Problem

Two equal positive charges are at opposite corner…

05:50

Problem 28 Medium Difficulty

A particle of mass $1.00 \times 10^{-9} \mathrm{~kg}$ and charge $3.00 \mathrm{~pC}$ is moving in a vacuum chamber where the electric field has magnitude $2.00 \times 10^{3} \mathrm{N} / \mathrm{C}$ and is directed straight upward. Neglecting other forces except gravity, calculate the particle's (a) acceleration and (b) velocity after $2.00 \mathrm{~s}$ if it has an initial velocity of $5.00 \mathrm{~m} / \mathrm{~s}$ in the downward direction.

Answer

a. $-3.8 \mathrm{m} / \mathrm{s}^{2}$
b. $-2.6 \mathrm{m} / \mathrm{s}$

Topics

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Book Cover for College Physics 2017

College Physics 2017

Chapter 15

Electric Forces and Fields

Discussion

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Video Transcript

Hello, Strains. Let's start our discussion. Suppose a particle off mask one into 10. This bar minus nine kg. Ah, And Josh, if you PICO cooler that is three into Henry's bar minus 12. According is moving in a vacuum chamber there. The magnitude off electric field is two in tow. Tennis part three. You didn't, but okay is directed in straight up for it means that electric field Listen straight upward election okay? No, we have to neglect all other forces. Except gravity means that gravity is in downward direction. It is like that Is mg downward forces acting. Okay, we have to find out that acceleration or that particle A particle. Okay, so basically, force will be que e minus MGI because electric field is in a poor direction and ah, forced to do gravity isn't downward direction. So let s elevation off the particle will depend on this, and that will be a me. So it will make you be by m g Exploration will be Q E minus MGI by. I am No. By putting these values, we will get on G is 9.8 meters per second squared. We will get on acceleration off minus 3.8. Made a poor second square. Okay, No letter. Suppose you have to find the velocity off the particle activities because to second And if initial lost your article is five minute apart Second. Okay, So these equals toe you plus 80 and you will be five meters per second. And a solution is minus three. Point it. After a full second, we have to find. So by solving this, would will get the value off minus 2.6 meters per second. This is the velocity off the particle. Well, this is all from before this video. I hope you will like the video tank.

Deepak K.

Other Schools

Topics

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Book Cover for College Physics 2017

College Physics 2017

Chapter 15

Electric Forces and Fields

Top Physics 102 Electricity and Magnetism Educators

Elyse G.

Cornell University

Christina K.

Rutgers, The State University of New Jersey

Jared E.

University of Winnipeg

Meghan M.

McMaster University