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JEE Main20268 April 2026Evening ShiftPhysicsElectrostaticsActual

Two point charges q₁=3 , C and q₂=-4 , C are placed at points (2 i +3 j +3 k ) and ( i + j + k ) respectively. Force on charge q₂ is ________ N. ( Take 1 4 ₀ = 9 10^9 SI Units )

Options

  1. A(12 i +24 j +24 k ) 10⁻³
  2. B(4 i +8 j +8 k ) 10⁻³
  3. C(3 i +6 j +6 k ) 10⁻³
  4. D(-4 i -8 j -8 k ) 10⁻³

Correct answer

B. (4 i +8 j +8 k ) 10⁻³

Step-by-step solution

Position vector of q₁ , r ₁ = 2 i + 3 j + 3 k Position vector of q₂ , r ₂ = i + j + k Vector from q₁ to q₂ is r ₂₁ = r ₂ - r ₁ = ( i + j + k ) - (2 i + 3 j + 3 k ) = - i - 2 j - 2 k Magnitude | r ₂₁| = (-1)^2 + (-2)^2 + (-2)^2 = 9 = 3 Force on q₂ due to q₁ is given by Coulomb's law in vector form: F ₂₁ = 1 4 ₀ q₁ q₂ | r ₂₁|^3 r ₂₁ Substituting the given values: F ₂₁ = 9 10^9 (3 10⁻⁶) (-4 10⁻⁶) 3^3 (- i - 2 j - 2 k ) F ₂₁ = -108 10⁻³ 27 (- i - 2 j - 2 k ) F ₂₁ = -4 10⁻³ (- i - 2 j - 2 k ) F ₂₁ = (4 i + 8 j + 8 k ) 1

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