حلول التدريبات العملية 2 في الكهروستاتيكا

103. 26.6

بما أن القوة الكهربائية فقط هي التي تبذل شغل الجسم، فإنه سوف يتحقق حفظ الطاقة الميكانيكية، نكتب معادلة حفظ الطاقة: 

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨20px¨»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»E«/mi»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»A«/mi»«/msub»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»=«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»E«/mi»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»B«/mi»«/msub»«mspace linebreak=¨newline¨/»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»E«/mi»«msub mathcolor=¨#0000FF¨»«mi mathvariant=¨bold¨»K«/mi»«mi mathvariant=¨bold¨»A«/mi»«/msub»«/msub»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»+«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»U«/mi»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»A«/mi»«/msub»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»=«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»E«/mi»«msub mathcolor=¨#0000FF¨»«mi mathvariant=¨bold¨»K«/mi»«mi mathvariant=¨bold¨»B«/mi»«/msub»«/msub»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»+«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»U«/mi»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»B«/mi»«/msub»«/mstyle»«/math»

يبدأ الجسم بالحركة من حالة السكون، وبالتالي فإن الطاقة الحركية للجسم في النقطة A تساوي صفرًا.

نعبر عن سرعة الجسم في النقطة B من معادلة حفظ الطاقة:

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نعوّض معطيات السؤال ونحسب سرعة الجسم في النقطة B :

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