Four charges, each charge $q$ coulomb are placed at points $(-1,0,0),(1,0,0),(0,-1,0)$ and $(0,1,0)$ in the $x y$-plane. The distance along the axes are measured in metres. The magnitude of the electric field at the point $z=1 \mathrm{~m}$ on the $Z$-axis is

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Aicharya · Accepted Answer

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 $ \begin{aligned} \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} \end{aligned} $ $E_{A P}$ and $E_{B P}$ are perpendicular to each other. Let $E^{\prime}$ be the resultant electric field of $E_{A P}$ and $E_{B P}$, then $ \begin{array}{rlr} E^{\prime} & =\sqrt{E_{A P}^2+E_{B P}^2}=\sqrt{2 E_{A P}^2} \quad\left[\because E_{A P}=E_{B P}\right] \\ & =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^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{2}} \end{array} $ In the similar way, net electric field due to $E_{C P}$ and $E_{D P}$ $ E^{\prime \prime}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\sqrt{2}} $ Again $E^{\prime}$ and $E^{\prime \prime}$ are directed in same direction i.e along $+z$ direction, hence net electric field, $ \begin{aligned} E_{\mathrm{net}} & =E^{\prime}+E^{\prime \prime}=2 E^{\prime} \\ & =2 \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{2}}=\frac{q}{2 \sqrt{2} \pi \varepsilon_0} \mathrm{~N} / \mathrm{C} \end{aligned} $

Four charges, each charge $q$ coulomb are placed at points $(-1,0,0),(1,0,0),(0,-1,0)$ and $(0,1,0)$ in the…

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