( ) 2. = kq 1 q 2 r 2. Analysis: F E. k q 1. Solution: F E % 8.99!10 9 ( (1.00!10 )4 C )(1.00!10 )5 C ) $ (2.00 m) 2 F E

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1 Section 7.: Coulomb s Law Tutorial 1 Practice, page Given: q C; q C; r.00 m; k N m /C Required: Analysis: Solution: # N m % ( ( )4 C )( )5 C ) C ' (.00 m).5 N Statement: The magnitude of the electric force between the two charges is.5 N.. Given: q 1 q; q q; r m; ; k N m /C Required: Analysis: Use to develop a quadratic equation to solve for. Solution: k q 1 q 3 k q q 3 3 q 1 r q 13 3 q r q ( 1+ ) 1+ + ( ) 0 1 Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-1

2 Solve the quadratic equation: 0 1 x b ± b 4ac a ( ) ± ( ) 4( 1) ( 1 ) ( ) ± ± 1± 1 Only the positive distance is necessary: 1+ m.414 m Statement: The third charge is.414 m to the left of q. 3. Given: q µc C; d 1 0 m; q 3.0 µc C; d 40.0 cm 0.40 m; q µc C; d cm 1.0 m; k N m /C Required: net at q Analysis: Solution: Determine the electric force due to q 1: 1 # N m % ( (.0 10 )6 C )() )6 C ) C ' (0.40 m ) 1 ) N (two extra digits carried) Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-

3 Determine the electric force due to q 3: 3 kq 3 q # N m % ( () )6 C )() )6 C ) C ' (1.0 m ) 0.40 m ) N (two extra digits carried) Determine the net force: net F E1 + F E N [left] N [left] net 0.55 N [left] Statement: The force on the 3.0 µc charge is 0.55 N [left]. Section 7. Questions, page Given: N; 3r 1 Required: Analysis: Determine how the force changes when the distance is tripled, then substitute for the original force,. Solution: (3r 1 ) 9r # kq 1 q r 1 % (0.080 N) '10 (3 N Statement: The new force is N. Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-3

4 . Given: N; 3r 1 ; q 1B 3q 1A Required: Analysis: Determine how the force changes when the distance and the charge are tripled, then substitute for the original force;. Solution: kq 1B q k(3q 1A )q (3r 1 ) 3kq 1A q 9r 1 1 3# kq 1A q r 1 % (0.080 N) 3.7 '10 ( N Statement: The new force is.7 10 N. 3. Given: q C; q C; r 0.10 nm m; k N m /C Required: Analysis: Solution: # N m % ( ( )19 C )( )19 C ) C ' ( )10 m ).310 )8 N Statement: The magnitude of the electric force between the two electrons is N Given: 1.50 r 1 Required: Analysis: Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-4

5 Solution: kq q 1 r 1 # 1.50%.5 kq q 1 # r 1 %.51 Statement: The magnitude of the electric force will increase by a factor of Given: q µc C; q 1.00 µc C; m 1.00 kg; g 9.8 m/s ; ; k N m /C Required: ; r Analysis: Rearrange the equation to solve for r. Then determine. r mg (1.00 kg)(9.8 m/s ) 9.8 N Solution: r # N m % C ( ' ( )6 C )( )6 C ) 9.8 N r m Statement: The distance between the charges is m. 6. (a) Given: m kg; m kg; r 1.0 nm m; G N m /kg Required: Analysis: Gm 1 m Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-5

6 Solution: Gm 1 m N # m ' ) ( kg )( kg ) % kg ( ( m ) N Statement: The magnitude of the gravitational force between the electron and the proton is N. (b) Given: q C; q C; r m; k N m /C Required: Analysis: Solution: # N m % ( ( )19 C )( )19 C ) C ' ( )9 m ).310 )10 N Statement: The magnitude of the electric force between the electron and proton is N. (c) If the distance were increased to 1.0 m, there would be no change because the ratios of the forces are independent of the separation distance. 7. Given: q 1 q; q 3q; r 1 50; 40; 13 3 Required: Analysis: Use to develop a quadratic equation to solve for. First determine r 1 : r 1 50 ( 40) 90 Solution: 13 3 kq 1 q 3 kq q 3 3 q 1 r q 13 3 q 3q ( ) 90 Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-6

7 ( 90 ) Solve the quadratic equation: x b ± b 4ac a ( ) ± ( 90 ) 4( 1) ( 4050 ) ( ) 90 ± ± ± 45 3 Only the positive distance is necessary: x x 7 Statement: The third charge is at x Given: q C; q C; r 1 10 cm 0.10 m; Required: Analysis: Use to develop a quadratic equation to solve for. Solution: kq 1 q 3 kq q 3 3 q 1 r q 13 3 Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-7

8 C ( ) C ( ) r ( ) Solve the quadratic equation: x b ± b 4ac a ( ) ± ( 0. ) 4( 1) ( 0.01 ) ( ) 1 0. ± ± ± Only the positive distance is necessary: 0.1 m m 0.4 m Statement: The third charge is 0.4 m, o4 cm, beyond the smaller charge. 9. (a) Given: q C; L 5 cm 0.5 m; k N m /C Required: net at q 3 Analysis: Determine the distance between particles using the Pythagorean theorem. Then use to determine the magnitude of the electric force between two particles. Solution: Determine the distance: r (0.5 m) + (0.5 m) 0.15 m r m (two extra digits carried) Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-8

9 Determine the magnitude of the electric force: # N m % ( (7.510 )6 C )(7.510 )6 C ) C ' ( m ) N (two extra digits carried) The x-components of the forces will add to zero, so calculate the y-components of the forces. net sin45 (4.044 N)sin 45 net 5.7 N [down] Statement: The net force on the charge at the bottom is 5.7 N [down]. (b) Given: q C; L 0.5 m; k N m /C ; N Required: net at q Analysis: Solution: Determine the magnitude of the force between the two particles on the x-axis: # N m % ( (7.510 )6 C )(7.510 )6 C ) C ' (0.5 m m ).03 N (two extra digits carried).0 N Determine the x- and y-components of the diagonal force: x sin45 y cos45 (4.044 N)sin45 x.860 N Add the horizontal forces:.860 N +.03 N N [to the right] The vertical force is.860 N [up]. Determine the magnitude of the net force: net (4.883 N) + (.860 N) 3.0 N N (two extra digits carried) net 5.7 N (4.044 N)cos45 y.860 N Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-9

10 Determine the direction of the net force: tan.860 N N #.860 N tan 1 % N ' ( 30 Statement: The net force on the charge on the right is 5.7 N [E 30 N]. (c) Given: q C; L 0.5 m; q e C; k N m /C Required: net at origin Analysis: kq q 1 ; the two charges on the x-axis have a net force of zero, so the only force is an attractive force from the charge at the bottom. Solution: # N m % ( (7.510 )6 C )( )19 C ) C ' (0.5 m ) )13 N (two extra digits carried) )13 N Statement: The net force on the electron is N [down]. 10. Given: m 5.00 g kg; L 1.00 m; θ 30.0 ; g 9.8 m/s ; k N m /C Required: q Analysis: The electric, gravitational, and tension forces on the pith balls give a net force of zero. Since the electric force is entirely horizontal and the gravitational force is entirely vertical, first determine the gravitational force, mg. Then use trigonometry to determine the tension force and the electric force. Use the situation. to determine the charge on each pith ball. Draw a sketch of Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-10

11 Solution: Determine the gravitational force on one pith ball: mg ( kg)(9.8 m/s ) N Determine the force of tension on one pith ball: F T cos 30.0 # % F F T g cos N cos15.0 F T N Determine the electric force on one pith ball: F T sin 30.0 # % ( N)sin '10 ( N Determine the distance between pith balls: r Lsin 30.0 # % (1.00 m)sin15.0 r m Determine the charge on the pith balls: kq q k q k ( N )( m ) N # m ' % ( ) C q C Statement: The charge on each pith ball is C. Copyright 01 Nelson Education Ltd. Chapter 7: Electric Fields 7.-11