Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1601

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

Short Answer

Expert verified
The kinetic energy of the charged particle after 't' seconds is \(K = \left[\left(E^2q^2t^2\right) / (2m)\right]\).

Step by step solution

01

Electromagnetic Force

The electromagnetic force acting on the charged particle in the electric field can be found using the formula: \( F = qE \) where F is the electromagnetic force, q is the charge of the particle, and E is the electric field strength.
02

Acceleration

To find the acceleration of the charged particle, use Newton's second law: \( F = ma \) where 'F' is the electromagnetic force and 'a' is the acceleration of the particle. Therefore, 'a' can be found as : \( a = \frac{F}{m} = \frac{qE}{m} \)
03

Velocity

The velocity 'v' of the charged particle after 't' seconds can be found using the formula: \( v = at \) where 'a' is the acceleration we found earlier, and 't' is the time. Put the found value of acceleration in this formula: \( v = \frac{qEt}{m} \)
04

Kinetic Energy

Finally, we can find the kinetic energy 'K' of the charged particle using the formula: \( K = \frac{1}{2} mv^2 \) Put the value of 'v' we found in this formula: \( K = \frac{1}{2} m \left(\frac{qEt}{m}\right)^2 = \frac{1}{2} \frac{q^2E^2t^2}{m} \) So, the kinetic energy of the charged particle after 't' seconds is: \( K = \left[\left(E^2q^2t^2\right) / (2m)\right] \) Hence, the correct option is (B).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 Charge \(q\) is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is $\ldots \ldots \ldots \ldots$ (A) \(\left(\mathrm{q} / \in_{0}\right)\) (B) (q / \(2 \in_{0}\) ) (C) \(\left(2 q / \epsilon_{0}\right)\) (D) Zero

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

Three particles, each having a charge of \(10 \mu \mathrm{c}\) are placed at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). The electrostatic potential energy of the system is (Given $\left.\left[1 /\left(4 \pi \epsilon_{0}\right)\right]=9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right)$ (A) \(100 \mathrm{~J}\) (B) \(27 \mathrm{~J}\) (C) Zero (D) Infinite

The plates of a parallel capacitor are charged up to \(100 \mathrm{~V}\). If $2 \mathrm{~mm}$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6 \mathrm{~mm}\) the dielectric constant of the plate is (A) 5 (B) 4 (C) \(1.25\) (D) \(2.5\)
See all solutions

Recommended explanations on English Textbooks

Essay Prompts

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Rhetoric

Read Explanation

Syntax

Read Explanation

Prosody

Read Explanation

Cues and Conventions

Read Explanation
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.

Sign-up for free