Chapter 16 – Problem 13

Given

Q=6.00x10-3Cr=1.00mr=1.00m 2 k=9.0x109 N2m2/C2

Find Force on the top

right charge

Chapter 16 – Problem 13

Given

Q=6.00x10-3Cr=1.00mr=1.00m 2 k=9.0x109 N2m2/C2

Find Force on the top

right charge

Chapter 16 – Problem 13

Given

Q=6.00x10-3Cr=1.00mr=1.00m 2 k=9.0x109 N2m2/C2

Find Force on the top

right charge

Chapter 16 – Problem 13

Given

Q=6.00x10-3Cr=1.00mr=1.00m 2 k=9.0x109 N2m2/C2

Find Force on the top

right charge

Chapter 16 – Problem 13

Given

Q=6.00x10-3Cr=1.00mr=1.00m 2 k=9.0x109 N2m2/C2

Find Force on the top

right charge

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

F = KQQ / r2 = 9.0x109 N2m2/C2(6.0x10-3)2 / (1.0m)2

Fx=3.24x105N for top left on top right Fy=3.24x105N for bottom right on top right

F=KQQ / r2=9.0x109 N2m2/C2(6.0x10-3)2/(1.0m 2 )2

F=1.62x105N Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Fx=3.24x105N for top left on top rightFx=1.62x105N cos45= 1.15x105N for bottom left on top rightTotal x = 4.39x105N

Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top rightTotal y = 4.39x105N

Chapter 16 – Problem 13

Total x = 4.39x105N

Total y = 4.39x105N

total (4.39x105N)2 + (4.39x105N)2

Total = 6.2x105N away from the center

Chapter 16 – Problem 13

Total x = 4.39x105N

Total y = 4.39x105N

total (4.39x105N)2 + (4.39x105N)2

Total = 6.2x105N away from the center

Chapter 16 – Problem 13

Total x = 4.39x105N

Total y = 4.39x105N

total (4.39x105N)2 + (4.39x105N)2

Total = 6.2x105N away from the center

Chapter 16 – Problem 13

Total x = 4.39x105N

Total y = 4.39x105N

total (4.39x105N)2 + (4.39x105N)2

Total = 6.2x105N away from the center

Chapter 16 – Problem 15

GivenQ=1.6x10-19C m=9.11x10-31kg r=.53x10-10m k=9.0x109 N2m2/C2

G=6.67.0x10-11 N2m2/kg2

FindFE

Fg

Chapter 16 Problem 15

F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2

(.53x10-10m)2

F=8.2x10-8N

F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 (.53x10-10m)2

F=3.6x10-47N

Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

Chapter 16 Problem 15

F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2

(.53x10-10m)2

F=8.2x10-8N

F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 (.53x10-10m)2

F=3.6x10-47N

Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

Chapter 16 Problem 15

F = KQQ / r2 = 9.0x109 N2m2/C2(1.6x10-19)2

(.53x10-10m)2

F=8.2x10-8N

F = Gmm / r2 = 6.67x10-11 N2m2/C2(9.11x10-31kg)2 (.53x10-10m)2

F=3.6x10-47N

Ratio=8.2x10-8N/3.6x10-47N=2.3x1039 times greater

Chapter 16 Problem 19

Given

The charge must be placed beyond one of the charges in order for the net force to equal zero

Chapter 16 Problem 19

GivenQ=5.7x10-6Cr=.25m+xQ2=3.5x10-6C

r2=x

k=9.0x109

N2m2/C2

FindX position a positive or negative particle would experience no force

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 19

F=KQQ / r2 K(5.7x10–6 C) Q / (.25m + x)2 =K(3.5x10–6 C) Q / x2

(5.7x10–6 C) / (.25m + x)2 =(3.5x10–6 C) / x2

X=.91m beyond the negative charge

The result is independent of the magnitude of the charge and sign of the particle to placed at .91 m

Chapter 16 Problem 21

Given q=1.6x10-19C m=9.11x10-31kg E=600 N/C

Find a=?m/s2

Direction=?

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 21

F=qE= ma F = (1.6x10-19C) (600N/C)=(9.11x10-31kg) a

a=1.04x1014m/s2

Since the charge on the electron is negative and the direction of force and thus the acceleration is opposite to the direction of the electric fieldThe direction of the acceleration is independent of the velocity

Chapter 16 Problem 27

Given q=1.6x10-19C m=9.11x10-31kg a=125m/s2

FindE=? N/C

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 27

F=qE= ma F = (1.6x10-19C) E=(9.11x10-31kg) (125m/s2) E=7.12x10-10N/C south

Because the charge on the electron is negative, the direction of force,and thus the acceleration, is opposite to the direction of the electric field, so the electric field is south

Chapter 16 Problem 30a

.05m

.10m

9.0x10-6C 9.0x10-6C

Find the Electric field strength at (0,.05m)

Given

Chapter 16 Problem 30a

GivenQ=9.0x10-6Ck=9.0x109 N2m2/C2

r= (.10m)2+(.05m)2

r=.112mtan= .05m/.10m = 26.6o

FindE at (0,.05m)

Problem 30a

E=kQ r2

E = 9.0x109 N2m2/C2 (9.0x10-6 C)(.112m)2

E=6.48x106 N/C

Chapter 16 Problem 30a

X-compThe sum of their x components will equal zero

Y comp The sum of their y

components equal twice the y component due to one of the charges

Chapter 16 Problem 30a

2Fy=2(6.48x106 N/C) sin 26.6

Total Fy =5.8x106N/C

Chapter 16 Problem 30b

.05m

.05m

9.0x10-6C 9.0x10-6C

Find the Electric field strength at (0,.05m)

Given

.15m

Chapter 16 Problem 30b

GivenQ=9.0x10-6Ck=9.0x109 N2m2/C2

r= (.05m)2+(.05m)2

r=.0707m = 45o

GivenQ=9.0x10-6Ck=9.0x109 N2m2/C2

r= (.15m)2+(.05m)2

r=.158m tan = .05m/.15m = 18.40

Problem 30b

E = 9.0x109 N2m2/C2 (9.0x10-6 C)(.0707m)2

E=1.62x107 N/C

Problem 30b

E = 9.0x109 N2m2/C2 (9.0x10-6 C)(.158m)2

E=3.24x106 N/C

Chapter 16 Problem 30b

Ey=(1.62x107 N/C) sin 45

EY=1.15x107N/C

Ey=(3.24x106N/C)sin18.4= 1.02x106N/CEy=1.15x107N/C+1.02x106N/C =

Ey=1.25x107N/C

Chapter 16 Problem 30b

Ex=(1.62x107 N/C) cos 45

Ex=1.15x107N/C

Ex=(3.24x106N/C)cos18.4=3.07x106N/CEx=1.15x107N/C-3.07x106N/C=

Ex=8.43x106N/C

Chapter 16 Problem 30b

Ex=9.97x106N/C

Ey=1.25x107N/C

E= (8.43x106N/C)2+(1.25x107N/C)2

E=1.5x107N/C

Problem 31

Problem 31

K=9.0x109N/CQ=+45.0x10-5CQ=-3.0x10-5C r=.60 m _2=.423m 2Find E field in the center

Problem 31

E =KQ 9.0x109N/C(+4.5x10-5C) r2 (.423m)2

E=2.26x103 N/CE =KQ 9.0x109N/C(+3.1x10-5C) r2 (.423m)2

E=5.96x105 N/CE=1.46x106N/C diagonally away from +

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given Find K.E. = 65.0 keV VB-VA=?kV q = +2e

F/R/S/A KE + PE = O conservation of energy KE + q(VB – VA) = 0

q(VB – VA) = KE

(VB – VA) = KE = -65.0 keV = -32.5kV q +2e

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given FindW = 25.0x10-4 J K.E. = 4.82 x 10 -4J VB-VA=?V

q = -7.5x10-6 C

F/R/S/A W = KE + PE

W = KE + q(VB – VA)

W - KE = q(VB – VA)

W - KE = (VB – VA)

q25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB)

=+269V (-7.50x10-6 C)

Chapter 17 Problem 7,9,13,15,19

Given: Find r = 15.0x10-2m V = ? J q = 4.00x10-6C C k = 9.0x109 N m2 / C2 F/R/S/A V= kq r V = 9.0x109 N m2 / C2 ( 4.00x10-6 C) = 2.40x105 V

15.0x10-2m

Chapter 17 Problem 7,9,13,15,19

Given: Find r = 15.0x10-2m V = ? J q = 4.00x10-6C C k = 9.0x109 N m2 / C2 F/R/S/A V= kq r V = 9.0x109 N m2 / C2 ( 4.00x10-6 C) = 2.40x105 V

15.0x10-2m

Chapter 17 Problem 7,9,13,15,19

Given: Find r = 15.0x10-2m V = ? J q = 4.00x10-6C C k = 9.0x109 N m2 / C2 F/R/S/A V= kq r V = 9.0x109 N m2 / C2 ( 4.00x10-6 C) = 2.40x105 V

15.0x10-2m

Chapter 17 Problem 7,9,13,15,19

Given: Find r = 15.0x10-2m V = ? J q = 4.00x10-6C C k = 9.0x109 N m2 / C2 F/R/S/A V= kq r V = 9.0x109 N m2 / C2 ( 4.00x10-6 C) = 2.40x105 V

15.0x10-2m

Chapter 17 Problem 7,9,13,15,19

Given: Find r = 15.0x10-2m V = ? J q = 4.00x10-6C C k = 9.0x109 N m2 / C2 F/R/S/A V= kq r V = 9.0x109 N m2 / C2 ( 4.00x10-6 C) = 2.40x105 V

15.0x10-2m

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J

Chapter 17 Problem 15 Q1 a b Q2

L d

Va = k[(Q1/r1a) + (Q2/r2a)]Va =9x109Nm2/C[((30x10-6C)/.16m)+ (30x10-6C)/.16m))]=3.38x106V

Vb =9x109Nm2/C[((30x10-6C)/.26m)+ (30x10-6C)/.06m))]=5.54x106V

Wab = KE+PE KE = 0

Wab = PE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)=

Wab = +1.08 J