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⇒ Say, both the charges are joined by a single line, with '10a ' as the distance between them.

⇒ now, say at a distance of 'x ' from charge '+q ' the electric field is zero; so, from '+9q ' the distance is '10a-x '.

⇒ now, we know electric field is (k*(q/r² 

⇒ and, we also know electrix field due to a charge at a point is the field experienced by a unit positive charge(test charge) at that point. say a unit positive charge is placed at point P, at a distance of 'x ' as specified earlier. 

⇒ now, calculate field, on the test charge, due to both charges, 

⇒ (k*(q/x²)=(k*(9q/(10a-x)²) 

⇒ solving, we get x=2.5a
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