Question

.............. is the minimum permissible magnitude of a charge​

Answer 1

Explanation:

Answer : (a) 2.3×10−28N

Explanation :

F=14πε0q1q2r2

F will be minimum when q1andq2 are minimum i.e.q1=q2=1.6×10−19

Also, given r=1m.

Fmin=9×109N×(1.6×10−10)2

Fmin=2.3×10−28N.