Question

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

• A. $20\mu C,6\mu C$
• B. $16\mu C,10\mu C$
• C. $18\mu C,8\mu C$
• D. $13\mu C,13\mu C$

Answer 1

The correct option is A $20\mu C,6\mu C$

Let charges be

$\begin{array}{c}\left(A\right)q_1+Q_2=26\\ q_1=26-q_2\\ F=\frac{K{Q}_1{q}_2}{r^2},K=9\times {10}^9\\ F=0.36\\ 4\times 0.36=9\times {10}^9\times q_1{q}_2\\ 0.16={10}^9\times q_1{q}_2\\ 0.16\times {10}^{-9}=q_1{q}_2\Rightarrow 0.16\times {10}^{-9}=\left(26-q_2\right)q_2\\ \Rightarrow 0.16\times {10}^{-9}=26q_2-q_2^2\Rightarrow q_2^2-26q_2+0.16\times {10}^{-9}\end{array}$

Solving quadratic

$\begin{array}{c}{q}_2\approx 6\\ q_1=26-6=20\end{array}$