How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33×10−21 n?

The number of excess electrons on each sphere is 759

Explanation:

Given that,

distance , d = 20 cm

                    = 0.20 m

let the number of electrons is n  

Electric force (F) = k × (n × e)² /d²

3.33 × = 9 × × (n × 1.602 × )² /0.2²

solving for n

n = 759

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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f is prop to m1xm2 double m1 double force