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.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $54 \times 10^6 NC^{-1}$
Electric field at D due to B, $|\vec{E_{DB}}|=\frac{1}{4πε_0}\frac{q_0}{(9\sqrt{2})^2}$
$=\frac{1}{4πε_0}\frac{2×10^{-6}}{5×10^{-4}}$
$|\vec{E_{DA}}|=\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}$
$|\vec{E_{DC}}|=\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}$
$E_D=\vec{E_{DA}}+\vec{E_{DB}}+\vec{E_{DC}}$
$=\frac{1}{4πε_0}\left(\frac{2×10^{-6}}{5×10^{-4}}\cos 45\hat j+\frac{2×10^{-6}}{5×10^{-4}}\sin 45\hat i\right)$
$+\left(\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}\hat j\right)$
$+\left(\frac{1}{4πε_0}\frac{\sqrt{2}×10^{-6}}{5×10^{-4}}\hat i\right)$
$≃54 \times 10^6 NC^{-1}$