A) \[\frac{\sqrt{3}q}{\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
B) \[\frac{3\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
C) \[\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
D) \[\frac{2\sqrt{3}q}{\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
Correct Answer: A
Solution :
Electric field due to charge +2q at centre O - \[{{\vec{E}}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2q}{{{d}^{2}}}\left[ \frac{+\sqrt{3}\hat{i}-\hat{j}}{2} \right]\] Due to -2q \[{{\vec{E}}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2q}{{{d}^{2}}}\left[ \frac{\sqrt{3}\hat{i}-\hat{j}}{2} \right]\] Due to \[-\,4\,q\] \[{{\vec{E}}_{3}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{4q}{{{d}^{2}}}\left[ \frac{\sqrt{3}\hat{i}+\hat{j}}{2} \right]\] Net electric field at point O \[{{\vec{E}}_{0}}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}=\frac{\sqrt{3}q}{\pi {{\varepsilon }_{0}}{{d}^{2}}}\hat{i}\]You need to login to perform this action.
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