Charges Q₁ and Q₂ are at points A and B of a right angle triangle OAB (see figure). The resultant electric field at point O is perpendicular to the hypotenuse, then
Q₁/Q₂ is proportional to :

Solution

Physics sir · 3 steps

Put O at the origin, write the field as a vector, and set its dot product with the hypotenuse to zero.

1. Set up axes. Take O as the origin, A at (0, x₁) and B at (x₂, 0). With both charges positive, Q₁ pushes the field at O along -y with size kQ₁/x₁². Q₂ pushes it along -x with size kQ₂/x₂².
2. Perpendicular condition. The hypotenuse runs from A to B, along the vector (x₂, -x₁). The resultant field E⃗ = (-kQ₂/x₂², -kQ₁/x₁²) is perpendicular to it when their dot product is zero.

   -kQ₂/x₂² x₂ + kQ₁/x₁² x₁ = 0 ⇒ Q₁/x₁ = Q₂/x₂

3. So. Q₁/Q₂ = x₁/x₂, which is option (c).

The idea to keep. For a direction condition on a resultant field, write the field in components and use a dot product. Perpendicular means the dot product is zero, and parallel means the component ratios match.

Answer from the NTA final answer key. Solution by PrepWiser.