Q. Two charges each equal to $\eta q\left( \eta^{- 1} < \sqrt{3} \right)$ are placed at corners of an equilateral triangle of side $a.$ The electric field at the third corner is $E_{3}$ then (where $E_{0} = q/4\pi\varepsilon_{0}a^{2}$)
A) $E_{3} = E_{0}$
B) $E_{3} < E_{0}$
C) $E_{3} > E_{0}$
D) $E_{3} \geq E_{0}$
Explanation
(c) $E_{1} = \frac{\eta q}{4\pi\varepsilon_{0}a^{2}},E_{2} = \frac{\eta q}{4\pi\varepsilon_{0}a^{2}}$. Therefore $E = {\overrightarrow{E}}_{1} + {\overrightarrow{E}}_{2}$
$= \sqrt{E_{1}^{2} + E_{2}^{2} + 2E_{1}E_{2}\cos{60{^\circ}}} = \frac{\sqrt{3}\eta q}{4\pi\varepsilon_{0}a^{2}}$
$Since\ \eta^{- 1} < \sqrt{3},1 < \sqrt{3}\eta,\sqrt{3}\eta > 1$
$\Rightarrow \frac{\sqrt{3}\eta q}{4\pi\varepsilon_{0}a^{2}} > \frac{q}{4\pi\varepsilon_{0}a^{2}} \Rightarrow E_{3} > E_{0}\left( E_{0} = \frac{q}{4\pi\varepsilon_{0}a^{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.