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‫ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ) ﺍﻟﻤﻅﻬﺭ ﺍﻟﻜﻬﺭﺒﺎﺌﻲ (‬ ‫‪ - 1‬ﻗﺎﻨﻭﻥ ﻜﻭﻟﻭﻡ‬ ‫>> ﺸﺤﻨﺘﺎﻥ ﻜﻬﺭﺒﺎﺌﻴﺘﺎﻥ ‪ q‬ﻭ'‪ q‬ﺘﻁﺒﻘﺎﻥ ﺘﻁﺒﻕ ﻜل ﺸﺤﻨﺔ ﻋﻠﻰ ﺍﻷﺨﺭﻯ ﻗﻭﺓ ﻜﻬﺭﺒﺎﺌﻴﺔ ‪ ،‬ﻗﻴﻤﺘﻬﺎ ﻤﺘﻨﺎﺴﺒﺔ‬ ‫ﻁﺭﺩﺍ ﻤﻊ ﻗﻴﻤﺔ ﺍﻟﺸﺤﻨﺘﻴﻥ ﻭ ﻋﻜﺴﺎ ﻤﻊ ﻤﺭﺒﻊ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻨﻬﻤﺎ ﺒﺤﻴﺙ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ‪:‬‬ ‫'‪q‬‬

‫‪q‬‬ ‫‪d2‬‬

‫* ‪F = F' = K‬‬

‫‪ F‬ﻭ '‪ : F‬ﺒﻭﺤﺩﺓ ﺍﻟﻨﻴﻭﺘﻥ )‪(N‬‬ ‫‪ : d‬ﺒﻭﺤﺩﺓ ﺍﻟﻤﺘﺭ ‪ q .‬ﻭ'‪ q‬ﺒﺎﻟﻜﻭﻟﻭﻡ )‪. (C‬‬

‫ﻟﻬﺎﺘﻴﻥ ﺍﻟﻘﻭﺘﻴﻥ ﻨﻔﺱ ﺍﻟﺤﺎﻤل ﻭ ﻨﻔﺱ ﺍﻟﺸﺩﺓ ﻭ ﻟﻜﻥ ‪:‬‬ ‫· ﻗﻭﺘﻲ ﺘﺠﺎﺫﺏ ﺇﺫﺍ ﻜﺎﻨﺕ ﻟـ ‪ q‬ﻭ'‪ q‬ﺇﺸﺎﺭﺘﺎﻥ ﻤﺨﺘﻠﻔﺘﺎﻥ ‪.‬‬ ‫· ﻗﻭﺘﻲ ﺘﻨﺎﻓﺭ ﺇﺫﺍ ﻜﺎﻨﺕ ﻟـ ‪ q‬ﻭ'‪ q‬ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ ‪.‬‬ ‫‪F‬‬

‫'‪q‬‬

‫‪q‬‬

‫'‪F‬‬

‫‪F‬‬

‫‪q‬‬

‫‪d‬‬

‫'‪F‬‬

‫'‪q‬‬

‫‪d‬‬

‫‪ : K‬ﺜﺎﺒﺕ ﻓﻴﺯﻴﺎﺌﻲ ﻴﻤﻴﺯ ﺍﻟﻭﺴﻁ ﺍﻟﺫﻱ ﺘﻭﺠﺩ ﻓﻴﻪ ﺍﻟﺸﺤﻨﺘﻴﻥ ﻗﻴﻤﺘﻪ ﻓﻲ ﺍﻟﻨﻅﺎﻡ ﺍﻟﺩﻭﻟﻲ‬ ‫‪K= 9 * 10+9 N.m2/C2‬‬ ‫‪ -2‬ﺍﻟﺘﺠﺎﺫﺏ ﺒﻴﻥ ﻋﺩﺓ ﺸﺤﻥ‬ ‫ﻟﺩﻴﻨﺎ ﺜﻼﺙ ﺸﺤﻥ ﻨﻘﻁﻴﺔ ‪ qA‬ﻭ ‪ qB‬ﻭ ‪ qC‬ﺘﻘﻌﺎﻥ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻨﻘﺎﻁ ‪ A‬ﻭ‪ B‬ﻭ ‪ C‬ﻭ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬ ‫ﺒﺤﻴﺙ ﺃﻥ ‪qA = qB = q = 1 *10-6 C :‬‬ ‫‪qc=-q=- 1*10-6 C‬‬ ‫‪AB = BC = d=25 cm‬‬ ‫‪AC=2 d‬‬ ‫ﺍﻟﺸﺤﻨﺔ‪ qA‬ﺘﺨﻀﻊ ﺇﻟﻰ ﻗﻭﺓ ﺘﻨﺎﻓﺭ‪ F B/A‬ﻤﻥ ﺍﻟﺸﺤﻨﺔ‪ qB‬ﻭ ﺇﻟﻰ ﻗﻭﺓ ﺘﺠﺎﺫﺏ‪ F C/A‬ﻤﻥ ﺍﻟﺸﺤﻨﺔ ‪. qC‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﻗﺎﻨﻭﻥ ﻜﻭﻟﻭﻡ ﻴﻤﻜﻥ ﺃﻥ ﻨﻜﺘﺏ ‪:‬‬ ‫‪qC‬‬

‫‪FC/A‬‬

‫‪qB‬‬

‫‪C‬‬

‫‪q2‬‬ ‫‪4d 2‬‬

‫*‪F C/A = K‬‬

‫‪B‬‬

‫ﻭ‬

‫ﻫﺫﻩ ﻟﻬﺎ ﻨﻔﺱ ﺍﻟﺤﺎﻤل ﻭ ﻟﻜﻥ ﺒﺎﺘﺠﺎﻫﻴﻥ ﻤﺘﻌﺎﻜﺴﻴﻥ‬ ‫‪F A = F B/A – F C/A‬‬ ‫‪q2‬‬ ‫‪4d 2‬‬ ‫‪q2‬‬ ‫‪3‬‬ ‫=‪FA‬‬ ‫‪K* 2‬‬ ‫‪4‬‬ ‫‪d‬‬

‫*‪- K‬‬

‫‪q2‬‬ ‫‪d2‬‬

‫‪q2‬‬ ‫‪d2‬‬

‫‪qA‬‬

‫*‪F B/A = K‬‬

‫*‪F A = K‬‬ ‫ﻭ ﻤﻨﻪ ‪F A =108 m N‬‬

‫‪d‬‬

‫‪A‬‬

‫‪FB/A‬‬

‫‪ -3‬ﻗﻭﻯ ﺍﻟﺘﺠﺎﺫﺏ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﻭ ﺜﺒﺎﺕ ﺍﻷﻨﻭﻴﺔ‬ ‫ﻨﻌﻠﻡ ﺃﻥ ﺜﺒﺎﺕ ﺍﻷﻨﻭﻴﺔ ﻤﺭﻫﻭﻥ ﺒﻘﻭﻯ ﻜﺒﻴﺭﺓ ﻤﻭﺠﻭﺩﺓ ﺒﻴﻥ ﺍﻟﻨﻭﻜﻠﻴﻭﻨﺎﺕ ﻭ ﻤﺤﻴﻁﻬﺎ‪ .‬ﻭ ﻜﺫﻟﻙ ﺘﻭﺠﺩ ﻗﻭﻯ‬ ‫ﻜﻬﺭ ﻭﺴﺎﻜﻨﺔ ﺘﻨﺎﻓﺭﻴﺔ ﺒﻴﻥ ﺒﺭﻭﺘﻭﻨﺎﺕ ﺍﻟﻨﻭﺍﺓ ﺍﻟﻭﺍﺤﺩﺓ ‪ .‬ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺘﺅﺩﻱ ﺒﺎﻷﻨﻭﻴﺔ ﺍﻟﺜﻘﻴﻠﺔ ) ﺍﻟﻐﻨﻴﺔ‬

‫ﺒﺎﻟﺒﺭﻭﺘﻭﻨﺎﺕ ( ﺇﻟﻰ ﻋﺩﻡ ﺍﻟﺜﺒﺎﺕ ‪ .‬ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻨﺒﻌﺎﺙ ﺠﺴﻴﻤﺎﺕ ﺃﻟﻔﺎ ) ﺫﺭﺓ ﺍﻟﻬﻠﻴﻭﻡ ( ﺃﻭ ﺍﻨﻘﺴﺎﻡ ﻨﻭﻭﻱ ‪.‬‬ ‫ﺍﻷﻨﻭﻴﺔ ﺍﻟﺨﻔﻴﻔﺔ ﻴﻜﻭﻥ ﻋﺩﺩ ﺍﻟﺒﺭﻭﺘﻭﻨﺎﺕ ﻤﺴﺎﻭﻱ ﺒﺎﻟﺘﻘﺭﻴﺏ ﻟﻌﺩﺩ ﺍﻟﻨﻴﻭﺘﺭﻭﻨﺎﺕ ‪ ،‬ﺃﻤﺎ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻸﻨﻭﻴﺔ ﺍﻟﺜﻘﻴﻠﺔ‬ ‫ﻓﺈﻥ ﻋﺩﺩ ﺍﻟﺒﺭﻭﺘﻭﻨﺎﺕ ﻴﺯﻴﺩ ﻋﻥ ﺍﻟﻨﻴﻭﺘﺭﻭﻨﺎﺕ ‪.‬‬

‫ﻻ ﺘﻭﺠﺩ ﺃﻨﻭﻴﻪ ﺜﺎﺒﺘﺔ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺃﻜﺜﺭ ﻤﻥ ‪ 83‬ﺒﺭﻭﺘﻭﻥ ‪.‬‬ ‫‪ -4‬ﺍﻟﺫﺭﺓ ﻭﺍﻟﺠﺯﻱﺀ ﻭ ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ‬ ‫ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﺍﻟﻨﺎﺸﺌﺔ ﻓﻲ ﺍﻟﺫﺭﺓ ﺒﻴﻥ ﺍﻟﻨﻭﺍﺓ ﺍﻟﻤﺸﺤﻭﻨﺔ ﺇﻴﺠﺎﺒﺎ ﻡ ﺍﻻﻟﻜﺘﺭﻭﻨﺎﺕ ﻓﻲ ﺤﺭﻜﺘﻬﺎ ﺍﻟﺩﺍﺌﻤﺔ ﺤﻭل‬ ‫ﺍﻟﻨﻭﺍﺓ ﻭ ﺍﻟﻤﺸﺤﻭﻨﺔ ﺴﻠﺒﺎ ﺘﺴﻤﻰ ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ‪ ،‬ﻜﺫﻟﻙ ﻫﺫﻩ ﺍﻷﻓﻌﺎل ﺘﻠﻌﺏ ﺩﻭﺭﺍ ﻓﻲ‬ ‫ﺍﻟﺭﻭﺍﺒﻁ ﺒﻴﻥ ﺍﻟﺫﺭﺍﺕ ﻭ ﺍﻟﺘﻲ ﺘﻔﺴﺭ ﻨﺸﺄﺓ ﺠﺯﻴﺌﺎﺕ ‪.‬‬ ‫ﻤﺜﺎل ‪ :‬ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ﻓﻲ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ‬

‫‪-‬‬

‫‪F'e‬‬

‫‪Fe‬‬

‫إﻟﻜﺘﺮون‬

‫‪r‬‬

‫‪+‬‬ ‫ﺑﺮوﺗﻮن‬

‫* ﻗﺎﻨﻭﻥ ﻜﻭﻟﻭﻡ ﻓﻲ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻴﻌﻁﻲ ‪:‬‬ ‫‪e2‬‬ ‫‪r2‬‬

‫* ‪ Fe = k‬ﻭ ﻤﻨﻪ‬

‫‪Fe =8.2*10-8 N‬‬

‫* ﻗﺎﻨﻭﻥ ﻨﻴﻭﺘﻥ ﻓﻲ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻴﻌﻁﻲ ‪:‬‬ ‫‪me * m p‬‬ ‫‪r2‬‬

‫*‪ Fg = G‬ﻭ ﻤﻨﻪ ‪Fg=3.61 *10-47 N‬‬

‫ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﺠﺎﺫﺒﺔ ﻓﻲ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻤﻬﻤﻠﺔ ﺃﻤﺎﻡ ﺍﻷﻓﻌﺎل ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ‪.‬‬ ‫‪ -5‬ﺜﻨﺎﺌﻲ ﻗﻁﺏ ﻜﻬﺭﺒﺎﺌﻲ‬

‫ﺜﻨﺎﺌﻲ ﺍﻟﻘﻁﺏ ﻤﻜﻭﻥ ﻤﻥ ﺯﻭﺝ ﻤﻥ ﺍﻟﺸﺤﻨﺎﺕ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﺍﻟﻨﻘﻁﻴﺔ ‪ q‬ﻭ'‪ q‬ﻗﺭﻴﺒﺘﻴﻥ ﺠﺩﺍ ﻤﻥ ﺒﻌﻀﻬﻤﺎ ﺍﻟﺒﻌﺽ ‪.‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل ‪:‬‬ ‫ﻗﻴﻤﺔ ﺸﺤﻨﺔ ‪ q=1*10-6 C‬ﺘﻘﻊ ﻋﻠﻰ ﺒﻌﺩ ‪ d=10 cm‬ﻤﻥ ﺸﺤﻨﺔ ﺃﺨﺭﻯ '‪ q‬ﺘﺨﻀﻊ ﺇﻟﻰ ﻗﻭﺓ ﻜﻬﺭ ﻭﺴﺎﻜﻨﺔ‬ ‫ﺸﺩﺘﻬﺎ ‪. F = 1 mN‬‬

‫ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺸﺤﻨﺔ '‪ q‬؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ ‪:‬‬ ‫ﺸﺩﺓ ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﻭ ﺴﺎﻜﻨﺔ ﺒﻴﻥ ﺸﺤﻨﺘﻴﻥ ﻤﺘﻤﺎﺜﻠﺘﻴﻥ ﻗﻴﻤﺔ ﻜل ﻤﻨﻬﺎ ‪ q=q'=100 nC‬ﻫﻲ ‪. F = 10 mN‬‬ ‫ﻤﺎ ﻫﻲ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﺍﻟﺸﺤﻨﺘﻴﻥ ؟‬

‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ ‪:‬‬

‫ﻓﻲ ﺠﺯﻱﺀ ﺜﻨﺎﺌﻲ ﺍﻟﻜﻠﻭﺭ ﺍﻟﺒﻌﺩ ﺒﻴﻥ ﺫﺭﺘﻴﻪ ﻴﺴﺎﻭﻱ ﺒﺎﻟﺘﻘﺭﻴﺏ ‪. 198 pm‬‬ ‫ﺃﺤﺴﺏ ﺍﻟﻘﻭﺓ ﺍﻟﻜﻬﺭﻭﺴﺎﻜﻨﺔ ﺒﻴﻥ ﺍﻟﺫﺭﺘﻴﻥ ‪.‬‬ ‫ﻨﻌﻁﻲ ‪ :‬ﺍﻟﻌﺩﺩ ﺍﻟﺫﺭﻱ ﻟﺫﺭﺓ ﺍﻟﻜﻠﻭﺭ ‪. Z= 17‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺭﺍﺒﻊ ‪:‬‬

‫ﺠﺴﻤﺎﻥ ﻨﻘﻁﻴﺎﻥ ﻜﺘﻠﺘﻴﻬﻤﺎ ‪ m = 1 mg‬ﺘﺤﻤﻼﻥ ﺸﺤﻨﺘﺎﻥ ﻨﻘﻁﻴﺘﺎﻥ ﻤﺘﻤﺎﺜﻠﺘﺎﻥ )‪ . (q‬ﻤﺤﺼﻠﺔ ﺍﻟﻘﻭﻯ ﺍﻟﺠﺎﺫﺒﺔ ﻭ‬

‫ﺍﻟﻜﻬﺭﻭﺴﺎﻜﻨﺔ ﺍﻟﻤﺅﺜﺭﺓ ﻋﻠﻰ ﺍﻟﺠﺴﻤﻴﻥ ﻤﻌﺩﻭﻤﺔ ‪.‬‬ ‫‪ -1‬ﻤﺜل ﻋﻠﻰ ﺭﺴﻡ ﻜل ﻤﻥ ﺍﻟﺠﺴﻤﺎﻥ ﻭ ﻤﺨﺘﻠﻑ ﺍﻟﻘﻭﻯ ﺍﻟﻤﺅﺜﺭﺓ ‪.‬‬ ‫‪ -2‬ﺍﺴﺘﻨﺘﺞ ﻗﻴﻤﺔ ﺍﻟﺸﺤﻨﺔ ﺍﻟﺘﻲ ﻴﺤﻤﻠﻬﺎ ﻜل ﺠﺴﻡ ‪.‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺨﺎﻤﺱ ‪:‬‬

‫ﺸﺤﻨﺘﺎﻥ ﻤﺘﻤﺎﺜﻠﺘﺎﻥ)‪ (q‬ﻭ ﺴﺎﻟﺒﺘﺎﻥ ﻤﻭﻀﻭﻋﺘﺎﻥ ﻓﻲ ﺍﻟﻨﻘﺎﻁ ‪ A‬ﻭ‪ B‬ﺍﻟﺒﻌﺩ ﺒﻴﻨﻬﻤﺎ ‪. AB = 2 a‬‬ ‫‪ -1‬ﺃﺤﺴﺏ ﻤﺤﺼﻠﺔ ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﻭﺴﺎﻜﻨﺔ ﺍﻟﻤﺅﺜﺭﺓ ﻋﻠﻰ ﺸﺤﻨﺔ ﺜﺎﻟﺜﺔ )‪ (q‬ﻤﻭﻀﻭﻋﺔ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪M‬‬ ‫ﻤﻨﺘﺼﻑ ﺍﻟﻘﻁﻌﺔ ]‪. [AB‬‬

‫‪ -2‬ﺃﻋﻁﻲ ﻋﺒﺎﺭﺓ ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﻭﺴﺎﻜﻨﺔ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺸﺤﻨﺔ ﺍﻟﺜﺎﻟﺜﺔ ﻭﺍﻗﻌﺔ ﻓﻲ ﻨﻘﻁﺔ ‪ P‬ﺒﺤﻴﺙ ‪. AP = 4 a :‬‬ ‫ﻨﻌﻁﻲ ‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺴﺎﺩﺱ ‪:‬‬ ‫ﺜﻨﺎﺌﻲ ﻗﻁﺏ ﻤﻜﻭﻥ ﻤﻥ ﺸﺤﻨﺘﻴﻥ ﻨﻘﻁﻴﺘﻴﻥ‪ q +‬ﻭ‪ q-‬ﺘﻘﻌﺎﻥ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ‪ A‬ﻭ ‪ B‬ﺒﺤﻴﺙ ﺃﻥ‬

‫‪ . AB = 2 a‬ﺸﺎﺭﺩﺓ ﺴﺎﻟﺒﺔ ﺘﻘﻊ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪ P‬ﻤﻨﺼﻑ ﺍﻟﻘﻁﻌﺔ ]‪ [AB‬ﻭ ﻋﻠﻰ ﺒﻌﺩ ‪ d‬ﻤﻥ ﻤﻨﺘﺼﻑ‬ ‫ﺍﻟﻘﻁﻌﺔ ]‪. [AB‬‬ ‫‪ -1‬ﻤﺎ ﻫﻲ ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﻭﺴﺎﻜﻨﺔ ﺍﻟﻤﺅﺜﺭﺓ ﻋﻠﻰ ﺜﻨﺎﺌﻲ ﺍﻟﻘﻁﺏ ؟‬ ‫‪ -2‬ﻤﺜل ﻫﺫﻩ ﺍﻟﻘﻭﻯ ﻋﻠﻰ ﺭﺴﻡ ‪.‬‬

‫‪ -3‬ﺃﻜﺘﺏ ﻋﺒﺎﺭﺓ ﻫﺫﻩ ﺍﻟﻘﻭﻯ ‪ .‬ﻤﺎ ﻫﻭ ﻓﻌل ﻫﺫﻩ ﺍﻟﻘﻭﻯ ﻋﻠﻰ ﺜﻨﺎﺌﻲ ﺍﻟﻘﻁﺏ ؟‬

‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺴﺎﺒﻊ ‪:‬‬ ‫‪ -1‬ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻘﻭﺓ ﺍﻟﺠﺎﺫﺒﺔ ﻭ ﺍﻟﻘﻭﺓ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﺒﻴﻥ ﻨﻭﺍﺓ ﺫﺭﺓ ﺍﻷﻭﻜﺴﺠﻴﻥ ﻭ ﺇﻟﻜﺘﺭﻭﻥ ﻋﻠﻤﺎ ﺃﻥ ﺍﻟﻤﺴﺎﻓﺔ‬ ‫ﺒﻴﻨﻬﻤﺎ ‪ . 5.8*10-11 m‬ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ؟‬ ‫ﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ‪. 9*10-31 kg :‬‬

‫ﻜﺘﻠﺔ ﺍﻟﻨﻭﻜﻠﻴﻭﻥ ‪. 1.6*10-27kg :‬‬ ‫ﺍﻟﺸﺤﻨﺔ ﺍﻟﻌﻨﺼﺭﻴﺔ ‪. 1.6*10-19 C :‬‬ ‫ﺫﺭﺓ ﺍﻷﻭﻜﺴﺠﻴﻥ ‪( A = 16 ; Z = 8 ) :‬‬

‫‪ -2‬ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﺍﻟﻨﻭﺍﺓ ﻭ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺤﺘﻰ ﺘﺼﺒﺢ ﻗﻴﻤﺔ ﺍﻟﻘﻭﺓ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﺘﺴﺎﻭﻱ ‪ 1/100‬ﻤﻥ‬

‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺤﺴﻭﺒﺔ ﺴﺎﺒﻘﺎ ؟‬ ‫‪ -3‬ﻨﻔﺱ ﺍﻟﺴﺅﺍل ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻘﻭﺓ ﺍﻟﺠﺎﺫﺒﺔ ‪.‬‬ ‫‪ -4‬ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺒﻌﺩ ﺒﻴﻥ ﺍﻟﻨﻭﺍﺓ ﻭ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺤﺘﻰ ﺘﺘﺴﺎﻭﻱ ﻗﻴﻤﺘﻲ ﺍﻟﻘﻭﺓ ﺍﻟﺠﺎﺫﺒﺔ ﻭ ﺍﻟﻘﻭﺓ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ؟‬

‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻤﻥ ‪:‬‬

‫ﺸﺎﺭﺩﺘﻴﻥ ﻟﻠﻜﻠﻭﺭ ﺸﺤﻨﺘﻴﻬﻤﺎ ) ‪ (-e‬ﺘﻘﻌﺎﻥ ﻋﻠﻰ ﻤﺤﻭﺭ ‪ X'OX‬ﻓﻲ ﺍﻟﻨﻘﺎﻁ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺘﻴﻥ )‪(-a) , (+a‬‬ ‫ﺤﻴﺙ ‪ . a = 0.282 nm‬ﻨﻀﻊ ﺸﺎﺭﺩﺓ ﺍﻟﺼﻭﺩﻴﻭﻡ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪ ، O‬ﺸﺤﻨﺘﻬﺎ ) ‪. (e+‬‬ ‫‪ -1‬ﻤﺎ ﻫﻲ ﻤﺤﺼﻠﺔ ﺍﻟﻘﻭﻯ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ ﺍﻟﺘﻲ ﺘﺅﺜﺭ ﺒﻬﺎ ﺸﺎﺭﺩﺘﻲ ﺍﻟﻜﻠﻭﺭ ﻋﻠﻰ ﺸﺎﺭﺩﺓ ﺍﻟﺼﻭﺩﻴﻭﻡ ؟‬ ‫‪ -2‬ﺃﺤﺴﺏ ﻤﺤﺼﻠﺔ ﻫﺫﻩ ﺍﻟﻘﻭﻯ ﻓﻴﻤﺎ ﺇﺫﺍ ﻭﻀﻌﻨﺎ ﺸﺎﺭﺩﺓ ﺍﻟﺼﻭﺩﻴﻭﻡ ﻓﻲ ﻨﻘﻁﺔ ‪M‬‬ ‫ﻓﺎﺼﻠﺘﻬﺎ ‪.X=± 0.0282nm‬‬

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