Charge $+ q, + q a n d + q$ are placed at the corners, $A B a n d C$ respectively of an equilateral triangle $A B C$ . 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 :

E a l o n g A O

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2 E a l o n g A O

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E a l o n g B O

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E a l o n g C O

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Solution

The correct option is A $E a l o n g A O$

$+ q$ , $+ q$ and $+ q$ at the corners $A B$ and $C$ respectively of an equilateral triangle $Δ A B C$ .
Electric field $= E$
Circumcentre $= 0$
the charge $= + q$
then the magnitude and direction of the resultant electric field at $0$ . If the charges are identical the field intensity would have been zero. Therefore, in the present case, the intensity at the circum centre is same as that of a point charge $q$ alone as shown.
$E = \frac{q}{4 π ϵ_{0} {(\frac{l}{\sqrt{3}})}^{2}} = \frac{3 q}{4 π ϵ_{0} l^{2}}$ along $A O$ .

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