What will be the magnitude of the electric fi | vedclass

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Question

(B) The electric field at point $O$ is the vector sum of the electric fields produced by each charge at the corners of the segments. Let $k = \frac{1}

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{2l^{2}}(2 \sqrt{2}-1)$

Vedclass · Answer

(B) The electric field at point $O$ is the vector sum of the electric fields produced by each charge at the corners of the segments. Let $k = \frac{1}{4 \pi \varepsilon_{0}}$. The charges are located at a distance $l$ from $O$. For the horizontal and vertical segments,the net electric field components can be calculated by considering the vector sum of fields from charges at distance $l$. Based on the symmetry and the given configuration,the net electric field $E$ at point $O$ is calculated as: $E = \frac{k q}{2l^{2}}(2 \sqrt{2}-1)$.

What will be the magnitude of the electric field at point $O$ as shown in the figure? Each side of the figure has length $l$ and segments are perpendicular to each other.

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