Three charges $–\mathrm{ }{\mathrm{q}}_1$, ${\mathrm{q}}_2$ and $–\mathrm{ }{\mathrm{q}}_3$ are placed as shown in the figure. The $\mathrm{x}$-component of the force on $–{\mathrm{q}}_1$ is proportional to

• A

  $\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}–\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }$
• B

  $\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}–\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$
• Correct answer
  C

  $\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }$
• D

  $\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$

Solution

$\mathrm{x}$ component of force on $–\mathrm{q}\Rightarrow$




${\mathrm{F}}_1\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}(90\mathrm{ }–\mathrm{\theta })\mathrm{ }+\mathrm{ }{\mathrm{F}}_2$

$=\mathrm{ }\frac{\mathrm{K}{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{a}}^2}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }+\frac{\mathrm{K}{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{b}}^2}$

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