At what distance of separation must two 1.00-microCoulomb charges be positioned in order for the repulsive force between them to be equivalent to the weight (on Earth) of a 1.00-kg mass?

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Answer 1 · 2015-11-03T14:55:32

#30,3mm#

From Coulomb's Law of Electrostatic forces between charged objects we get that

#F=k(Q_1Q_2)/r^2#

#therefore r = sqrt((kQ_1Q_2)/F)#

#=sqrt((9xx10^9)(1xx10^(-6))^2)/(9,8)#

#=30,3mm#

(Note that the weight force is given by #W=mg=1xx9,8=9,8N#)