Electrostatics Question 375
Question: 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 [MP PMT 1997]
Options:
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}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ |E _{A}|=|E _{B}|=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}} $
therefore $ E _{net}=\frac{\sqrt{3}q}{4\pi {\varepsilon _{0}}a^{2}} $
