Q. 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
A) $E\ $along $AO$
B) $2E$ along $AO$
C) $E$ along $BO$
D) $E$ along $CO$
Explanation
(a) The situation is as shown below. As shown, we resolve E along $CO$ and also $BO$ into two perpendicular components.
[image:1_image317.emf]
The horizontal components cancel each other.
The vertical (cosine) components add up along $OA$ to give $2E\ \cos\ 60{^\circ}.$
[image:1_image318.emf]
Resultant field along $AO$
$= 2E - 2E\cos\ 60{^\circ}$
$= 2E - 2E \times \frac{1}{2}$
$= 2E - E = E$
Hence, resultant field is $E\ along\ AO$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.