Two charges each equal to ηqη−1<3 are placed at the corners of an equilateral triangle of side a. The electric field at the third corner is E3

E1=ηq4πε0a2,E2=ηq4πε0a2 Therefore E3→=E→1+E→2 ⇒E3=E12+E22+2E1E2cos⁡60∘=3ηq4πε0a2 Since η−1<3⇒1<3η⇒3η>1⇒ 3ηq4πε0a2>q4πε0a2⇒E3>E0                here E0=q4πε0a2