Q. 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
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) $\frac{q}{2\pi\varepsilon_{0}a^{2}}$
Explanation
(c) $\left| E_{A} \right| = \left| E_{B} \right| = k.\frac{q}{a^{2}}$
So, $E_{net} = \sqrt{E_{A}^{2} + E_{B}^{2} + 2E_{A}E_{B}\cos 60^{o}}$
$= \frac{\sqrt{3}k.q}{a^{2}}$
$\Rightarrow E_{net} = \frac{\sqrt{3}\ q}{4\pi\varepsilon_{0}a^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.