Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 21, Problem 30P
To determine

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

Expert Solution & Answer
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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^ .

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Chapter 21 Solutions

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