Two charges each equal to $η q (η^{- 1} < \sqrt{3})$ are placed at the corners of an equilateral triangle of side a. The electric field at the third corner is E₃

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$E_{3} = E_{0}$

$E_{3} > E_{0}$

$E_{3} < E_{0}$

$E_{3} < E_{0}$ $where (E_{0} = q / 4 π ε_{0} a^{2})$

answer is C.

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Detailed Solution

$\begin{array}{l} E_{1} = \frac{η q}{4 π ε_{0} a^{2}}, E_{2} = \frac{η q}{4 π ε_{0} a^{2}} \\ Therefore \vec{E_{3}} = {\vec{E}}_{1} + {\vec{E}}_{2} \\ \Rightarrow E_{3} = \sqrt{E_{1}^{2} + E_{2}^{2} + 2 E_{1} E_{2} \cos 60^{\circ}} = \frac{\sqrt{3} η q}{4 π ε_{0} a^{2}} \\ Since η^{- 1} < \sqrt{3} \\ \Rightarrow 1 < \sqrt{3} η \\ \Rightarrow \sqrt{3} η > 1 \\ \Rightarrow \frac{\sqrt{3} η q}{4 π ε_{0} a^{2}} > \frac{q}{4 π ε_{0} a^{2}} \\ \Rightarrow E_{3} > E_{0} (h e r e E_{0} = \frac{q}{4 π ε_{0} a^{2}}) \end{array}$