question_answer

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 \[\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\frac{N\text{-}{{m}^{2}}}{{{C}^{2}}} \right]\] [EAMCET (Med.) 2000]

A)            \[9\times {{10}^{4}}\]V

B)                                      \[18\times {{10}^{4}}\]V

C)            \[36\times {{10}^{4}}\]V

D)                                      \[36\times {{10}^{-4}}\]V

Correct Answer: C

Solution :

                     Potential at C \[=\left( 9\times {{10}^{9}}\times \frac{4\times {{10}^{-6}}}{0.2} \right)\times 2\]\[=36\times {{10}^{4}}V\]