As shown in the figure, three charges are placed at three corners of a square of side $\sqrt{5}$ cm. If $q_0=\sqrt{2} \times 10^{-6} C$, the magnitude of electric field at point D is.

$54 \times 10^6 NC^{-1}$

The correct answer is Option (1) → $54 \times 10^6 NC^{-1}$

Electric field at D due to B, $|\vec{E_{DB}}|=\frac{1}{4πε_0}\frac{q_0}{(9\sqrt{2})^2}$

$=\frac{1}{4πε_0}\frac{2×10^{-6}}{5×10^{-4}}$

$|\vec{E_{DA}}|=\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}$

$|\vec{E_{DC}}|=\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}$

$E_D=\vec{E_{DA}}+\vec{E_{DB}}+\vec{E_{DC}}$

$=\frac{1}{4πε_0}\left(\frac{2×10^{-6}}{5×10^{-4}}\cos 45\hat j+\frac{2×10^{-6}}{5×10^{-4}}\sin 45\hat i\right)$

$+\left(\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}\hat j\right)$

$+\left(\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}\hat i\right)$

$≃54 \times 10^6 NC^{-1}$