Two identical small charged plastic balls are dropped into a vertical tube, one after the other. The repulsive electrostatic force acting on the top$-$most ball is just enough to counterbalance the gravitational force acting on it at a distance $12\mathrm{.}65mm$ above the lower ball $(d$ is the center to center distance between the balls$),$ resulting in an equilibrium state. If we were to double the charge on each ball, what would be the new separation at which the upper ball would be in equilibrium? Give your answer in $mm.$

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The electrostatic repulsion between two charged objects is given by Coulombs Law F k q1q2 d2 where F ... View the full answer