Answer:r=12.02 cm
Explanation:
Given
[tex]q_1=31 mC=31\times 10^{-3}C[/tex]
Placed at x=8 cm
[tex]q_2=-70 mc=-70\times 10^{-3} C[/tex]
placed at a distance, suppose r
Electric field due to positive charge will be away from the origin and electric field due to negative charge is towards the origin
Thus net effect will be zero
[tex]E=\frac{kq}{r^2}[/tex]
[tex]E_1=\frac{9\times 10^{9}\times 31\times 10^{-3}}{0.08^2}[/tex]
[tex]E_2=\frac{9\times 10^{9}\times 70\times 10^{-3}}{r^2}[/tex]
Equate [tex]E_1=E_2[/tex]
[tex]\frac{31}{8^2}=\frac{70}{r^2}[/tex]
[tex]r=8\times \sqrt{\frac{70}{31}}[/tex]
r=12.02 cm
