A) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OB
B) \[\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]along OA
C) \[\frac{4q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]along OC
D) zero
Correct Answer: B
Solution :
(b)\[\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] along OA If the charge q at A is replaced by -q, it is equivalent to adding charge -2q at Therefore, electric field at O. \[{{E}_{2}}=\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]along OA.
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