Q. Equal charges $q$ each are placed at the vertices $A$ and $B\ $of an equilateral triangle $ABC$ of side$a$. The magnitude of electric intensity 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) $\frac{2q}{4\pi\varepsilon_{0}a^{2}}$
Explanation
(c) $E_{1} = E_{2} = \frac{q \times 1}{4\pi\varepsilon_{0}a^{2}}$ , acting at 60$^{0}$.
$\therefore$ Resultant intensity
$E = \sqrt{E_{1}^{2} + E_{2}^{2} + 2E_{1}E_{2}\cos\theta}$
$= \sqrt{E_{1}^{2} + E_{1}^{2} + 2E_{1}^{2}\cos 60^{0}}$
${= E}_{1}\sqrt{3} = \frac{q\sqrt{3}}{4\pi\varepsilon_{0}a^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.