Equal charges q are placed at the vertices A and B of an eq…

Physics Electrostatics Potential & Capacitance Electric Field and Potential,Defference,Energy and Dipole Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of sidea. The magnitude of electric field at the point C is

$\frac{q}{4 \pi \varepsilon_0 a^2}$

$\frac{\sqrt{2 a q}}{4 \pi \varepsilon_0 a^2}$

$\frac{\sqrt{3}q}{4 \pi \varepsilon_0 a^2}$

$\frac{q}{2 \pi \varepsilon_0 a^2}$

Text Solution

Verified by Experts

The correct answer is:

$|E_A| = |E_B| = k \cdot \frac{q}{a^2}$

So, $E_{net} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos 60^\circ}$

$= \frac{\sqrt{3} k \cdot q}{a^{2}}$

⇒ ⇒ $E_{\text{net}} = \frac{\sqrt{3} \, q}{4 \pi \epsilon_0 a^2}$

Previous Question Next Question

Get practice questions, solutions, and test series in one place.

Share your experience with this question and solution.

Send your comment, doubt, correction, or feedback to admin.

What students say about this solution

No reviews yet. Be the first to write a review.