question_answer

Equal charges \[q\] are placed at the vertices \[A\] and \[B\] of an equilateral triangle \[ABC\] of side \[a\]. The magnitude of electric field at the point \[C\] is                                                   [MP PMT 1997]

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 \[{{E}_{net}}=\frac{\sqrt{3}\,q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]