Charges `+2q, +q and +q` are placed at the corners 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

Options

A. E along AO
B. 2E along AO
C. E along BO
D. E along CO

Answer

A

Explanation

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 cos `60^(@)`.

Resultant field along AO
`=2E-2Ecos60^(@)`
`=2E-2Exx(1)/(2)`
`=2E-E=E`
Hence, resultant field is E along AO.