Three equal charges are placed on the three corners of a square. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$, then the ratio of magnitudes $\left(\frac{F_{13}}{F_{12}}\right)$ is

• Correct answer
  A

  $\frac{1}{2}$
• B

  $\frac{2}{3}$
• C

  $\frac{1}{\sqrt{2}}$
• D

  $\frac{1}{3}$

Solution

The given situation is shown in the figure.



$\begin{array}{ll} & \left|q_1\right|=\left|q_2\right|=\left|q_3\right|=q \\ \therefore \quad & F_{12}=\frac{k q_1 q_2}{A B^2}=\frac{k q \cdot q}{a^2} \\ & F_{12}=\frac{k q^2}{a^2} \\ & F_{13}=\frac{k \cdot q_1 q_3}{A C^2}=\frac{k \cdot q \cdot q}{(\sqrt{2} a)^2}\end{array}$
$\begin{aligned} & {\left[\because A C=\sqrt{A B^2+B C^2}=\sqrt{a^2+a^2}=\sqrt{2} a\right]} \\ & \quad=\frac{k q^2}{2 a^2}\end{aligned}$
$\therefore \quad \frac{F_{13}}{F_{12}}=\frac{\frac{k q^2}{2 a^2}}{\frac{k q^2}{a^2}}=\frac{1}{2}$

Hence, the correct answer is option: 'A'.