A) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
B) \[\frac{\sqrt{2}\,q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
C) \[\frac{\sqrt{3}\,q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
D) \[\frac{q}{2\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
Correct Answer: C
Solution :
\[|{{E}_{A}}|\,=\,|{{E}_{B}}|=k.\frac{q}{{{a}^{2}}}\] So, \[{{E}_{net}}=\sqrt{E_{A}^{2}+E_{B}^{2}+2{{E}_{A}}{{E}_{B}}\cos 6{{0}^{o}}}\] \[=\frac{\sqrt{3}\,k.\,q}{{{a}^{2}}}\] Þ \[{{E}_{net}}=\frac{\sqrt{3}\,q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]You need to login to perform this action.
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