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