Answer:
a=0.47 m/s^2
Explanation:
Assuming
Two 1.3 kg masses are 1.1 m apart (center to center) on a friction less table. Each has + 9.1 μC of charge
Q=Q1=Q2=9.1 uC
From Coloumb's law ,The electric force on one of the masses is given by
[tex]F=KQ^2/r^2 =(9\times10^9)(9.1\times10^{-6})^2/1.12[/tex]
F=0.616 N =0.62 N (approx)
Initial acceleration of the mass
a=F/m =0.62/1.3
a=0.47 m/s^2
