COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 16, Problem 51QAP
To determine

The net force exerted on each charge by other charges.

Expert Solution & Answer
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Answer to Problem 51QAP

Net force on charge A is 36.3 N at angle 24.4° below negative x -axis. Net force on charge B is 43.45 N at angle 11° below the positive x -axis. Net force on charge C is 25.2 N at angle 67.5° above the negative x -axis.

Explanation of Solution

Given:

Charge A, charge B and charge C are placed in a coordinate system as shown below.

  COLLEGE PHYSICS-ACHIEVE AC (1-TERM), Chapter 16, Problem 51QAP

Formula used:

From Coulomb's law

  Fq1 on q2=k| q 1 q 2 |r2where,r=seperationbetweencharge1andcharge2

Calculation:

For charge A,

x component of net force on charge A :

  F net on A,x=FB on A,x+FC on A,x=FB on A,x+0=k| q B q A |r AB2=( 8.99× 10 9 N m 2 C 2 )|6.00× 10 6C||3.00× 10 6C| ( 7.00× 10 2 m )2=33.02N

Because charges A and B have the same sign, the force on charge A is repulsive and acts in the

negative x direction.

Thus, the net force on charge A in the x -direction is - 33.0 N.

The y component of net force on charge A :

  F net on A,y=FB on A,y+FC on A,y=0+FC on A,y=k| q C q A |r AC2F net on A,y=( 8.99× 10 9 N m 2 C 2 )|2.00× 10 6C||3.00× 10 6C| ( 6.00× 10 2 m )2F net on A,y=14.98N

Because charges A and C have the same sign, the force on charge A is repulsive and acts in the

negative y -direction.

Thus, the net force on charge A in the y -direction is -15.0 N

The magnitude of the net force on charge A :

  F net on A= ( F  net on A,x )2+ ( F  net on A,y )2F net on A= (33.02N)2+ (14.98N)2F net on A=36.3N

The direction of the net force on charge A :

  θA=tan1(14.98N33.02N)=24.4°

For charge B,

x component of net force on charge B :

   F net on B,x = F A on B,x + F ConB,x = k| q A || q B | r AB 2 + k| q C q B | r CB 2 cos θ CB

   F net on B,x = k| q A q B | r AB 2 + k| q C q B | ( r AB 2 + r AC 2 ) ( r AB r AB 2 + r AC 2 )

   F net on B,x = k| q A q B | r AB 2 + k| q C q B | r AB ( r AB 2 + r AC 2 ) 3/2

   F net on B,x = ( 8.99× 10 9 N m 2 C 2 )| 3.00× 10 6 C || 6.00× 10 6 C | ( 7.00× 10 2 m ) 2

   + ( 8.99× 10 9 N m 2 C 2 )| 2.00× 10 6 C || 6.00× 10 6 C |( 7.00× 10 2 m ) [ ( 7.00× 10 2 m ) 2 + ( 6.00× 10 2 m ) 2 ] 3/2

   F net on B,x =42.66N

y component of net force on charge B :

  F net on B,y=FA on B,y+FConB,y=0+k| q C q B |r BC2sinθCB=k| q C q B |r AB2+r AC2( r AC r AB 2 + r AC 2 )=k| q C q B | r AC ( r AB 2 + r AC 2 ) 3/2=( 8.99× 10 9 N m 2 C 2 )|2.00× 10 6C||6.00× 10 6C|( 6.00× 10 2 m) [ ( 7.00× 10 2 m ) 2 + ( 6.00× 10 2 m ) 2 ] 3/2=8.260N

Because charges C and B have the same sign, the force on charge B is repulsive and acts in the negative y-direction. Thus, the net force on charge B in the y -direction is -8.26 N

The magnitude of the net force on charge B :

  F net on B= ( F  net on B,x )2+ ( F  net on B,y )2F net on B= (42.66N)2+ (8.26N)2F net on B=43.45N

The direction of the net force on charge B :

  θB=tan1(8.260N42.66N)=11.0°

For charge C,

x component of net force on charge C :

  F net on c,x=FA on C,x+FB on C,x=0+k| q B q C |r BC2cosθBCF net on c,x=k| q B q C |r AB2+r AC2( r AB r AB 2 + r AC 2 )F net on c,x=k| q B q C | r AB ( r AB 2 + r AC 2 ) 3/2F net on c,x=( 8.99× 10 9 N m 2 C 2 )|6.00× 10 6C||2.00× 10 6C|( 7.00× 10 2 m) [ ( 7.00× 10 2 m ) 2 + ( 6.00× 10 2 m ) 2 ] 3/2F net on c,x=9.636N(directionofforceistowardsnegativex-axis)

Because charges C and B have the same sign, the force on charge C is repulsive and acts in the

negative x direction. Thus, the net force on charge C in the x-direction is -9.64 N

y component of net force on charge C :

   F netonC,y = F AonC,y + F B on C,y = k| q A q C | r AC 2 + k| q B || q C | r BC 2 sin θ BC

   F netonC,y = k| q A q C | r AC 2 + k| q B q C | r AB 2 + r AC 2 ( r AC r AB 2 + r AC 2 )

   F netonC,y = k| q A || q C | r AC 2 + k| q B | q C | r AC ( r AB 2 + r AC 2 ) 3/2

   F netonC,y = ( 8.99× 10 9 N m 2 C 2 )| 3.00× 10 6 C || 2.00× 10 6 C | ( 6.00× 10 2 m ) 2

   + ( 8.99× 10 9 N m 2 C 2 )| 6.00× 10 6 C || l2.00× 10 6 C |( 6.00× 10 2 m ) [ ( 7.00× 10 2 m ) 2 + ( 6.00× 10 2 m ) 2 ] 3/2

   F netonC,y =23.24N

The magnitude of the net force on charge C :

  F net on C= ( F  net on C,x )2+ ( F  net on C,y )2F net on C= (-9.636N)2+ (23.24N)2F net on C=25.2N

The direction of the net force on charge C :

  θC=tan1(23.24N9.636N)=67.5°

Conclusion:

Net force on charge A is 36.3 N at angle 24.4° below negative x -axis. Net force on charge B is 43.45 N at angle 11° below the positive x -axis. Net force on charge C is 25.2 N at angle 67.5° above the negative x -axis.

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Chapter 16 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

Ch. 16 - Prob. 11QAPCh. 16 - Prob. 12QAPCh. 16 - Prob. 13QAPCh. 16 - Prob. 14QAPCh. 16 - Prob. 15QAPCh. 16 - Prob. 16QAPCh. 16 - Prob. 17QAPCh. 16 - Prob. 18QAPCh. 16 - Prob. 19QAPCh. 16 - Prob. 20QAPCh. 16 - Prob. 21QAPCh. 16 - Prob. 22QAPCh. 16 - Prob. 23QAPCh. 16 - Prob. 24QAPCh. 16 - Prob. 25QAPCh. 16 - Prob. 26QAPCh. 16 - Prob. 27QAPCh. 16 - Prob. 28QAPCh. 16 - Prob. 29QAPCh. 16 - Prob. 30QAPCh. 16 - Prob. 31QAPCh. 16 - Prob. 32QAPCh. 16 - Prob. 33QAPCh. 16 - Prob. 34QAPCh. 16 - Prob. 35QAPCh. 16 - Prob. 36QAPCh. 16 - Prob. 37QAPCh. 16 - Prob. 38QAPCh. 16 - Prob. 39QAPCh. 16 - Prob. 40QAPCh. 16 - Prob. 41QAPCh. 16 - Prob. 42QAPCh. 16 - Prob. 43QAPCh. 16 - Prob. 44QAPCh. 16 - Prob. 45QAPCh. 16 - Prob. 46QAPCh. 16 - Prob. 47QAPCh. 16 - Prob. 48QAPCh. 16 - Prob. 49QAPCh. 16 - Prob. 50QAPCh. 16 - Prob. 51QAPCh. 16 - Prob. 52QAPCh. 16 - Prob. 53QAPCh. 16 - Prob. 54QAPCh. 16 - Prob. 55QAPCh. 16 - Prob. 56QAPCh. 16 - Prob. 57QAPCh. 16 - Prob. 58QAPCh. 16 - Prob. 59QAPCh. 16 - Prob. 60QAPCh. 16 - Prob. 61QAPCh. 16 - Prob. 62QAPCh. 16 - Prob. 63QAPCh. 16 - Prob. 64QAPCh. 16 - Prob. 65QAPCh. 16 - Prob. 66QAPCh. 16 - Prob. 67QAPCh. 16 - Prob. 68QAPCh. 16 - Prob. 69QAPCh. 16 - Prob. 70QAPCh. 16 - Prob. 71QAPCh. 16 - Prob. 72QAPCh. 16 - Prob. 73QAPCh. 16 - Prob. 74QAPCh. 16 - Prob. 75QAPCh. 16 - Prob. 76QAPCh. 16 - Prob. 77QAPCh. 16 - Prob. 78QAPCh. 16 - Prob. 79QAPCh. 16 - Prob. 80QAPCh. 16 - Prob. 81QAPCh. 16 - Prob. 82QAPCh. 16 - Prob. 83QAPCh. 16 - Prob. 84QAPCh. 16 - Prob. 85QAPCh. 16 - Prob. 86QAPCh. 16 - Prob. 87QAPCh. 16 - Prob. 88QAPCh. 16 - Prob. 89QAPCh. 16 - Prob. 90QAPCh. 16 - Prob. 91QAP
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