Asked by brooke
An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electric force acting on the alpha particle when the alpha particle is 1.7 multiplied by 10-14 m from the gold nucleus?
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Answers
Answered by
bobpursley
coulombs law...
F=kq1*q2/r^2
F=kq1*q2/r^2
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