AP EAPCET 06-Jul-2022 Shift 2 Paper

The given situation is shown below.According to given situation we have to find electric field at point $P(0,0,1)$.Distance of point $P$ from each charges is equal to $\sqrt{1^2+1^2}=\sqrt{2}$ $E_{A P}\;=E_{B P}=E_{C P}=E_{D P}=\;\frac{1}{4\pi\varepsilon_0} \cdot \;\frac{q}{(\sqrt{2})^2}$$\;=\;\frac{1}{4\pi\varepsilon_0} \cdot \;\frac{q}{2}$ $E_{A P}$ and $E_{B P}$ are perpendicular to each other. Let $E'$ be the resultant electric field of $E_{A P}$ and $E_{B P}$, then$E'\;=\sqrt{E_{A P}^2+E_{B P}^2}=\sqrt{2 E_{A P}^2} . \; \; [\because E_{A P}=E_{B P}]$$\;=E_{A P} \sqrt{2}$$\;=\;\frac{1}{4\pi\varepsilon_0} \;\frac{q\sqrt{2}}{2}=\;\frac{1}{4\pi\varepsilon_0} \;\frac{q}{\sqrt{2}}$$E'\;=\;\frac{1}{4\pi\varepsilon_0} \;\frac{q}{\sqrt{2}}$ In the similar way, net electric field due to $E_{C P}$ and $E_{p p}$$E''=\;\frac{1}{4\pi\varepsilon_0} \cdot \;\frac{q}{\sqrt{2}}$ Again $E'$ and $E''$ are directed in same direction i.e along $+z$ direction, hence net electric field, $E_{\;\text{net }\;}=E'+E''=2 E'$$[\because E'=E'']$ $=2\;\frac{1}{4\pi\varepsilon_0}\;\frac{q}{\sqrt{2}}=\;\frac{q}{2\sqrt{2} \pi \varepsilon_0} \mathrm{N} / \mathrm{C}$