Three point charges +4e , +2e and +e are kept at each vertex of an equilateral triangle ABC , each of side a . The net electric intensity acting on charge +e is ( 60^ = 1/2 and k is constant )

Step-by-step solution

Let the charge +e be placed at vertex C of the equilateral triangle ABC, while +4e and +2e are placed at vertices A and B respectively. The electric intensity (electric field) at C due to the charge +4e at A is: E₁ = k(4e) a^2 The electric intensity at C due to the charge +2e at B is: E₂ = k(2e) a^2 Since ABC is an equilateral triangle, the angle between the electric field vectors E ₁ and E ₂ is 60^ . The net electric intensity at C is given by the vector sum of E ₁ and E ₂ : E_ net = E₁^2 + E₂^2 + 2E₁ E₂ 60^ Subst