899 mm, which is small compared to its natural length of 10 cm.

Step by step solution

01

Understand the Forces in the System

Each particle with charge exerts a repulsive electrostatic force on the other due to Coulomb's Law. The spring between the particles will exert a restoring force when it is extended. These two forces need to balance for the system to be in equilibrium.
02

Calculate the Electrostatic Force

The electrostatic force \( F_e \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:\[F_e = \frac{k_e q_1 q_2}{r^2}\]where \( k_e = 8.99 \times 10^9 \mathrm{~N} \, \mathrm{m}^2/\mathrm{C}^2 \) is Coulomb's constant. Since \( q_1 = q_2 = 2.0 \times 10^{-8} \mathrm{C} \) and the distance becomes \( l + x \) (natural length plus extension), we'll need \( r = 0.1 + x \).
03

Calculate the Restoring Force of the Spring

The restoring force \( F_s \) of a spring extended by a length \( x \) is given by Hooke's Law:\[F_s = k_s x\]where \( k_s = 100 \mathrm{~N/m} \) is the spring constant. This force acts to restore the spring to its natural length.
04

Set Up the Equilibrium Equation

In equilibrium, the magnitudes of the electrostatic force and the spring's restoring force are equal:\[\frac{8.99 \times 10^9 \times (2.0 \times 10^{-8})^2}{(0.1 + x)^2} = 100x\]This equation equates the electrostatic repulsion to the spring's restoring force.
05

Solve for the Extension x

Simplify and solve the equilibrium equation for \( x \). Using the equation:\[\frac{8.99 \times 10^9 \times 4.0 \times 10^{-16}}{(0.1 + x)^2} = 100x\]we substitute known quantities and assume \( x \) is small compared to 0.1, allowing us to approximate:\[(0.1 + x)^2 \approx 0.01 \implies x = 8.99 \times 10^{-4} \, \text{m or} \, 0.899 \, \text{mm}\]
06

Justify Assumption of Small x

The calculated extension \( x = 0.899 \times 10^{-3} \mathrm{m} \) is much smaller than the natural length of the spring \( 0.1 \text{ m} \), justifying that the assumption of the extension being small is valid.

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.

Electrostatic force
Electrostatic force is a fundamental concept in physics that describes the interactive force between electrically charged particles. According to Coulomb's Law, the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
This can be mathematically described using the formula: \[ F_e = \frac{k_e q_1 q_2}{r^2} \]Where:
  • \( F_e \) is the electrostatic force,
  • \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \mathrm{~N} \mathrm{~m}^2/\mathrm{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the charges,
  • \( r \) is the distance between the charges.
This law explains the repulsive or attractive force you can calculate between charged particles.
Such forces are key in understanding the behavior of particles when they are close to one another, and in exercises like ours, it helps in evaluating the forces acting on the spring connected between the two charged particles.
Hooke's Law
Hooke's Law is a principle of physics that states the restoring force exerted by a spring is directly proportional to the extension or compression of the spring. This fundamental concept helps in defining the behavior of springs when they are under deformation due to external forces.
The law is expressed with the formula:\[ F_s = k_s x \]Where:
  • \( F_s \) is the restoring force,
  • \( k_s \) is the spring constant, a measure of the spring's stiffness,
  • \( x \) is the extension or compression of the spring from its natural length.
This means when a spring is extended or compressed, it will exert a force to return to its original shape.
In our exercise, Hooke’s Law helps compute the force exerted by the spring when the charged particles either draw it out or compress it, balancing with the electrostatic force to maintain equilibrium.
Equilibrium in physics
Equilibrium in physics refers to a state where a system experiences no net force, resulting in no change in the motion of the particles involved. This concept can manifest in different types, such as static equilibrium where an object remains at rest, and dynamic equilibrium, where an object moves at a constant velocity.
In our context of a spring system with charged particles, achieving equilibrium means the repulsive electrostatic force between the charges must be equal and opposite to the restoring force of the spring.

Setting Up Equilibrium

To find the extension in the system as per the laws discussed, we set the electrostatic force \( F_e \) equal to the spring’s restoring force \( F_s \):\[\frac{8.99 \times 10^9 \times (2 \times 10^{-8})^2}{(0.1 + x)^2} = 100x\]This equation demonstrates how the forces balance, leading to the system's equilibrium. Solving this equation lets you determine the small extension of the spring required to balance these forces effectively.In conclusion, understanding equilibrium is crucial, as it ensures the forces in the system do not induce acceleration, resulting in a balanced and stable arrangement.

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 electric field \(\vec{E}=\vec{i} A x\) exists in the space, where \(A=10 \mathrm{~V} \mathrm{~m}^{-2}\). Take the potential at \((10 \mathrm{~m}, 20 \mathrm{~m})\) to be zero. Find the potential at the origin.

Three equal charges, \(2.0 \times 10^{-6} \mathrm{C}\) each, are held fixed at the three corners of an equilateral triangle of side \(5 \mathrm{~cm}\). Find the Coulomb force experienced by one of the charges due to the rest two.

An electric field of \(20 \mathrm{~N} \mathrm{C}^{-1}\) exists along the \(x\) -axis in space. Calculate the potential difference \(V_{B}-V_{A}\) where the points \(A\) and \(B\) are given by, (a) \(A=(0,0) ; B=(4 \mathrm{~m}, 2 \mathrm{~m})\) (b) \(A=(4 \mathrm{~m}, 2 \mathrm{~m}) ; B=(6 \mathrm{~m}, 5 \mathrm{~m})\) (c) \(A=(0,0) ; B=(6 \mathrm{~m}, 5 \mathrm{~m})\) Do you find any relation between the answers of parts (a), (b) and (c) ?

A ball of mass \(100 \mathrm{~g}\) and having a charge of \(4.9 \times 10^{-5} \mathrm{C}\) is released from rest in a region where a horizontal electric field of \(2.0 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}\) exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball ? (c) Where will the ball be at the end of \(2 \mathrm{~s}\) ?

Consider a circular ring of radius \(r\), uniformly charged with linear charge density \(\lambda\). Find the electric potential at a point on the axis at a distance \(x\) from the centre of the ring. Using this expression for the potential, find the electric field at this point

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.