question_answer

Charges\[+2q,+q\]and\[+q\]are placed at the comers A, B and C of an equilateral triangle ABC. If E is the electric field at the circumcentre O of the triangle, due to the charge\[+q\], then the magnitude and direction of the resultant electric field at O is

A)  E along AO   

B)         2 E along AO

C)  E along BO         

D)         E along CO

E)  zero

Correct Answer: A

Solution :

The situation is as shown below. As shown we resolve E along CO and also along BO into two perpendicular components. The horizontal components cancel each other. The vertical (cosine) components add up along OA to give\[2E\text{ }cos\text{ }60{}^\circ \]. Resultant field along AO                 \[=2E-2E\cos 60{}^\circ \]                 \[=2E-2E\times \frac{1}{2}\]                 \[=2E-E=E\] Hence, resultant field is E along AO.