Question

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Answer 1

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)=$k q1q2/r^2$

    $k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,    

        $F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

         $F= 0.6 N$