Question

Two charges 2.0 × 10⁻⁶ C and 1.0 × 10⁻⁶ C are placed at a separation of 10 cm. Where should a third charge be placed, such that it experiences no net force due to these charges?

Answer 1

Given:
$$\begin{gathered} q_1=2.0\times {10}^{-6}\mathrm{C} \\ {q}_2=1.0\times {10}^{-6}\mathrm{C} \end{gathered}$$
Let the third charge, q, be placed at a distance of x cm from charge q_1, as shown in the figure.



By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{Q_1{Q}_2}{r^2}$
Force on charge q due to q₁,
$F=\frac{9\times {10}^9\times 2.0\times {10}^{-6}\times q}{x^2}$
Force on charge q due to q₂,
$F\text{'}=\frac{9\times {10}^9\times {10}^{-6}\times q}{{\left(10-x\right)}^2}$
According to the question,
$F-F\text{'}=0$
$$\begin{gathered} \Rightarrow F=F\text{'} \\ \Rightarrow \frac{9\times {10}^9\times 2\times {10}^{-6}\times q}{x^2}=\frac{9\times {10}^9\times {10}^{-6}\times q}{{\left(10-x\right)}^2} \\ \Rightarrow x^2=2{\left(10-x\right)}^2 \\ \Rightarrow x^2-40x+200=0 \\ \Rightarrow x=20\pm 10\sqrt{2} \\ \Rightarrow x=5.9\mathrm{cm}(\because x\ne 20+10\sqrt{2}) \end{gathered}$$
So, the third charge should be placed at a distance of 5.9 cm from q₁.