Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 21, Problem 30P
To determine

The force exerted by the three point charges on the fourth charge.

Expert Solution & Answer
Check Mark

Explanation of Solution

Given: The magnitude of the three point charges is 3.00nC .

The length of the side of the square is 5.00cm .

Formula used:

Write the expression for the coulomb law.

  F=kq1q2r2

Here, F is the force, k is the constant, q1 is the point charge, q2 is the second point charge and r is the distance between the two point charges.

The representation of the four charges at the vertices of the square is shown below:

  Physics for Scientists and Engineers, Chapter 21, Problem 30P

Here, F1,4 is the force on the fourth charge due to first charge, F3,4 is the force on the fourth charge due to third charge, F2,4 is the force fourth charge due to second charge.

Write the expression for the net force acting on the fourth charge.

  F4=F1,4+F2,4+F3,4   ...... (1)

Write the expression for the force exerted on fourth charge due to first charge.

  F1,4=kq1q4r1,42j^   ...... (2)

Here, r1,4 is the distance between the first charge and the fourth charge.

Write the expression for the force exerted by the second charge on the fourth charge.

  F2,4=kq2q4r2,42i^   ...... (3)

Here, r2,4 is the distance between the second point charge and the fourth point charge.

Write the expression for the force exerted by the third charge on the fourth charge.

  F3,4=kq3q4r3,42r^3,4   ...... (4)

Here, r^3,4 is the unit vector that points from q3 to q4 , r3,4 is the distance between the third charge and the fourth charge.

Write the expression for r3,4 in terms of distance between the third and the first charge and between the first and the fourth charge.

  r3,4=r3,1+r1,4   ...... (5)

Write the expression for the unit vector.

  r^3,4=r3,4|r3,4|

     ...... (6)

Here, |r3,4| is the magnitude of position vector.

Calculation:

Substitute 8.988×109Nm2/C2 for k , 3.00nC for q2 , 3.00nC for q4 , 0.05m for r1,4 in equation (2).

  F1,4=( 8.988× 10 9 Nm 2 / C 2 ) [ 3.00nC]2 ( 0.05m )2j^F1,4=(3.23× 10 5N)j^

Substitute 8.988×109Nm2/C2 for k , 3.00nC for q1 , 3.00nC for q4 , 0.05m for r1,4 in equation (3).

  F2,4=( 8.988× 10 9 Nm 2 / C 2 ) [ 3.00nC]2 ( 0.05m )2i^F2,4=(3.23× 10 5N)i^

Substitute (0.05m)i^ for r3,1 and (0.05m)j^ for r1,4 in equation (5).

  r3,4=(0.05m)i^+(0.05m)j^

Substitute (0.05m)i^+(0.05m)j^ for r3,4 in equation (6).

  r^3,4=( 0.05m)i^+( 0.05m)j^| ( 0.05m ) 2+ ( 0.05m ) 2|r^3,4=0.707i^+0.707j^

Substitute 8.988×109Nm2/C2 for k , 3.00nC for q3 , 3.00nC for q4 , 0.707i^+0.707j^ for r^3,4

  0.052m for r3,4 in equation (4).

  F3,4=( 8.988× 10 9 Nm 2 / C 2 )3.00nC( 3.00nC) ( 0.05 2 m )20.707i^+0.707j^F3,4=(1.14× 10 5N)i^(1.14× 10 5N)j^

Substitute (3.23×105N)j^ for F1,4 , (3.23×105N)i^ for F2,4 and (1.14×105N)i^(1.14×105N)j^ for F3,4 in equation (1).

  F4=(3.23× 10 5N)j^+(3.23× 10 5N)i^(1.14× 10 5N)i^(1.14× 10 5N)j^F4=(2.5× 10 5N)i^+(2.5× 10 5N)j^

Conclusion:

The force exerted by the three point charges on the fourth charge is (2.5×105N)i^+(2.5×105N)j^ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Two point charges q1 and q2 are 3.15 m apart, and their total charge is 22 µC. Note that you may need to solve a quadratic equation to reach your answer. (a) If the force of repulsion between them is 0.10 N, what are the two charges (in µC)? smaller value  µCl arger value  µC (b) If one charge attracts the other with a force of 0.457 N, what are the two charges (in µC)? smaller value  µC larger value  µC
A square with sides of length L has a point charge at each ofits four corners. Two corners that are diagonally opposite havecharges equal to +3.25 mC; the other two diagonal corners havecharges Q. Find the magnitude and sign of the charges Q such thateach of the +3.25@mC charges experiences zero net force
Three point charges 1.0uC and 10 uC are placed on a straight line and net force on 0.10 uC is zero. Find x in cm.

Chapter 21 Solutions

Physics for Scientists and Engineers

Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Electric Fields: Crash Course Physics #26; Author: CrashCourse;https://www.youtube.com/watch?v=mdulzEfQXDE;License: Standard YouTube License, CC-BY