Two electrons are moving towards each other,each with velocity of 10^6 m/S .What will be the closest distance of approach between them?
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Answered by
46
at the closest distance of approach the kinetic energy of both the electron will be converted into the electrostatic potential between the two electron at that moment.
=2.(1/2 .mass of an electron . (106)2) = 1/(4.π.ε) (charge of electron)2 /(r)
Solve the above equation for r. here ε is permittivity constant.
Hope may help you
=2.(1/2 .mass of an electron . (106)2) = 1/(4.π.ε) (charge of electron)2 /(r)
Solve the above equation for r. here ε is permittivity constant.
Hope may help you
Answered by
17
kE =1/2×m×vsquare
pE= 1/4×pi×9×10power9
according to the question,(given closest distance of approach) so,
PE=KE
put mass of electron as 9.1 ×10power-31
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