Physics q about electric fields?

Question

Two charges, -31 µC and +13 µC, are fixed in place and separated by 1.9 m.

(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does not necessarily lie between the two charges.)

(b) What would be the force on a charge of +40 µC placed at this spot?

Answer 1

♣ Coulomb says strength of electric field is vector

E=q*r/(4pi*eps*|r|^3), where eps=8.854e-12, |r| is distance and r is vector to test point;  

(♠♠) here q1=13µC, q2=-31µC;

x is relative to positive charge and directed from q1 to q2.

So eqn is 0= E1+E2 =

= q1*x/(4pi*eps*|x|^3) + q2*(x-1.9)/(4pi*eps*|x-1.9|^3);

♦deleting 4pi*eps and simplifying we get

0 =13*/x^2 –31*/(x-1.9)^2; or;

13x^2 –2*13*1.9x +13*1.9^2 –31x^2 =0;

18x^2 +2*13*1.9x -13*1.9^2 =0, hence

x=(-13*1.9 ±√((13*1.9)^2 +18*13*1.9^2))/18;

x1=-3.50m, x2=0.75m;

♦ inserting x1 into (♠♠) eqn we get

13*x1/|x1|^3 –31*(x1-1.9)/|x1-1.9|^3 = 0; OK!  

♦ inserting x2 into (♠♠) eqn we get

13*x2/|x2|^3 –31*(x2-1.9)/|x2-1.9|^3 = 46.62 ≠ 0;

thus the only answer is x=-3.50m;