Transcript Chapter 13
Chapter 16 – Problem 13 Given Q=6.00x10-3C r=1.00m r=1.00m 2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13 Given Q=6.00x10-3C r=1.00m r=1.00m 2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13 Given Q=6.00x10-3C r=1.00m r=1.00m 2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13 Given Q=6.00x10-3C r=1.00m r=1.00m 2 k=9.0x109 N2m2/C2 Find Force on the top right charge Chapter 16 – Problem 13 Given Q=6.00x10-3C r=1.00m r=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 right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total y = 4.39x105N Chapter 16 – Problem 13 Fx=3.24x105N for top left on top right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total y = 4.39x105N Chapter 16 – Problem 13 Fx=3.24x105N for top left on top right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total y = 4.39x105N Chapter 16 – Problem 13 Fx=3.24x105N for top left on top right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total y = 4.39x105N Chapter 16 – Problem 13 Fx=3.24x105N for top left on top right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total y = 4.39x105N Chapter 16 – Problem 13 Fx=3.24x105N for top left on top right Fx=1.62x105N cos45= 1.15x105N for bottom left on top right Total x = 4.39x105N Fy=3.24x105N for bottom right on top right Fy=1.62x105N sin45= 1.15x105N for bottom left on top right Total 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 Given Q=1.6x10-19C m=9.11x10-31kg r=.53x10-10m k=9.0x109 N2m2/C2 G=6.67.0x10-11 N2m2/kg2 Find FE 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 Given Q=5.7x10-6C r=.25m+x Q2=3.5x10-6C r2=x k=9.0x109 N2m2/C2 Find X 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 field The 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 field The 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 field The 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 field The 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 field The 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 field The 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 Find E=? 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 Given 9.0x10-6C .05m 9.0x10-6C .10m Find the Electric field strength at (0,.05m) Chapter 16 Problem 30a Given Q=9.0x10-6C k=9.0x109 N2m2/C2 r= (.10m)2+(.05m)2 r=.112m tanf = .05m/.10m f = 26.6o Find E 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-comp The 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 Given .05m 9.0x10-6C .05m 9.0x10-6C .15m Find the Electric field strength at (0,.05m) Chapter 16 Problem 30b Given Q=9.0x10-6C k=9.0x109 N2m2/C2 r= (.05m)2+(.05m)2 r=.0707m f = 45o Given Q=9.0x10-6C k=9.0x109 N2m2/C2 r= (.15m)2+(.05m)2 r=.158m tan f = .05m/.15m f = 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/C Ey=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/C Ex=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/C Q=+45.0x10-5C Q=-3.0x10-5C r=.60 m _2=.423m 2 Find E field in the center Problem 31 E =KQ 9.0x109N/C(+4.5x10-5C) r2 (.423m)2 E=2.26x103 N/C E =KQ 9.0x109N/C(+3.1x10-5C) r2 (.423m)2 E=5.96x105 N/C E=1.46x106N/C diagonally away from + Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given D K.E. = 65.0 keV q = +2e Find VB-VA=?kV F/R/S/A D KE + DPE = O conservation of energy D KE + q(VB – VA) = 0 q(VB – VA) = D KE (VB – VA) = D KE = -65.0 keV = -32.5kV q +2e Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 4.82x10-4J = (VB-VA) = -269 V therefore (VA-VB) =+269V (-7.50x10-6 C) Chapter 17 Problem 7,9,13,15,19 Given W = 25.0x10-4 J D K.E. = 4.82 x 10 -4J q = -7.5x10-6 C Find VB-VA=?V F/R/S/A W = D KE + DPE W = D KE + q(VB – VA) W - D KE = q(VB – VA) W - D KE = (VB – VA) q 25.0x10-4J – 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 L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J Chapter 17 Problem 15 Q1 a L b Q2 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 = DKE+DPE DKE = 0 Wab = DPE = q(Vb-Va) = .50x10-6C(5.54x106V-3.38x106V)= Wab = +1.08 J