In the figure particle 1 of charge q1 = 1.05 μC

Question

In the figure particle 1 of charge q1 = 1.05 μC and particle 2 of charge q2 = -2.95 μC, are held at separation L = 9.7 cm on an

satis · Answer

Coulomb's law, force of attraction/repulsion 
 F = kQ₁Q₂/r² ,
  
where Q₁ and Q₂ are the charges in coulombs, r is separation in meters 
  
 Since the particles are of opposite charge, the field between them cannot be zero (the field from both particles is in the same direction there).   
It has to be on the x axis. 
  Call q3 = 1µC 
 let x be location of q3, with q1 at x=0 
 given the polarities, q3 has to be located to the right of q2,  so it's force

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In the figure particle 1 of charge q1 = 1.05 μC

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