Answer:
[tex]x=s*\frac{q_{1} +-\sqrt{q_{1}q_{2}}}{q_{1}-q_{2}}[/tex]
we choose the sign, in order to x<s
Explanation:
x=distance from q1
[tex]E_{1} =E_{2}[/tex]
[tex]k*q_{1}/x^{2}=k*q_{2}/(s-x)^2}[/tex]
[tex]q_{1}*(s^{2}-2sx+x^{2})-q_{2}x^{2}=0[/tex]
[tex](q_{1}-q_{2})x^{2}-2sq_{1}x+q_{1}*s^{2}=0[/tex]
this is a quadratic equation:
[tex]x=\frac{-b+-\sqrt{b^{2}-4ac} }{2a}=\frac{2sq_{1} +-\sqrt{(-2sq_{1})^{2}-4(q_{1}-q_{2})q_{1}s^{2}}} {2(q_{1}-q_{2})}[/tex]
[tex]x=s*\frac{q_{1} +-\sqrt{q_{1}q_{2}}}{q_{1}-q_{2}}[/tex]
