Point charges $q$, $-q$, $2Q$ and $Q$ are placed in order at the corners $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$ of a square of side $2b$. If the field at the midpoint of $\mathrm{CD}$ is zero, then $\frac{q}{Q}$ is

• A

  $1$
• B

  $2$
• C

  $\frac{2\sqrt{2}}{5}$
• Correct answer
  D

  $\frac{5\sqrt{5}}{2}$

Solution

$E=E_C-E_D-E_A\cos \theta -E_B\cos \theta$
$\frac{k\left(2Q\right)}{b^2}-\frac{kQ}{b^2}=2\times \frac{kqb}{{\left(b^2+4b^2\right)}^{\frac{3}{2}}}$
$\Rightarrow Q=\frac{2q}{5\sqrt{5}}$

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