Two point charges Q₁=2 C and Q₂=1 C are placed as shown. The coordinates of the point P are (2 ~cm , 1 ~cm ) . The electric intensity vector at P subtends an angle with the positive X -axis. The value of is given by

Step-by-step solution

Given, Q₁=2 C =2 10⁻⁶ C Q₂=1 C =1 10⁻⁶ C The electric field intensity at a point P due to a point charge is given by aligned E &= 1 4 ₀ Q r² & E Q r² aligned Ratio of electric fields due to Q₁ and Q₂ at point P is aligned E₁ E₂ &= Q₁ r₁² r₂² Q₂ &= 2 10⁻⁶ (2 10⁻² )² (1 10⁻² )² 1 10⁻⁶ = 1 2 aligned Here, E₁ is along X -axis and E₂ is along Y -axis. So, the resultant field E have angle with X -axis as = E₂ E₁ =2