JEE Mains · Physics · STD 12 - 1. Electric charges and fields
What will be the magnitude of electric field at point \(O\) as shown in figure ? Each side of the figure is \(I\) and perpendicular to each other.

- A \(\frac{q}{4 \pi \varepsilon_{0}(l)^{2}}\)
- B \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(2l^{2})}(2 \sqrt{2}-1)\)
- C \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{l^{2}}\)
- D \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 {q}}{2l^{2}}(\sqrt{2})\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(2l^{2})}(2 \sqrt{2}-1)\)
Step-by-step Solution
Detailed explanation
Electric field for point charge \(E_{1}=\frac{k q}{\ell^{2}}=E_{2}\) \(E_{3}=\frac{k q}{(\sqrt{2} \ell)^{2}}=\frac{k q}{2 \ell^{2}}\) \(E=\frac{\sqrt{2} k q}{\ell^{2}}-\frac{k q}{2 \ell^{2}}=\frac{k q}{2 \ell^{2}}(2 \sqrt{2}-1)\)
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