Two positive charges separated by a distance $2 m$ each other with a force of $0.36 N.$ If the combined charge is $26\mu \mathrm{C}$ the charges are

Given data: Two positive charges with a 2m separation and a 0.36N force are in opposition to one another.   

Concept used: conventional current 

Detailed solution:

 $F=0.36 \mathrm{N}$ 

 $q_1+q_2=26\mu \mathrm{C}$

 $F=\frac{k{q}_1{q}_2}{r^2}$ 

 $0.36=\frac{9\times {10}^9\times q_1\left(26-q_1\times {10}^{-12}\right)}{2^2}$ 

$160=26q_1-q_1^2$

 $q_1=\frac{26\pm \sqrt{(-26{)}^2-4\times 160}}{2}$

 $=\frac{26\pm 6}{2}=16\mu \mathrm{C}\text{ and }10\mu \mathrm{C}$