KE=PE
mv²/2=k•q1•q2/r
r=2•k•q1•q2/ mv² =
=2•9•10⁹•2.73•10⁻⁶•9.56•10⁻⁶/9.5•10⁻³•97.2²=
=5.23•10⁻³ m
A charge of -2.73 μC is fixed in place. From a horizontal distance of 0.0465 m, a particle of mass 9.50 x 10-3 kg and charge -9.56 μC is fired with an initial speed of 97.2 m/s directly toward the fixed charge. What is the distance of closest approach?
1 answer