/*! 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 42 A particle of mass \(m\) and cha... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle of mass \(m\) and charge \(q\) is placed at rest in a uniform electric field \(E\) and then released. The \(\mathrm{KE}\) attained by the particle after moving a distance \(y\) is (a) \(q E y^{2}\) (b) \(q E^{2} y\) (c) \(q E y\) (d) \(q^{2} E y\)

Short Answer

Expert verified
(c) \( q E y \)

Step by step solution

01

Understand the Problem

We have a charged particle with mass \( m \) and charge \( q \) initially at rest in a uniform electric field \( E \). The particle moves a distance \( y \) and we need to determine the kinetic energy (\( \mathrm{KE} \)) it attains.
02

Use the Work-Energy Principle

According to the work-energy principle, the work done on the particle by the electric field will convert into its kinetic energy. This can be expressed as \( W = \Delta \mathrm{KE} \).
03

Calculate the Work Done by the Electric Field

The work done by an electric field on a charge is given by the formula \( W = qEd \), where \( q \) is the charge, \( E \) is the electric field strength, and \( d \) is the distance moved in the direction of the field. Here, \( d = y \), so \( W = qEy \).
04

Equate Work to Kinetic Energy

Set the work done equal to the kinetic energy the particle gains. Therefore, \( \Delta \mathrm{KE} = qEy \).
05

Choose the Correct Option

Now that we have derived that \( \mathrm{KE} = qEy \), we can match it with the given options. The correct answer is (c) \( q E y \).

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Key Concepts

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

Work-Energy Principle
The Work-Energy Principle is an essential concept in physics that helps us understand how forces acting on an object can change its energy state. This principle states that the work done by the net force on an object is equal to the change in its kinetic energy. In mathematical terms, this can be expressed as: \[ W = \Delta KE \] Where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy. To apply this principle, consider a particle in an electric field. As it moves through the field over a distance, the electric field does work on the charged particle.
  • The work done by the electric field can be calculated using the expression \( W = qE d \), where \( q \) is the charge, \( E \) is the electric field strength, and \( d \) is the distance moved in the field’s direction.
  • This work leads to a change in the particle's kinetic energy, transforming the potential energy into motion or KE.
The understanding of this principle is crucial for solving problems where forces result in an object's motion, like in the given problem scenario.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It’s an invaluable concept in physics because it connects with the forces acting on a body, such as in situations involving electric fields. The formula for kinetic energy is given by: \[ KE = \frac{1}{2}mv^2 \] Where \( m \) is the mass of the object, and \( v \) is its velocity. However, when solving problems involving electric forces, you often find the kinetic energy by assessing the work done.
  • In the context of the electric field, when a particle moves through a field, the work done by the field changes the particle’s kinetic energy.
  • This change is what propels the particle, increasing its motion speed as it absorbs the field's energy.
The expression derived in the exercise showcases that the change in kinetic energy can be directly calculated as \( KE = qEy \), using the work done by the uniform electric field. Understanding such derivations is vital for predicting how a particle will move under the influence of forces like electricity.
Charge and Mass of a Particle
The characteristics of a particle, specifically its charge and mass, have significant roles in how it behaves when subjected to forces, especially in an electric field.
  • Charge (\( q \)): This property determines the particle's interaction with electric fields. A higher charge implies a stronger force exerted by the electric field, leading to more work done and hence, more kinetic energy gained by the particle.
  • Mass (\( m \)): Mass affects how easily the particle can be accelerated. According to Newton's second law of motion, force equals mass times acceleration \( F = ma \). Thus, a greater mass means more force is needed to achieve the same acceleration or kinetic energy.
These properties affect equations like \( KE = qEy \), where \( KE \) is derived from the force due to charge that moves the mass across a distance in the field. Understanding the interplay of charge and mass helps explain how particles speed up or slow down in fields and forces.

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Most popular questions from this chapter

The region surrounding an oscillating electric dipole has (a) electric field only (b) magnetic field only (c) electric as well as magnetic field (d) nothing can be predicted

It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss' theorem because (a) Gauss' law fails in this case (b) this problem does not have spherical symmetry (c) Coulomb's law is more fundamental than Gauss' law (d) spherical Gaussian surface will alter the dipole moment

An electron moves through a small distance in a uniform electric field. The magnitude of electric field is \(2 \times 10^{4} \mathrm{NC}^{-1}\). Now, if the direction of field is reversed keeping the magnitude same and a proton moves through the same distance, then which of the following options is correct? (a) The time of fall will be more in case of electron (b) The time of fall will be more in case of proton (c) The time of fall will be same in both cases (d) The time of fall will be independent of charge

The electric flux \(\varphi\) through a hemisphere surface of radius \(R\), placed in a uniform electric field of intensity \(E\) parallel to the axis of its circular plane is (a) \(2 \pi R E\) (b) \(2 \pi R^{2} E\) (c) \(\pi R^{2} E\) (d) \((4 / 3) \pi R^{3} E\)

A unit positive charge has to be brought from infinity to a mid-point between two charges \(20 \mu \mathrm{C}\) and \(10 \mu \mathrm{C}\) separated by a distance of \(50 \mathrm{~m}\). How much work will be required? (a) \(10.8 \times 10^{4} \mathrm{~J}\) (b) \(10.8 \times 10^{3} \mathrm{~J}\) (c) \(1.08 \times 10^{6} \mathrm{~J}\) (d) \(0.54 \times 10^{5} \mathrm{~J}\)

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