Electrostatics Question 406
Question: Two charges of $ 4\mu C $ each are placed at the corners A and B of an equilateral triangle of side length 0.2 m in air. The electric potential at C is $ [ \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}\frac{N\text{-}m^{2}}{C^{2}} ] $ [EAMCET (Med.) 2000]
Options:
A) $ 9\times 10^{4} $ V
B) $ 18\times 10^{4} $ V
C) $ 36\times 10^{4} $ V
D) $ 36\times {{10}^{-4}} $ V
Show Answer
Answer:
Correct Answer: C
Solution:
Potential at C $ =( 9\times 10^{9}\times \frac{4\times {{10}^{-6}}}{0.2} )\times 2 $
$ =36\times 10^{4}V $
