A) \[\frac{2.56Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\] towards + ve x-axis
B) \[\frac{6.93Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\] towards + ve x-axis
C) \[\frac{6.93Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\] towards - ve x-axis
D) Zero
Correct Answer: B
Solution :
The given, \[{{q}_{1}}=2Q,\,{{q}_{2}}=-2Q,\,{{q}_{3}}=-4Q\] The net electric field E at the origin \[E=\left( \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}}{{{d}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{2}}}{{{d}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{3}}}{{{d}^{2}}} \right)\cos \theta \] \[=\left( \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2Q}{{{d}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2Q}{{{d}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{4Q}{{{d}^{2}}} \right)\cos \theta \] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2\times 4Q}{{{d}^{2}}}\times \frac{\sqrt{3}}{2}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{8\times 1.732Q}{{{d}^{2}}\times 2}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{6.928Q}{{{d}^{2}}}\] \[=\frac{6.932Q}{4\pi {{\varepsilon }_{0}}.\,\,{{d}^{2}}}\]towards + ve x-axis.
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