Two electrons are moving towards each other,each with a velocity of 106m/s.What will be losest distance of approach between them?

the closest distance of approach is when its total kinectic energy is concerted into potential energy.
K.E = P.E2×12mv2 = k×e×er       here m = mass of electron , k = 9×109N-m2/C2 = r =k×e2mv2= 9×109×(1.6×10-19)29.1×10-31×(106)2                           = 9×2.56×109×10-389.1×10-31×1012 =2.53×10-2910-19                            =2.53×10-29+19                              =2.53×10-10 meter.

  • 163

It should be nearly 5.12 x10-10m.

  • -37

plz explain

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