question_answer

Three charged particles A, B and C wit charges -4q, 2q and -2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is                                                          [JEE MAIN Held On 08-01-2020 Morning]

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 \[{{\vec{E}}_{0}}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}=\frac{\sqrt{3}q}{\pi {{\varepsilon }_{0}}{{d}^{2}}}\hat{i}\]