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Electric Charges And Fields

Question
CBSEENPH12037231

The electrostatic force of repulsion between two positively charged ions carrying equal charge is 3.7 x 10–9 N, when they are separated by a distance of 5A . How many electrons are missing from each ion?

Solution

Here given,
           Electrostatic force of repulsion -F = 3.7 × 10-9N

 Let us say charge is q1 = q2 = q distance between two charges-r = 5 Å = 5 × 10-10 m,To find the number of electrons missing - n = ?

Using Coulomb's law,
                   
                  F = 14π 0q1 q2r2
         
      3.7 × 10-9 = 9 × 109 qq(5 × 10-10)2          q2 = 3.7 × 10-9 × 25 × 10-209 × 109                    = 10.28 × 10-38            q = 3.2 × 10-19 coulomb

As           q = ne
            n=qe=3.2 × 10-191.6 × 10-19 = 2