Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point C is
q4πε0a2
2q4πε0a2
3q4πε0a2
q2πε0a2
EA=EB=k⋅qa2
So, Enet =EA2+EB2+2EAEBcos60∘ =3k⋅qa2 ⇒Enet=3q4πε0a2