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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$E_{3} = E_{0}$

$E_{3} > E_{0}$

$E_{3} < E_{0}$

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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