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shambhavi mishra Grade: 11
        Find out the value of electrostatic potential energy of 2 electrons separated by 

3.0 angstrom in vacuum.
6 years ago

Answers : (1)

ABHISHEK JAIN AskiitiansExpert-IITD
20 Points
										

Dear Shambhavi,


 


since P.E. = Kq1q2\r


charge on electron = 1.6 × 10^ -19


distance b/w two electrons = 3Å = 3 × 10^ -10 meter


coulamb's constant k = 9 × 10^9


put the values in above formula:


P.E. = 9 × 10^9 × 1.6 × 10^-19 × 1.6× 10^-19 ÷ 3 × 10^-10


      = 7.676 × 10^-19 J




 


 

6 years ago
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