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				   Find out the value of electrostatic potential energy of 2 electrons separated by 

3.0 angstrom in vacuum.

6 years ago


Answers : (1)


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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