Figure shows three charges `-q,+q,` and `-2q` kept at points A,B and C respectivley. OA =OB=OC=r. The electric field inteinsity at O has a magnitude of

Options

A. `1/(4pi epsilon_(0)) q/(r^(2))`
B. `1/(4 pi epsilon_(0))(sqrt(3)q)/(r^(2))`
C. `1/(4 p epsilon_(0)) (2q)/(r^(2))`
D. `1/(4 pi epsilon_(0)) (sqrt(3)q)/(2r^(2))`

Answer

C

Explanation

The field at O due to charges at A and B are equal and their resultant is `2q//r^(2)`. The field due ot the chage at C is alwo `2q//r^92)` towards C. The angle between the two vectors is `120^(@)`.