A) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OA
B) \[\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OA
C) \[\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]along OA
D) \[\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OA
Correct Answer: A
Solution :
\[\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OA When a charge q from corner A is removed, electric field at O is \[{{E}_{1}}=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]along OA
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