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) Resolve $E$ along $CO$ and $BO$ into two perpendicular components
[image:1_image241.emf]
The sine components cancel each other
The cosine components add up along $OA$ to give $2E\cos 60^{o}$
$\therefore$ Resultant field along $AO = 2E - 2E\cos{60{^\circ}}$
$= 2E - E = E$
$\therefore$ 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.